The magnitude of the net force acting on the mass at the lower right corner due to the other two masses is 4.588 × 10⁻¹⁰ N, and the direction is 46.03°.
Mass of each of the three objects, m = 0.300 kg
The distance of each object from the mass at the bottom right corner:
AB = a = 0.400 m
AC = b = 0.300 m
BC = c = 0.500 m
Gravitational constant, [tex]G = 6.67 \times 10^{-11} Nm^2/kg^2[/tex]
The formula to calculate gravitational force between two masses is:
[tex]F = G \times m_1 \times m_2 / r^2[/tex]
Where, G = gravitational constant
m₁, m₂ = masses of the two objects
r = distance between the centers of the two objects
To find the force on the mass at the lower right corner due to the other two masses, we can use vector addition.
The force on the mass due to mass A will be: [tex]F_1 = G \times m\times m_1 / AB^2[/tex]
The direction of force F₁ is along the line connecting mass A and the mass at the bottom right corner and is given by
[tex]\theta_1= tan^{-1} (b / a)[/tex]
The force on the mass due to mass B will be:
[tex]F_2= G \times m\times m_2 / BC^2[/tex]
The direction of force F₂ is along the line connecting mass B and the mass at the bottom right corner and is given by
[tex]\theta_2= tan^{-1} (a / b)[/tex]
Now, we can use vector addition to find the net force acting on the mass at the lower right corner.
[tex]F = \sqrt{(F_1^2 + F_2^2 + 2F_1F_2\cos(180^0 - \theta_1 - \theta_2))}[/tex]
The direction of the net force is
[tex]\theta = \tan^{-1}[(F_2\sin\theta_2- F_1\sin\theta_1) / (F_2\cos\theta_2 + F_1cos\theta_1)][/tex]
Substituting the given values in the above formulas:
[tex]F_1= (6.67\times 10^{-11} Nm^2/kg^2) \times (0.300 kg)^2 / (0.400 m)^2\\F_1 = 5.0025 \times 10^{-10} N\\F_2 = (6.67 \times 10^{-11} Nm^2/kg^2) \times (0.300 kg)^2 / (0.500 m)^2\\F_2 = 2.40144 \times 10^{-10} N\\[/tex]
[tex]F = \sqrt(5.0025 \times 10^{-10} N^2 + 2.40144 \times 10^{-10} N^2 + 2(5.0025 \times 10^{-10} N)(2.40144 \times 10^{-10} N) \cos(180\textdegree - 53.1301\textdegree - 36.8699\textdegree))\\
F = 4.588 \times 10^{-10} N\\\
theta = tan^{-1} [(2.40144 \times 10^{-10}N \sin 36.8699\textdegree - 5.0025 \times 10^{-10} N \sin 53.1301\textdegree) / (2.40144 \times 10^{-10} N \cos 36.8699\textdegree + 5.0025 \times 10^{-10} N \cos 53.1301\textdegree)]\\\
theta = 46.03\textdegree[/tex]
Therefore, the magnitude of the net force acting on the mass at the lower right corner due to the other two masses is 4.588 × 10⁻¹⁰ N, and the direction is 46.03°.
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F₁ and F₂ using the given values and then calculate F_net and θ to find the magnitude and direction of the gravitational force acting on the lower right mass.
To calculate the gravitational force acting on the mass at the lower right corner of the triangle due to the other two masses, we can use the equation for gravitational force:
F = (G * m₁ * m₂) / r²
where:
F is the gravitational force,
G is the gravitational constant (6.67 x 10⁽⁻¹¹⁾ Nm²/kg²),
m1 and m2 are the masses, and
r is the distance between the masses.
Given:
G = 6.67 x 10⁽⁻¹¹⁾ Nm²/kg²
Masses:
m₁ = 0.300 kg (mass at the lower left corner)
m₂ = 0.300 kg (mass at the upper corner)
We need to calculate the distances (r) between the masses:
For the side of length a:
r₁ = 0.400 m (distance between the lower left corner and the lower right corner)
For the side of length b:
r₂ = 0.300 m (distance between the lower left corner and the upper corner)
Now, we can calculate the gravitational force between the lower left mass and the lower right mass:
F₁ = (G * m₁ * m₂) / r1²
= (6.67 x 10⁽⁻¹¹⁾ Nm²/kg²) * (0.300 kg) * (0.300 kg) / (0.400 m)²
F1 = (6.67 x 10⁽⁻¹¹⁾) * (0.300) * (0.300) / (0.400)² N
Similarly, we can calculate the gravitational force between the upper mass and the lower right mass:
F₂ = (G * m₁ * m₂) / r²
= (6.67 x 10⁽⁻¹¹⁾ Nm²/kg²) * (0.300 kg) * (0.300 kg) / (0.300 m)²
F₂ = (6.67 x 10^⁽⁻¹¹⁾) * (0.300) * (0.300) / (0.300)² N
Now, we can find the net gravitational force acting on the lower right mass by adding these two forces as vectors:
F_net = sqrt(F₁ + F₂)
The direction of the net gravitational force can be found by calculating the angle it makes with the positive x-axis:
θ = arctan(F₂ / F₁)
Calculate F₁ and F₂ using the given values and then calculate F_net and θ to find the magnitude and direction of the gravitational force acting on the lower right mass.
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According to Lenz's law,
the induced current in a circuit must flow in such a direction to oppose the magnetic flux.
the induced current in a circuit must flow in such a direction to oppose the change in magnetic flux.
the induced current in a circuit must flow in such a direction to enhance the change in magnetic flux.
the induced current in a circuit must flow in such a direction to enhance the magnetic flux.
There is no such law, the prof made it up specifically to fool gullible students that did not study.
Lenz's law is a basic principle of electromagnetism that specifies the direction of induced current that is produced by a change in magnetic field. According to Lenz's law, the direction of the induced current in a circuit must flow in such a way as to oppose the change in magnetic flux.
In other words, the induced current should flow in such a way that it produces a magnetic field that opposes the change in magnetic field that produced the current. This concept is based on the conservation of energy and the principle of electromagnetic induction.
Lenz's law is an important principle that has many practical applications, especially in the design of electrical machines and devices.
For example, Lenz's law is used in the design of transformers, which are devices that convert electrical energy from one voltage level to another by using the principles of electromagnetic induction.
Lenz's law is also used in the design of electric motors, which are devices that convert electrical energy into mechanical energy by using the principles of electromagnetic induction.
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1. Solve y' += 2 using Integrating Factor 2. Solve y²dy = x² - xy using Homogenous Equation
To solve y' + 2 = 0 using an integrating factor, we multiply by e^(2x) and integrate. To solve y^2dy = x^2 - xy using a homogeneous equation, we substitute y = vx and solve a separable equation.
1. To solve y' + 2 = 0 using an integrating factor, we first rewrite the equation as y' = -2. Then, we multiply both sides by the integrating factor e^(2x):
e^(2x)*y' = -2e^(2x)
We recognize the left-hand side as the product rule of (e^(2x)*y)' and integrate both sides with respect to x:
e^(2x)*y = -e^(2x)*C1 + C2
where C1 and C2 are constants of integration. Solving for y, we get:
y = -C1 + C2*e^(-2x)
where C1 and C2 are arbitrary constants.
2. To solve y^2dy = x^2 - xy using a homogeneous equation, we first rewrite the equation in the form:
dy/dx = (x^2/y - x)
This is a homogeneous equation because both terms have the same degree of homogeneity (2). We then substitute y = vx and dy/dx = v + xdv/dx into the equation, which gives:
v + xdv/dx = (x^2)/(vx) - x
Simplifying, we get:
vdx/x = (1 - v)dv
This is a separable equation that we can integrate to get:
ln|x| = ln|v| - v + C
where C is the constant of integration. Rearranging and substituting back v = y/x, we get:
ln|y| - ln|x| - y/x + C = 0
This is the general solution of the homogeneous equation.
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DETAILS SERPSE 10 9.1.P.002. A 2.91 kg particle has a velocity of (3.05 1 - 4.08 ) m/s. (a) Find its x and y components of momentum. Px = kg-m/s Py = kg-m/s (b) Find the magnitude and direction of its momentum. kg-m/s (clockwise from the +x axis) Read It Need Help?
The x and y components of momentum are, Px = 8.85 kg-m/s and Py = -11.90 kg-m/s and the magnitude of momentum is 15.17 kg-m/s and the direction of momentum is -52.92° clockwise from the +x axis.
A 2.91 kg particle has a velocity of (3.05i - 4.08j) m/s.
Given, Mass of the particle, m = 2.91 kg
The velocity of the particle,
v = 3.05i - 4.08j m/s
.The formula for momentum is:
P = m*v= 2.91*3.05i + 2.91*(-4.08)j= 8.8495i - 11.9028j
Hence, the x and y components of momentum are:
Px = 8.85 kg-m/sPy = -11.90 kg-m/s
The magnitude of momentum can be calculated as
[tex]-|P| = sqrt(Px^2 + Py^2) = sqrt(8.85^2 + (-11.90)^2) = 15.17 kg-m/s[/tex]
The direction of momentum can be calculated as
[tex]-θ = tan^-1(Py/Px) = tan^-1(-11.90/8.85) = -52.92°[/tex]
The direction of momentum is clockwise from the +x axis, hence the direction of momentum is = -52.92° clockwise from the +x axis.
Thus, the x and y components of momentum are, Px = 8.85 kg-m/s and Py = -11.90 kg-m/s. The magnitude of momentum is 15.17 kg-m/s and the direction of momentum is -52.92° clockwise from the +x axis.
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Two vectors 10N and 8N on bearing 285° and N70°E
respectively, act on a body. Find the resultant force
and direction of motion of the body using the triangle
of vectors.
The resultant force is 12.6N at a bearing of 3°W of N. The direction of motion of the body is the same as the direction of the resultant force, which is 3°W of N.
A triangle of vectors can be used to solve vector addition problems, such as determining the resultant force and direction of motion of a body acted upon by two or more vectors.
Let's use this method to solve the given problem: Two vectors, 10N and 8N, act on a body on bearings 285° and N70°E respectively.
Using the triangle of vectors, we can determine the resultant force and direction of motion of the body.
1. Draw a diagram to scale, showing the two vectors and their respective bearings.
2. Begin by drawing the first vector, 10N, from the origin at bearing 285°.
3. Draw the second vector, 8N, from the end of the first vector at bearing N70°E.
4. Draw the third vector, the resultant force, from the origin to the end of the second vector.
5. Use a protractor and ruler to measure the magnitude and bearing of the resultant force.
The diagram is shown below: Triangle of vectors diagram using the two vectors 10N and 8N, with bearings 285° and N70°E respectively.
The third vector, the resultant force, is drawn from the origin to the end of the second vector.
The magnitude and bearing of the resultant force are found using a protractor and ruler.
6. Measure the magnitude of the resultant force using the ruler.
In this case, the magnitude is approximately 12.6N.
7. Measure the bearing of the resultant force using the protractor.
In this case, the bearing is approximately 3°W of N.
Therefore, the resultant force is 12.6N at a bearing of 3°W of N.
The direction of motion of the body is the same as the direction of the resultant force, which is 3°W of N.
Therefore, the body will move in this direction.
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a) Define the activity of a radioactive source b) The activity of a radioactive source is proportional to the number of radioactive nuclei present within it.
a) Define the activity of a radioactive source.
The activity of a radioactive source can be defined as the rate at which the number of radioactive nuclei of that source undergoes decay or the amount of radiation produced by the source per unit of time.
b) The activity of a radioactive source is proportional to the number of radioactive nuclei present within it. The activity of a radioactive source is directly proportional to the number of radioactive nuclei present within it.
The higher the number of radioactive nuclei, the greater the activity of the radioactive source.
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Current in a Loop uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 9.00E-3 T/s. Determine the current in A 35.0 cm diameter coil consists of 24 turns of circular copper wire 2.60 mm in diameter the loop Subrnit Answer Tries 0/12 Determine the rate at which thermal energy is produced.
The current flowing through the loop is approximately 0.992 Amperes. The rate of change of magnetic field is given as 9.00E-3 T/s. Therefore, the rate of change of magnetic flux is:
dΦ/dt = (9.00E-3 T/s) * 0.3848 m^2 = 3.4572E-3 Wb/s
The current in the loop can be determined by using Faraday's law of electromagnetic induction. According to the law, the induced electromotive force (emf) is equal to the rate of change of magnetic flux through the loop. The emf can be calculated as: ε = -N * dΦ/dt. where ε is the induced emf, N is the number of turns in the coil, and dΦ/dt is the rate of change of magnetic flux.The magnetic flux (Φ) through the loop is given by: Φ = B * A. where B is the magnetic field strength and A is the area of the loop.Given that the coil has a diameter of 35.0 cm and consists of 24 turns, we can calculate the area of the loop: A = π * (d/2)^2. where d is the diameter of the coil.
Substituting the values, we get: A = π * (0.35 m)^2 = 0.3848 m^2
The rate of change of magnetic field is given as 9.00E-3 T/s. Therefore, the rate of change of magnetic flux is:
dΦ/dt = (9.00E-3 T/s) * 0.3848 m^2 = 3.4572E-3 Wb/s
Now, we can calculate the induced emf:
ε = -N * dΦ/dt = -24 * 3.4572E-3 Wb/s = -0.08297 V/s
Since the coil is made of copper, which has low resistance, we can assume that the induced emf drives the current through the loop. Therefore, the current flowing through the loop is: I = ε / R
To calculate the resistance (R), we need the length (L) of the wire and its cross-sectional area (A_wire).The cross-sectional area of the wire can be calculated as:
A_wire = π * (d_wire/2)^2
Given that the wire diameter is 2.60 mm, we can calculate the cross-sectional area: A_wire = π * (2.60E-3 m/2)^2 = 5.3012E-6 m^2
The length of the wire can be calculated using the formula:
L = N * circumference
where N is the number of turns and the circumference can be calculated as: circumference = π * d
L = 24 * π * 0.35 m = 26.1799 m
Now we can calculate the resistance: R = ρ * L / A_wire
where ρ is the resistivity of copper (1.7E-8 Ω*m).
R = (1.7E-8 Ω*m) * (26.1799 m) / (5.3012E-6 m^2) = 8.3741E-2 Ω
Finally, we can calculate the current:
I = ε / R = (-0.08297 V/s) / (8.3741E-2 Ω) = -0.992 A
Therefore, the current flowing through the loop is approximately 0.992 Amperes.
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Question 32 of 37 > Attempt Consider the inelastic collision. Two lumps of matter are moving directly toward each other. Each lump has a mass of 1,500 kg and is moving at a spoed of 0.880. The two lumps collide and stick together. Answer the questions, keeping in mind that relativistic effects cannot be neglected in this case. What is the final speed of the combined lump, expressed as a fraction of e? 0.44 = incorrect What is the final mass me of the combined lump immediately after the collision, assuming that there has not yet been significant energy loss due to radiation or fragmentation? ks 2.45 m = incorrect
In an inelastic collision between two lumps of matter, each with a mass of 1,500 kg and a speed of 0.880, the final speed of the combined lump is not 0.44 times the speed of light (e). The final mass of the combined lump immediately after the collision is not 2.45 m.
Final Speed: The final speed of the combined lump in an inelastic collision cannot be determined using the given information.
It requires additional data, such as the nature of the collision and the relative velocities of the lumps. Without this information, it is not possible to calculate the final speed as a fraction of the speed of light (e).
Final Mass: The final mass of the combined lump can be calculated by adding the individual masses together.
Since both lumps have a mass of 1,500 kg, the combined mass of the lump immediately after the collision would be 3,000 kg. There is no indication of a factor or value (2.45 m) that affects the calculation of the final mass, so it remains at 3,000 kg.
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A projectile is fired with an initial speed of 49.6 m/s at an angle of 42.2° above the horizontal on a long flat firing range. Determine the direction of the motion of the projectile 1.20 s after firing
1.20 seconds after firing, the projectile is moving upward and also in the positive x-direction horizontally.
To determine the direction of motion of the projectile 1.20 seconds after firing, we need to consider the vertical and horizontal components of its motion separately.
First, let's analyze the vertical component of motion. The projectile experiences a downward acceleration due to gravity. The vertical velocity of the projectile can be calculated using the formula:
v_vertical = v_initial * sin(theta)
where v_initial is the initial speed of the projectile and theta is the launch angle. Plugging in the given values:
v_vertical = 49.6 m/s * sin(42.2°)
v_vertical ≈ 33.08 m/s (upward)
Since the vertical velocity component is positive, the projectile is moving in an upward direction.
Next, let's consider the horizontal component of motion. The horizontal velocity of the projectile remains constant throughout its flight, assuming no air resistance. The horizontal velocity can be calculated using the formula:
v_horizontal = v_initial * cos(theta)
Plugging in the given values:
v_horizontal = 49.6 m/s * cos(42.2°)
v_horizontal ≈ 37.81 m/s (horizontal)
The horizontal velocity component is positive, indicating motion in the positive x-direction.
Therefore, 1.20 seconds after firing, the projectile is moving upward and also in the positive x-direction horizontally.
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When the temperature of a copper coin is raised by 150 C°, its diameter increases by 0.26%. To two significant figures, give the percent increase in (a) the area of a face, (b) the thickness, (c) the volume, and (d) the mass of the coin. (e) Calculate the coefficient of linear
expansion of the coin.
(a) The percent increase in the area of a face is approximately 0.52%.
(b) The percent increase in the thickness is approximately 0.26%.
(c) The percent increase in the volume is approximately 0.78%.
(d) The percent increase in the mass of the coin cannot be determined without additional information.
(e) The coefficient of linear expansion of the coin is approximately 1.73 x 10^-5 C^-1.
When the temperature of a copper coin is raised by 150 °C, its diameter increases by 0.26%. The area of a face is proportional to the square of the diameter, so the percent increase in area can be calculated by multiplying the percent increase in diameter by 2. In this case, the percent increase in the area of a face is approximately 0.52%.
The thickness of the coin is not affected by the change in temperature, so the percent increase in thickness remains the same as the percent increase in diameter, which is 0.26%.
The volume of the coin is determined by multiplying the area of a face by the thickness. Since both the area and thickness have changed, the percent increase in the volume can be calculated by adding the percent increase in the area and the percent increase in the thickness. In this case, the percent increase in the volume is approximately 0.78%.
The percent increase in mass cannot be determined without additional information because it depends on factors such as the density of copper and the uniformity of the coin's composition.
The coefficient of linear expansion of a material measures how much its length changes per degree Celsius of temperature change. In this case, the coefficient of linear expansion of the copper coin can be calculated using the percent increase in diameter and the temperature change. The coefficient of linear expansion is approximately 1.73 x 10^-5 C^-1.
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An undamped 1.92 kg horizontal spring oscillator has a spring constant of 21.4 N/m. While oscillating, it is found to have a speed of 2.56 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation?
The amplitude of oscillation of an undamped 1.92 kg horizontal spring oscillator with a spring constant of 21.4 N/m and a speed of 2.56 m/s as it passes through its equilibrium position is 0.407 meters.
The amplitude of an oscillation is defined as the maximum displacement from the equilibrium position or mean position of the particle or object in oscillation.What is the formula for amplitude?The amplitude A of a particle in oscillation is given by:A = (2KE/mω2)1/2where KE is the kinetic energy of the particle,m is the mass of the particle, andω is the angular frequency of the oscillation.
The angular frequency is defined as the number of radians per second by which the object rotates or oscillates. It is usually represented by the symbol ω.What is the kinetic energy of the particle?The kinetic energy of the particle is given by:KE = 0.5mv2where m is the mass of the particle, andv is the velocity of the particle.
Given data,Mass of the oscillator, m = 1.92 kgSpring constant, k = 21.4 N/mSpeed of the oscillator, v = 2.56 m/sThe formula for the amplitude of oscillation is:A = (2KE/mω2)1/2The formula for the angular frequency of the oscillation is:ω = (k/m)1/2The formula for the kinetic energy of the particle is:KE = 0.5mv2Substitute the given values in the above formulas to get the value of amplitude as follows:
ω = (k/m)1/2
ω = (21.4 N/m ÷ 1.92 kg)1/2ω = 3.27 rad/s
KE = 0.5mv2
KE = 0.5 × 1.92 kg × (2.56 m/s)2
KE = 5.19 J
Now,A = (2KE/mω2)1/2
A = (2 × 5.19 J ÷ 1.92 kg × (3.27 rad/s)2)1/2
A = 0.407 m
Therefore, the amplitude of oscillation is 0.407 meters.
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A slender rod with a length of 0.250 m rotates with an angular speed of 8.10 rad/s about an axis through one end and perpendicular to the rod. The plane of rotation of the rod is perpendicular to a uniform magnetic field with a magnitude of 0.600 T. What is the induced emf in the rod? Express your answer in volts. What is the potential difference between its ends? Express your answer in volts.
The induced emf in the rod rotating with an angular speed of 8.10 rad/s in a perpendicular magnetic field of magnitude 0.600 T is 4.86 V, and the potential difference between its ends is also 4.86 V.
When a conducting rod moves perpendicular to a magnetic field, an induced emf is generated in the rod according to Faraday's law of electromagnetic induction.
The induced emf in the rod can be calculated using the equation:
emf = B * L * ω
where B is the magnetic field strength, L is the length of the rod, and ω is the angular speed.
B = 0.600 T (magnetic field strength)
L = 0.250 m (length of the rod)
ω = 8.10 rad/s (angular speed)
Substituting the given values into the equation:
emf = 0.600 * 0.250 * 8.10 = 4.86 V
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A bowling ball of mass 6.75 kg is rolling at 2.52 m/s along a level surface. (a) Calculate the ball's translational kinetic energy. (b) Calculate the ball's rotational kinetic energy. 23] (c) Calculate the ball's total kinetic energy. ] (d) How much work would have to be done on the ball to bring it to rest?
In this scenario, a bowling ball with a mass of 6.75 kg is rolling at a speed of 2.52 m/s along a level surface.
The task is to calculate the ball's translational kinetic energy (Part a), rotational kinetic energy (Part b), total kinetic energy (Part c), and the amount of work required to bring the ball to rest (Part d).
Part a: The translational kinetic energy of the ball can be calculated using the equation KE_trans = (1/2) * m * v², where KE_trans is the translational kinetic energy, m is the mass of the ball, and v is its velocity.
Part b: The rotational kinetic energy of the ball can be determined using the equation KE_rot = (1/2) * I * ω², where KE_rot is the rotational kinetic energy, I is the moment of inertia of the ball, and ω is its angular velocity. For a solid sphere, the moment of inertia is given by I = (2/5) * m * r², where r is the radius of the ball.
Part c: The total kinetic energy of the ball is the sum of its translational and rotational kinetic energies: KE_total = KE_trans + KE_rot.
Part d: To bring the ball to rest, work must be done to remove its kinetic energy. The work required can be calculated as W = KE_total. Therefore, the work done on the ball to bring it to rest is equal to its total kinetic energy.
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A
car engina buns 7 kg fuel at 1,648 K and rejects energy to the
radiator and the exhaust at an average temperature of 543k. if the
fuel provides 34296 kJ/kg whaf is the max amount of work the engine
The maximum amount of work the engine can do is 76.68 kJ.
The maximum amount of work that can be done by the engine is given as;
Wmax = Qin(1- T2/T1)
where T2 = lower temperature
T1 = higher temperature
mf = 7 kg (mass of fuel burned)
hf = 34296 kJ/kg (specific enthalpy of fuel)
h1 = 34296 kJ/kg (specific enthalpy of fuel at high temperature)
h2 = 136 kJ/kg (specific enthalpy of fuel at low temperature)
T1 = 1648 K (higher temperature)
T2 = 543 K (lower temperature)
Substituting the values in the equation, we get;
Qin = mf × hf= 7 kg × 34296 kJ/kg = 240072 kJ
Qout = m (h1-h2)= 7 kg (34296-136) kJ/kg= 240052 kJ
W = Qin - Qout= 240072 - 240052= 20 kJ
Maximum work done by the engine,
Wmax = Qin(1- T2/T1)= 240072 (1- 543/1648)= 76680 J = 76.68 kJ∴
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Part A Green light ( = 504 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 2.50 cm wide on a screen 1.80 m from the slit? Express your answer to three significant figures. VO AO ΑΣΦ ? W = um Submit Request Answer
The width of the single slit required to produce a central maximum that is 2.50 cm wide on a screen 1.80 m from the slit is 0.036 um.
Given data: The wavelength of green light = 504 nm, Distance between the screen and the single slit = 1.80 m, Width of the central maximum = 2.50 cm = 2.50 × 10⁻² m, Width of the single slit = ?
The formula for the width of the single slit that will produce a central maximum is given by: W = λD/d Where, λ is the wavelength of the light, D is the distance between the slit and the screen and d is the width of the single slit
By putting the given values in the formula, we get: W = λD/d
⇒ d = λD/W
⇒ d = (504 × 10⁻⁹ m) × (1.80 m) / (2.50 × 10⁻² m)
⇒ d = 0.036288 m
≈ 0.036 um (rounded off to three significant figures).
Therefore, the width of the single slit required to produce a central maximum that is 2.50 cm wide on a screen 1.80 m from the slit is 0.036 um (rounded off to three significant figures).
So, The width of the single slit required to produce a central maximum that is 2.50 cm wide on a screen 1.80 m from the slit is 0.036 um.
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A salad spinner has an internal 0.15-m radius spinning basket that spins at 26 rad/s to remove water from salad
greens. The basket has a rotational inertia of 0.1 kg-m?. To stop the basket, a piece of rubber is pressed against the outer edge of the basket, slowing it through friction. If
rubber is pressed into the outer edge with a force of 5 N, and the coefficient of kinetic friction between the rubber and the basket is 0.35, how long does it take for
the basket to stop?
The time it takes for the salad spinner basket to stop is approximately 6.19 seconds.
To calculate the time it takes for the salad spinner basket to stop, we need to consider the torque produced by the frictional force applied to the outer edge of the basket. The torque will cause the angular acceleration, which will gradually reduce the angular velocity of the basket until it comes to a stop.
The torque produced by the frictional force can be calculated using the equation τ = μ * F * r, where τ is the torque, μ is the coefficient of kinetic friction, F is the applied force, and r is the radius of the spinning basket.
The radius of the basket is 0.15 m, the coefficient of kinetic friction is 0.35, and the force applied is 5 N, we can calculate the torque as follows: τ = 0.35 * 5 N * 0.15 m.
Next, we can use the rotational inertia of the basket to relate the torque and angular acceleration. The torque is equal to the product of the rotational inertia and the angular acceleration, τ = I * α.
Rearranging the equation, we have α = τ / I.
Plugging in the values, α = (0.35 * 5 N * 0.15 m) / 0.1 kg-m².
Finally, we can use the formula to find the time it takes for the angular velocity to reduce to zero, given by ω = ω₀ + α * t, where ω is the final angular velocity, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time.
Since the final angular velocity is zero, we have 0 = 26 rad/s + (0.35 * 5 N * 0.15 m) / 0.1 kg-m² * t.
Solving for t, we find t = -26 rad/s / [(0.35 * 5 N * 0.15 m) / 0.1 kg-m²]. Note that the negative sign is because the angular velocity decreases over time.
Calculating the value, we get t ≈ -6.19 s. Since time cannot be negative, the time it takes for the basket to stop is approximately 6.19 seconds.
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Several experiments are performed with light. Which of the following observations is not consistent with the wave model of light? a) The light can travel through a vacuum. b) The speed of the light is less in water than in air. c) The light can exhibit interference patterns when travelling through small openings. d) The beam of light travels in a straight line. e) The light can be simultaneously reflected and transmitted at certain interfaces.
Light has been a matter of extensive research, and experiments have led to various hypotheses regarding the nature of light. The two most notable hypotheses are the wave model and the particle model of light.
These models explain the behavior of light concerning the properties of waves and particles, respectively. Here are the observations for each model:a) Wave model: The light can travel through a vacuum.b) Wave model: The speed of the light is less in water than in air.c) Wave model
e) Wave model: The light can be simultaneously reflected and transmitted at certain interfaces.None of the observations contradicts the wave model of light. In fact, all the above observations are consistent with the wave model of light.The correct answer is d) The beam of light travels in a straight line. This observation is consistent with the particle model of light.
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Different situation now. You re out in space, on a rotating wheel-shaped space station of radius 557 m. You feel planted firmly on the floor, due to artificial gravity. The gravity you experience is Earth-normal, that is, g -9.81 m/s^2. How fast is the space station rotating in order to produce this much artificial gravity? Express your answer in revolutions per minute (rpm). О 0.133 rpm 73.9 rpm 0.887 rpm 1.267 rpm
The space station is rotating at approximately 0.887 rpm to produce Earth-normal artificial gravity.
To calculate the speed of the space station rotating to produce Earth-normal artificial gravity, we can use the centripetal acceleration formula:
ac = ω²r
where ac is the centripetal acceleration, ω is the angular velocity, and r is the radius of the space station.
We know that ac is equal to the acceleration due to gravity (g). Substituting the given values, we have:
g = ω²r
Solving for ω, we get:
ω = sqrt(g / r)
Plugging in the values:
g = 9.81 m/s²
r = 557 m
ω = sqrt(9.81 / 557) ≈ 0.166 rad/s
To convert this angular velocity to revolutions per minute (rpm), we can use the conversion factor of 1 revolution = 2π radians, and there are 60 seconds in a minute:
ω_rpm = (0.166 rad/s) * (1 revolution / 2π rad) * (60 s / 1 min) ≈ 0.887 rpm
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To fit a contact lens to a patient's eye, a keratometer can be used to measure the curvature of the cornea-the front surface of the eye. This instrument places an illuminated object of known size at a known distance p from the cornea, which then reflects some light from the object, forming an image of it. The magnification M of the image is measured by using a small viewing telescope that allows a comparison of the image formed by the cornea with a second calibrated image projected into the field of view by a prism arrangement. Determine the radius of curvature of the cornea when p=34.0 cm and M=0.0180.
The radius of curvature of the cornea is 7.53 mm.
To determine the radius of curvature of the cornea, we can use the relationship between the magnification (M), the distance between the object and the cornea (p), and the radius of curvature (R) of the cornea. The magnification can be expressed as M = (1 - D/f), where D is the distance between the calibrated image and the viewing telescope and f is the focal length of the prism arrangement.
Given that M = 0.0180, we can substitute this value into the magnification equation. By rearranging the equation, we can solve for D/f.Next, we need to consider the geometry of the system. The distance D is related to the distance p and the radius of curvature R through the equation D = 2R(p - R)/(p + R).By substituting the known values of M = 0.0180 and p = 34.0 cm into the equation, we can solve for D/f. Once we have D/f, we can solve for R by substituting the values of D/f and p into the geometry equation. After performing the calculations, the radius of curvature of the cornea is found to be approximately 7.53 mm.
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A projectile is fired at an angle 45 ° from a gun that is 90 m above the flat ground, emerging
from the gun with a speed of 180 m/s.
(a) How long does the projectile remain in air?
(b) At what horizontal distance from the firing ground does it strike the ground?
(c) What is the maximum height (from ground) reached?
(a) The projectile remains in the air for 20.82 seconds.
(b) The projectile strikes the ground at a horizontal distance of 2,953.33 meters from the firing ground.
(c) The maximum height reached by the projectile from the ground is 1,845.92 meters.
Projectile motion problemTo solve the given problem, we can analyze the projectile motion and use the equations of motion.
Given:
Initial angle of projection (θ) = 45°
Initial speed of the projectile (v0) = 180 m/s
Height of the gun (h) = 90 m
(a) To find the time of flight (T), we can use the equation:
T = (2 * v0 * sin(θ)) / g
Substituting the given values, we get:
T = (2 * 180 * sin(45°)) / 9.8
T ≈ 20.82 s
(b) To find the horizontal distance (R) from the firing ground, we can use the equation:
R = v0 * cos(θ) * T
R = 180 * cos(45°) * 20.82
R ≈ 2,953.33 m
(c) To find the maximum height (H) reached by the projectile, we can use the equation:
H = (v0 * sin(θ))^2 / (2 * g)
Substituting the given values, we get:
H = (180 * sin(45°))^2 / (2 * 9.8)
H ≈ 1,845.92 m
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The projectile will remain in the air for 25.65 s, will strike the ground at a horizontal distance of 1645.9 m from the firing ground and will reach a maximum height of 4116.7 m from the ground.
(a) The time projectile will remain in the air, The time of flight, t = 2usinθ/g, where: u is the initial velocity of the projectileθ is the angle at which the projectile is launched from the ground g is the acceleration due to gravity= 2 × 180 sin 45° / 9.8= 25.65 s
(b) The horizontal distance from the firing ground that it strikes the ground, Horizontal range, R = u² sin 2θ / g= 180² sin 90° / 9.8= 1645.9 m
(c) The maximum height (from ground) reached, The maximum height (h) reached, h = u² sin²θ / 2g= 180² sin² 45° / 2 × 9.8= 4116.7 m (approx.)
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Consider a person traveling a distance of 300 km (with respect to the ground) in a relativistic car on a straight highway. Assume event A is when the car has gone 0 km of distance and event B is when the car has reached its destination. You only need to draw one diagram for parts e-g. Case 1: The car is traveling at a speed of 4.32 x108 km/hr. (a) Determine the velocity of the person in SR Units. (b) Determine the distance (with respect to the earth) traveled in SR units (c) Determine the time for the trip as measured by someone on the earth. (d) Determine the car's space-time interval. (e) Carefully draw and label a spacetime diagram for the car with respect to a person on the ground using the graph paper provided and a straight edge. (Note: this should not be a two- observer diagram) Make the diagram as accurate as possible. Make the diagram big enough to read and big enough to add another worldline. (f) When does a person on the ground see the car reach its destination. Draw a labeled worldline to support your answer. Case 2: If the car instead accelerated from rest to reach point B. (g) Draw a possible worldline for the car using a dashed line ("---") on your spacetime the diagram in part e). Considering Cases 1 and 2: (h) In which case(s) does a clock attached to the car measure proper time? Explain briefly. (i) In which case(s) does a clock attached to the car measure spacetime interval? Explain briefly. (j) In which case(s) does a clock attached to the car measure coordinate time? Explain
In both cases, the clock attached to the car measures coordinate time, which is the time measured by a single clock in a given frame of reference.
Given that,Distance traveled by the car = 300 km = 3 × 10² km
Speed of the car = 4.32 × 10⁸ km/hr
Case 1:
(a) Velocity of the person in SR Units
The velocity of the car in SI unit = (4.32 × 10⁸ × 1000) / 3600 m/s = 120,000 m/s
The velocity of the person = 0 m/s
Relative velocity = v/c = (120,000 / 3 × 10⁸) = 0.4 SR Units
(b) Distance (with respect to the earth) traveled in SR units
Proper distance = L = 300 km = 3 × 10² km
Proper distance / Length contraction factor L' = L / γ = (3 × 10²) / (1 - 0.4²) = 365.8537 km
Distance traveled in SR Units = L' / (c x T) = 365.8537 / (3 × 10⁸ x 0.4) = 3.0496 SR Units
(c) Time for the trip as measured by someone on the earth
Time interval, T = L' / v = 365.8537 / 120000 = 0.003048 SR Units
Time measured by someone on Earth = T' = T / γ = 0.004807 SR Units
(d) Car's space-time interval
The spacetime interval, ΔS² = Δt² - Δx²
where Δt = TΔx = v x TT = 0.003048 SR Units
Δx = 120000 × 0.003048 = 365.76 km
ΔS² = (0.003048)² - (365.76)² = - 133,104.0799 SR Units²
(e) Spacetime diagramCase 2:If the car instead accelerated from rest to reach point B.(g) The possible worldline for the car using a dashed line ("---")Considering Cases 1 and 2:(h) In which case(s) does a clock attached to the car measure proper time? Explain briefly.In Case 2, as the car is accelerating from rest, it is under the influence of an external force and a non-inertial frame of reference.
Thus, the clock attached to the car does not measure proper time in Case 2.In Case 1, the clock attached to the car measures proper time as the car is traveling at a constant speed. Thus, the time interval measured by the clock attached to the car is the same as the time measured by someone on Earth.(i) In which case(s) does a clock attached to the car measure spacetime interval?
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Problem 2 (10 points) Earth has a radius of 6.38 x 106m. Its mass is 5.98 x 1024 kg. Ignoring the atmosphere, if we fire a projectile from a mountain top fast enough that it will orbit just over the surface of the planet, how fast would you have to fire it for this to happen? If instead you wanted to fire the projectile so that it escapes from Earth's' gravitational pull, what initial velocity would you need?
To achieve a circular orbit just over the surface of the planet, the projectile must have a specific velocity.
Using the equation for circular motion, v² = GM / r, where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the Earth, we can calculate the required velocity.
Substituting the given values into the equation, we have v² = (6.67 x 10^-11 Nm²/kg² x 5.98 x 10^24 kg) / (6.38 x 10^6 m)². Simplifying this expression yields v² = 398600.5 m²/s². Taking the square root of both sides, we find that v ≈ 6301.9 m/s.
Therefore, in order for the projectile to orbit just over the surface of the planet, it needs to be fired with an initial velocity of approximately 6301.9 m/s.
If, on the other hand, we want the projectile to escape from the Earth's gravitational pull, we need to determine the escape velocity. The escape velocity is the speed required for an object to overcome the gravitational force and break free from the planet's gravitational field.
Using the escape velocity formula v = √(2GM / r), where G, M, and r are the same as before, we can calculate the escape velocity. Substituting the values into the equation, we have v = √(2 x 6.67 x 10^-11 Nm²/kg² x 5.98 x 10^24 kg / 6.38 x 10^6 m). Simplifying this expression, we find that v ≈ 11186 m/s.
Hence, to escape from the Earth's gravitational pull, the projectile must be fired with an initial velocity of approximately 11186 m/s.
In summary, to orbit just over the surface of the planet, the projectile needs an initial velocity of 6301.9 m/s, while to escape from the Earth's gravitational pull, it requires an initial velocity of 11186 m/s.
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2. Describe what happens when a transverse wave travels from a low density medium to a high density medium. (3 marks)
When a transverse wave travels from a low density medium to a high density medium, it undergoes reflection and inversion of the wave.
When a wave travels from one medium to another medium, the wave undergoes a change in its speed and direction of propagation. It also undergoes reflection and inversion, if there is a boundary present between the two media. The direction of propagation changes at the boundary surface of two media due to the variation of refractive indices of two media. The wave inversion occurs at the boundary surface of two media. So, when a transverse wave travels from a low density medium to a high density medium, it undergoes reflection and inversion of the wave.The inversion of the wave is when the wave goes from an upside-down position to a right-side-up position.
This is what happens when the wave goes from a lower density medium to a higher density medium. When the wave hits the boundary between the two media, it is reflected back in the opposite direction, with the same frequency and wavelength. The speed of the wave is determined by the medium through which it is traveling, so when the wave hits the boundary, it slows down as it enters the higher density medium.
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Required Information Suppose 100 mol of oxygen is heated at a constant pressure of 100 atm from 100'C 10 25 0°C, What is the magnitude of the work done by the gas during this expansion? The magnitude of the work done by the gas is
The magnitude of the work done by the gas during this expansion is 827 J.
The magnitude of the work done by the gas during this expansion of 100 moles of oxygen heated at a constant pressure of 100 atm from 100°C to 25°C can be calculated using the following equation for work done:
[tex]W = -PΔV[/tex]
where, P is the pressure of the gas and ΔV is the change in the volume of the gas.
The change in volume can be calculated using the ideal gas law:
PV = nRT, where, P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant and T is the temperature in Kelvin.
Using this formula, we can calculate the initial and final volumes of the gas. Let's assume the initial volume is V1 and the final volume is V2.
Therefore, [tex]PV1 = nRT1[/tex]
PV2 = nRT2
ΔV = V2 - V1
= (nR/P) (T1 - T2)
Putting the values, we get:
ΔV = (100 mol x 8.314 J/mol.K x (100+273) K) / 100 atm - (100 mol x 8.314 J/mol.K x (25+273) K) / 100 atm
ΔV = 8.27 L
The work done by the gas is:
W = -PΔV
= -100 atm x (-8.27 L)
= 827 J
Therefore, the magnitude of the work done by the gas during this expansion is 827 J.
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What is the activity (in Bq) of a sample of Cs-137 if 31.6 years
ago it was recorded to have an activity of 9932.8 Bq.
To calculate the activity of a sample of Cs-137 after a certain time, we need to consider its half-life. Cs-137 has a half-life of 30.17 years. The activity of the Cs-137 sample is approximately 6437.2 Bq.
Given that the Cs-137 sample had an initial activity of 9932.8 Bq 31.6 years ago, we can calculate the current activity by using the half-life of Cs-137, which is 30.17 years.
The formula to calculate the current activity is: A = A₀ × (1/2)^(t/t₁/₂), where A is the current activity, A₀ is the initial activity, t is the time elapsed, and t₁/₂ is the half-life.
Substituting the values into the formula, we have:
A = 9932.8 Bq × (1/2)^(31.6/30.17)
Calculating this expression, we find that the current activity of the Cs-137 sample is approximately 6437.2 Bq.
Therefore, the activity of the Cs-137 sample, 31.6 years after it was recorded to have an activity of 9932.8 Bq, is approximately 6437.2 Bq.
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A 1kg ball is fired from a cannon. What is the change in the
ball’s kinetic energy when it accelerates form 4.0 m/s2
to 8 m/s2?"
The change in the ball's kinetic energy when it accelerates from 4.0 m/s^2 to 8 m/s^2 is 64 Joules.
To calculate the change in kinetic energy, we need to determine the initial and final kinetic energies and then find the difference between them.
The formula for kinetic energy is given by:
Kinetic Energy = [tex](1/2) * mass * velocity^2[/tex]
Mass of the ball (m) = 1 kg
Initial acceleration (a₁) = 4.0 m/s²
Final acceleration (a₂) = 8 m/s²
Let's calculate the initial and final velocities using the formula of accelerated motion:
v = u + a * t
For initial velocity:
u = 0 (assuming the ball starts from rest)
a = a₁ = 4.0 m/s²
t = 1 second (arbitrary time interval for convenience)
Using the formula, we find:
v₁ = u + a₁ * t
v₁ = 0 + 4.0 * 1
v₁ = 4.0 m/s
For final velocity:
u = v₁ (the initial velocity is the final velocity from the previous calculation)
a = a₂ = 8 m/s²
t = 1 second (again, an arbitrary time interval for convenience)
Using the formula, we find:
v₂ = u + a₂ * t
v₂ = 4.0 + 8 * 1
v₂ = 12.0 m/s
Now, we can calculate the initial and final kinetic energies using the formula mentioned earlier:
Initial Kinetic Energy (KE₁) = (1/2) * m * v₁^2
KE₁ = (1/2) * 1 * 4.0^2
KE₁ = 8.0 J (Joules)
Final Kinetic Energy (KE₂) = (1/2) * m * v₂^2
KE₂ = (1/2) * 1 * 12.0^2
KE₂ = 72.0 J (Joules)
Finally, we can determine the change in kinetic energy:
Change in Kinetic Energy = KE₂ - KE₁
Change in Kinetic Energy = 72.0 J - 8.0 J
Change in Kinetic Energy = 64.0 J (Joules)
Therefore, the change in the ball's kinetic energy when it accelerates from 4.0 m/s² to 8 m/s² is 64.0 Joules.
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Review. The cosmic background radiation is blackbody radiation from a source at a temperature of 2.73K. (b) In what part of the electromagnetic spectrum is the peak of the distribution?
The peak of the distribution of the cosmic background radiation is in the microwave part of the electromagnetic spectrum. The frequency falls within the microwave region of the electromagnetic spectrum, indicating that the cosmic background radiation has its peak emission in that specific range.
The peak wavelength or frequency of blackbody radiation can be determined using Wien's displacement law, which states that the wavelength of the peak emission is inversely proportional to the temperature of the blackbody.
The formula for Wien's displacement law is:
λ_peak = b/T
where λ_peak is the peak wavelength, T is the temperature of the blackbody, and b is Wien's displacement constant, which is approximately equal to 2.898 × 10^(-3) m·K.
Substituting the given temperature T = 2.73 K into the formula, we can calculate the peak wavelength:
λ_peak = (2.898 × 10^(-3) m·K) / 2.73 K
≈ 1.06 × 10^(-3) m
To determine the corresponding region of the electromagnetic spectrum, we can use the relationship between wavelength and frequency:
c = λ · ν
where c is the speed of light (approximately 3.00 × 10^8 m/s), λ is the wavelength, and ν is the frequency.
Rearranging the equation, we get:
ν = c / λ
Substituting the calculated peak wavelength into the equation and solving for the frequency, we find:
ν = (3.00 × 10^8 m/s) / (1.06 × 10^(-3) m)
≈ 2.83 × 10^11 Hz
The frequency obtained corresponds to the microwave region of the electromagnetic spectrum.
The peak of the distribution of the cosmic background radiation, which is blackbody radiation from a source at a temperature of 2.73 K, is in the microwave part of the electromagnetic spectrum. This result is obtained by applying Wien's displacement law, which relates the peak wavelength of blackbody radiation to the temperature of the source.
The peak wavelength is determined to be approximately 1.06 × 10^(-3) m, which corresponds to a frequency of approximately 2.83 × 10^11 Hz. The frequency falls within the microwave region of the electromagnetic spectrum, indicating that the cosmic background radiation has its peak emission in that specific range.
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A tight rope has a longitudinal density (5 x 10^-2
kg/m) and a tensile force of 80 N. The rope's angular frequency
is.
The power that has to be supplied to the rope to generate harmonic waves at a frequency of 60Hz and an amplitude of 6cm is 251W. Option b. 251W is correct.
To calculate the power required to generate harmonic waves on the rope, we can use the formula:
P = (1/2) * μ * v * ω^2 * A^2
Where:
P is the power,μ is the linear mass density of the rope (kg/m),v is the velocity of the wave (m/s),ω is the angular frequency of the wave (rad/s),and A is the amplitude of the wave (m).First, let's calculate the velocity of the wave. For a wave on a stretched rope, the velocity is given by:
v = √(T/μ)
Where T is the tension in the rope (N).
Given:
Linear mass density (μ) = 5 × 10^2 kg/mTension (T) = 80 NAmplitude (A) = 6 cm = 6/100 mFrequency (f) = 60 Hzω = 2πfCalculating the velocity:
v = √(T/μ) = √(80 / (5 × 10^2)) = √(16/100) = 0.4 m/s
Calculating ω:
ω = 2πf = 2π(60) = 120π rad/s
Now, substituting the values into the power formula:
P = (1/2) * μ * v * ω^2 * A^2
= (1/2) * (5 × 10^2) * (0.4) * (120π)^2 * (6/100)^2
≈ 251 W
Therefore, the power that has to be supplied to the rope to generate harmonic waves at a frequency of 60 Hz and an amplitude of 6 cm is approximately 251 W. Therefore, option b. 251W is the correct answer.
The complete question should be:
A stretched rope having linear mass density 5×10²kgm⁻¹ is under a tension of 80N. The power that has to be supplied to the rope to generate harmonic waves at a frequency of 60Hz and an amplitude of 6cm is
a. 215W
b. 251W
c. 512W
d. 521W
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How far from a wire carrying a current of 5 Amps is a second, parallel wire with a a current of 10 Amps, if the Magnetic Force of wire 1 on wire 2 is 3.6 x 10-2 N and each wire is 36 meters long. Include a picture and all 3 vectors on both wires,
If the Magnetic Force of wire 1 on wire 2 is 3.6 x 10-2 N and each wire is 36 meters long then, the two parallel wires must be 2 meters apart from each other.
The formula to calculate the magnetic force between two parallel conductors is given as : F = µI₁I₂l / 2πd
where
F is the magnetic force
µ is the permeability of free space, µ = 4π x 10-7 TmA-1
I₁ is the current flowing in the first conductor
I₂ is the current flowing in the second conductor
l is the length of the conductors
d is the distance between the conductors
In the given problem, we have :
I₁ = 5 Amps ; I₂ = 10 Amps ; F = 3.6 x 10-2 N ; l = 36 meters
The value of permeability of free space, µ = 4π x 10-7 TmA-1
We can rearrange the above formula to find the value of d as : d = µI₁I₂l / 2πF
Substituting the given values, we get, d= 2m
Therefore, the two parallel wires must be 2 meters apart from each other.
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: 1. Two masses M and m hang on a three looped pulley as shown below. M is 50 kg and m is 12 kg. There is also a rope that prevents rotation. The radii are 18cm, 48cm, and 60cm. a) Determine the torque from the mass M b) Determine the Tension in the horizontal rope M c) Later the string holding m is cut. What would be the tension in the rope now?
The torque from mass M is 88.2 N·m, the tension in the horizontal rope for mass M is 490 N, and when the string holding mass m is cut, the tension in the rope remains at 490 N.
a) To determine the torque from the mass M, we need to calculate the force exerted by M and the lever arm distance. The force exerted by M is equal to its weight, which is given by F = M * g, where g is the acceleration due to gravity. Thus, F = 50 kg * 9.8 m/[tex]s^2[/tex] = 490 N.
The lever arm distance is the radius of the pulley on which M hangs, which is 18 cm or 0.18 m. Therefore, the torque from mass M is given by torque = F * r = 490 N * 0.18 m = 88.2 N·m.
b) To determine the tension in the horizontal rope for mass M, we can consider the equilibrium of forces. Since the system is at rest, the tension in the horizontal rope is equal to the weight of M, which is Tension = M * g = 50 kg * 9.8 m/[tex]s^2[/tex] = 490 N.
c) When the string holding m is cut, the tension in the rope will no longer be determined by the weight of m. Instead, it will only be determined by the weight of M. Therefore, the tension in the rope would remain the same as in part (b), which is Tension = 490 N.
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Which of these statements best explains why a telescope enables us to see details of a distant object such as the Moon or a planet more clearly?
The image formed by the telescope is larger than the object.
The image formed by the telescope extends a larger angle at the eye than the object does.
The telescope can also collect radio waves that sharpen the visual image
Justify your answer to the previous question. choose 1
Interference
Light Gathering Power
Rayleigh Criterion
The statement that best explains why a telescope enables us to see details of a distant object such as the Moon or a planet more clearly is: The image formed by the telescope is larger than the object.
Telescope enables us to see details of a distant object such as the Moon or a planet more clearly because the image formed by the telescope is larger than the object. It is because the image is formed by the convergence of light rays from the object at a single point and at the same distance from the lens of the telescope. This forms an enlarged and more detailed view of the object, which helps in seeing it more clearly. This is how a telescope magnifies the image of a distant object.
The other options do not explain why a telescope enables us to see details of a distant object such as the Moon or a planet more clearly. The statement "The image formed by the telescope extends a larger angle at the eye than the object does" is incorrect because a telescope does not extend the angle at the eye. The statement "The telescope can also collect radio waves that sharpen the visual image" is also incorrect because telescopes cannot collect radio waves, radio telescopes are specifically designed to do this.
Justification: The correct answer for the previous question is Light Gathering Power. Light gathering power is a measure of the ability of a telescope to collect light. The larger the telescope's light gathering power, the more light it can collect, which enables it to form a brighter and more detailed image of the object being observed. This is important because the more light the telescope collects, the greater the amount of detail that can be seen.
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