Answer:
The value is [tex]\lambda = 2 \ m[/tex]
Explanation:
From the question we are told that
The distance of the speaker from the second speaker to the east is [tex]d = 3 \ m[/tex]
The distance of the speaker from the listener to the south is [tex]a = 4 \ m[/tex]
Generally given that if the speaker move in any direction, their sound become louder , it then mean that the position of the listener of minimum sound (i.e a position of minima ) ,
Generally the path difference of the sound produce by both speaker at a position of minima is mathematically represented as
[tex]y = \frac{\lambda}{2}[/tex]
Generally considering the orientation of the speakers and applying Pythagoras theorem we see that distance from the second speaker to the listener is mathematically represented as
[tex]b = \sqrt{d^ 2 + a^2 }[/tex]
=> [tex]b = \sqrt{3^ 2 + 4^2 }[/tex]
=> [tex]b = 5[/tex]
Generally the path difference between the two speaker with respect to the listener is
[tex]y = b - a[/tex]
=> [tex]y = 5 - 4[/tex]
=> [tex]y = 1[/tex]
So
[tex]1 = \frac{\lambda}{2}[/tex]
=> [tex]\lambda = 2 \ m[/tex]
A boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, away from shore. As a consequence, the raft (a) remains stationary, (b) moves away from the shore, or (c) moves toward the shore. (Hint: Use Conservation of Momentum)
Answer:
(c) moves toward the shore.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = mass * velocity [/tex]
The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.
In this scenario, a boy standing at one end of a floating raft that is stationary relative to the shore walks to the opposite end of the raft, away from shore. As a consequence, the raft moves toward the shore because the momentum of the closed system (boy and raft) has a zero magnitude and would remain constant.
A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the ydirection of ay = 7.30 m/s2. The engines turn off after firing for 675 s, at which point the spacecraft has velocity components of vx = 3630 m/s and vy = 4276 m/s.What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on?Express the direction as an angle measured counterclockwise from the x -axis.
Answer:
v₀ = 677.94 m / s , θ = 286º
Explanation:
We can solve this exercise using the kinematic expressions, let's work on each axis separately.
X axis
has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation
vₓ = v₀ₓ + aₓ t
v₀ₓ = vₓ - aₓ t
let's calculate
v₀ₓ = 3630 - 5.10 675
v₀ₓ = 187.5 m / s
Y Axis
[tex]v_{y}[/tex] = v_{oy} - a_{y} t
v_{oy} = v_{y} - a_{y} t
let's calculate
v_{oy} = 4276 - 7.30 675
v_{oy} = -651.5 m / s
we can give the speed starts in two ways
a) v₀ = (187.5 i ^ - 651.5 j ^) m / s
b) in the form of module and angle
Let's use the Pythagorean theorem
v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]
v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]
v₀ = 677.94 m / s
we use trigonometry
tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]
θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }
θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])
θ = -73.94º
This angle measured from the positive side of the x-axis is
θ‘ = 360 - 73.94
θ = 286º
3.A bridge usually has expansion joints. They allow the bridge to become slightly longer when it
experiences thermal expansion. Look at the diagram below of the bridge joint. When the weather
becomes cool, the "teeth" of the joint move away from each other. When the weather becomes
warm, they move toward each other. Which statement is true about the particles that make up the
bridge?
A.When the sides of the joint are close together, the particles have more kinetic energy
than they do when the sides are farther apart.
B.When the sides of the joint are far apart, the particles have more kinetic energy than
they do when the sides are closer together.
C.The particles contain the same amount of kinetic energy no matter how much the
bridge has expanded.
D. The kinetic energy of the particles changes, but the amount it changes does not
depend on the temperature of the bridge.
Answer:
a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.
Explanation:
Starting at t = 0 a net external force in the +x-direction is applied to an object that has mass 5.00 kg. A graph of the force as a function of time is a straight line that passes through the origin and has slope 5.00 N/s.
If the object is at rest at t = 0 what is the magnitude of the force when the object has reached a speed of 9.00 m/s?
Express your answer with the appropriate units.
Answer:
15√2 N
Explanation:
The acceleration is given by ...
a = F/m = 5t/5 = t . . . . meters/second^2
The velocity is the integral of acceleration:
v = ∫a·dt = (1/2)t^2
This will be 9 m/s when ...
9 = (1/2)t^2
t = √18 . . . . seconds
And the force at that time is ...
F = 5(√18) = 15√2 . . . . newtons
A student is performing an experiment that involves the charge on a metal sphere that is attached to a charged electroscope. A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more. The rod is then removed, and the leaves return to their initial separated position. The student repeats the procedure, but this time the electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves again end up separated. What can be concluded about the charge on the separated leaves of the electroscope
Answer:
The leaves have a charge in each experiment, but the sign of the charge cannot be determined.
Explanation:
In the first experiment, A charged rod is brought near the sphere without touching it. As a result the leaves of the electroscope separate more.
Thus indicates that there are charges involved. Now, like charges would repel like what is happening here but we don't know if they are both positive or negative because in both cases, they will still repel.
Now for the second experiment, electroscope is grounded and the ground is removed before the rod is removed from near the sphere. The leaves end up being separated again.
Similar to the first time, it's clear there are charges but the charges repel. Thus, they are the same sign charges but we don't know if they are both positive or negative.
Thus, in both cases we can conclude that the leaves have charges but we don't know their signs.
The diameter of 0-gauge copper wire is 8.252 mm. Find the resistance of a 1.00-km length of such wire used for power transmission.
Answer:
The value is [tex]R = 0.321 \ \Omega[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 8.252 \ mm = 0.008252 \ m[/tex]
The length of the wire is [tex]l = 1.0 \ km = 1000 \ m[/tex]
Generally the cross sectional area of the copper wire is mathematically represented as
[tex]A = \pi * \frac{d^2}{4}[/tex]
=> [tex]A = 3.142 * \frac{ 0.008252^2}{4}[/tex]
=> [tex]A = 5.349 *10^{ - 5} \ m^2[/tex]
Generally the resistance is mathematically represented as
[tex]R = \frac{\rho * l }{A }[/tex]
Here [tex]\rho[/tex] is the resistivity of copper with the value [tex]\rho = 1.72*10^{-8} \ \Omega \cdot m[/tex]
=> [tex]R = \frac{1.72 *10^{-8} * 1000 }{5.349 *10^{ - 5} }[/tex]
=> [tex]R = 0.321 \ \Omega[/tex]
A 0.046 kg golf ball hit by a driver can accelerate from rest to 67 m/s in 1 ms while the driver is in contact with the ball. How much average force does the golf ball experience
Answer:
Average force = 67 mn
Explanation:
Given:
Initial velocity u = 0 m/s
Final velocity v = 67 m/s
Time t = 1 ms = 0.001 sec.
Computation:
Using Momentum theory
Change in momentum = F × Δt
(v-u)/t = F × Δt
F × 0.001 = (67 - 0)/0.001
F= 67,000,000
Average force = 67 mn
Vary the sled’s height and mass. Observe the effect of each change on the potential energy of the sled.
a. How does potential energy change when height is increased?
b. How does potential energy change when mass is increased?
c. Compare a sled’s potential energy at 10 m to its potential energy at 20 m. How does doubling height affect potential energy?
d. Compare the potential energy of a 100-kg sled and a 200-kg sled at the same height. How does doubling mass affect potential energy?
Answer:
a. Potential energy of the sled is increased when height of sled is increased.
b. Potential energy of the sled is increased when height of sled is increased.
c. P.E₂₀ = 2 P.E₁₀
d. P.E₂₀₀ = 2 P.E₁₀₀
Explanation:
The potential energy of the sled can be given by the following:
[tex]Potential\ Energy = P.E = mgh\\[/tex]
where,
m = mass of sled
g = acceleration due to gravity
h = height of sled
a.
It is clear from the formula that potential energy of sled is directly proportional to the height of sled.
Therefore, potential energy of the sled is increased when height of sled is increased.
b.
It is clear from the formula that potential energy of sled is directly proportional to the mass of sled.
Therefore, potential energy of the sled is increased when mass of sled is increased.
c.
[tex]P.E\ at\ 10\ m:\\P.E_{10} = 10mg\\P.E\ at\ 20\ m:\\P.E_{20} = 20mg\\\frac{P.E_{20}}{P.E_{10}} = \frac{20mg}{10mg}[/tex]
P.E₂₀ = 2 P.E₁₀
d.
[tex]P.E\ at\ 100\ kg:\\P.E_{100} = 100gh\\P.E\ at\ 200\ m:\\P.E_{200} = 200gh\\\frac{P.E_{200}}{P.E_{100}} = \frac{200gh}{100gh}[/tex]
P.E₂₀₀ = 2 P.E₁₀₀
Figure shows four paths along which objects move from a starting point to a final point, all im the same time interval. The paths pass over a grid of equally spaced straight lines. Rank the paths according to the average velocity of the objects.
Answer:
12345678901234567890
A child blows a leaf from rest straight up in the air. the leaf has a constant upward acceleration of magnitude 1.0 m by s square. how much time does it take the leaf to displace 1.0m upwards?
Answer:
√2
Explanation:
From the question, we're given that the
Acceleration of the leaf is 1 m/s²
Change in displacement of the leaf is 1 m/s.
Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest
Now, to solve this, we don't the equation of motion to ur
S = ut + 1/2at², substituting the whole parameters, we then have
1 = 0 * t + 1/2 * 1 * t²
1 = 1/2 * t²
t²/2 = 1
t² = 2
t = √2 seconds
Therefore the time it takes the leaf to dislodge is 2 seconds
Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
Complete Question
Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light
Answer:
The diameter is [tex]D = 0.59 \ m[/tex]
Explanation:
From the question we are told that
The best resolution is [tex]\theta = 0.3 \ arcsecond[/tex]
The wavelength is [tex]\lambda = 700 \ nm = 700 *10^{-9 } \ m[/tex]
Generally the
1 arcminute = > 60 arcseconds
=> x arcminute => 0.3 arcsecond
So
[tex]x = \frac{0.3}{60 }[/tex]
=> [tex]x = 0.005 \ arcminutes[/tex]
Now
60 arcminutes => 1 degree
0.005 arcminutes = > z degrees
=> [tex]z = \frac{0.005}{60 }[/tex]
=> [tex]z = 8.333 *10^{-5} \ degree[/tex]
Converting to radian
[tex]\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian[/tex]
Generally the resolution is mathematically represented as
[tex]\theta = \frac{1.22 * \lambda }{ D}[/tex]
=> [tex]D = \frac{1.22 * \lambda }{\theta }[/tex]
=> [tex]D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }[/tex]
=> [tex]D = 0.59 \ m[/tex]
A 50.0-kg child stands at the rim of a merry-go-round of radius 2.25 m, rotating with an angular speed of 3.30 rad/s.. What is the child's centripetal acceleration?
Answer: the child's centripetal acceleration=24.50 m/s²
Explanation:
Given that mass of child= 50 kg
radius of merry go round= 2.25m
angular speed = 3.30 rad/s
Centripetal Acceleration = v²/ r
But V= ωr
So Centripetal Acceleration = v²/ r = (ωr)²/ r
=(3.30)² x (2.25)²/ 2.25 = (3.30)² x 2.25
=24.5025m/s²
=24.50 m/s²
If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 29.1 m/s?
A. 54 m
B. 50 m
C. 64 m
D. 60 m
Answer:
54m is the answer......a boulder with a mass of about 1.5 x 10^5 kg falls and strikes the ground at 70 m/s how much kinetic energy dies the boulder deliver to the ground PLEASE HELP
Answer:
k. e= 1/2 mv^2
Ke = 1/2 * 1.5 * 10^5 * 70^2
3.675 *10^8 joules
What was your train of thought as you navigated the picture of the candle?
Answer:
Where is the picture
Explanation:
WHERE IS THE PICTURE
How does temperature rise and impurities affect the surface tension of water
(2 mks)
Answer:
Surface tension is the downward force acting on the surface of liquid due to presence of inter molecular forces or cohesive forces between the particles of liquid.
Surface tension decreases with increase in temperature as the forces among particles decrease due to increase in kinetic energy and thus the cohesive nature decreases and thus surface tension also decreases.
Surface tension may decrease or increase with increase in soluble impurities .Insoluble impurities decrease the surface tension.
a planet has been detected in a circular orbit around the star Rho1 Cancri with an orbital radius equal to 1.65 x 10^10 m. the orbital period of this planet is approximately 14.5 days which is the approximate mass of the star pho1 cancri
Answer:
Approximately [tex]1.69 \times 10^{30}\; \rm kg[/tex].
Explanation:
Deduction of the formulaLet [tex]M[/tex] and [tex]m[/tex] denote the mass of the star and the planet, respectively.
Let [tex]G[/tex] denote the constant of universal gravitation ([tex]G \approx 6.67408 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1}\cdot s^{-2}[/tex].)
Let [tex]r[/tex] denote the orbital radius of this planet (assuming that [tex]r\![/tex] is constant.) The question states that [tex]r = 1.65 \times 10^{10}\; \rm m[/tex].
The size of gravitational attraction of the star on this planet would be:[tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].
If attraction from the star is the only force on this planet, the net force on this planet would be [tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].
Let [tex]\omega[/tex] denote the angular velocity of this planet as it travels along its circular orbit around the star. The size of [tex]\omega\![/tex] could be found from the period [tex]T[/tex] of each orbit: [tex]\omega = (2\, \pi) / T[/tex].
In other words, this planet of mass [tex]m[/tex] is in a circular motion with radius [tex]r[/tex] and angular velocity [tex]\omega[/tex]. Therefore, the net force on this planet should be equal to [tex]m \cdot \omega^2 \cdot r[/tex].
Hence, there are two expressions for the net force on this planet:
[tex]\text{Net Force} = \displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex] from universal gravitation, and[tex]\displaystyle \text{Net Force} = m \cdot \omega^2 \cdot r = {\left(\frac{2\pi}{T}\right)}^{2} m \cdot r[/tex] from circular motion.Equate the right-hand side of these two equations:
[tex]\displaystyle \frac{G \cdot M \cdot m}{r^2} = {\left(\frac{2\pi}{T}\right)}^{2}\, m \cdot r[/tex].
Simplify this equation and solve for [tex]M[/tex], the mass of the star:
[tex]\displaystyle M = \frac{{(2\pi / T)}^2 \cdot r^3}{G}[/tex].
Notice that [tex]m[/tex], the mass of the planet, was eliminated from the equation. That explains why this question could be solved without knowing the exact mass of the observed planet.
Actual CalculationsConvert the orbital period of this star to standard units:
[tex]\begin{aligned}T &= 14.5\; \text{day} \times \frac{24\; \text{hour}}{1\; \text{day}} \times \frac{3600\; \text{second}}{1\; \text{hour}} \\ & = 1.2528 \times 10^{6}\; \rm \text{second}\end{aligned}[/tex].
Calculate the mass of the star:
[tex]\begin{aligned} M &= \frac{{(2\pi / T)}^2 \cdot r^3}{G} \\ &\approx \frac{\displaystyle {\left(\frac{2\pi}{1.2528 \times 10^{6}\; \rm s}\right)}^{2} \times \left(1.65 \times 10^{10}\; \rm m\right)^{3}}{6.67408 \times 10^{-11}\; \rm m^{3}\cdot kg^{-1} \cdot s^{-2}}\\ &\approx 1.69 \times 10^{30}\; \rm kg\end{aligned}[/tex].
You are driving at 25 m/s when an ambulance passes you and pulls into your lane, just in front of you, and speeds ahead at 35 m/s. The ambulance driver hears a siren sound of 850 Hz. What frequency do you hear
Answer:
The value is [tex]f_1 = 828 \ Hz[/tex]
Explanation:
From the question we are told that
The speed is [tex]v = 25 \ m/s[/tex]
The speed of the ambulance is [tex]u = 35 \ m/ s[/tex]
The frequency of the siren is [tex]f = 850 \ Hz[/tex]
Generally from Doppler effect equation we have that
[tex]f_1 = \frac{v_s - v }{ v_s + u } * f[/tex]
Here [tex]v_s[/tex] is the velocity of sound with the value [tex]v_s = 343 \ m/s[/tex]
=> [tex]f_1 = \frac{343 - 25 }{ 343 + 35 } * 850[/tex]
=> [tex]f_1 = 828 \ Hz[/tex]
Show that a 2,500,000-J change in kinetic energy occurs for an airplane that is moved 500 m in takeoff by a sustained force of 5000 N.
Answer:
The answer to your question is given below
Explanation:
To solve this problem, we'll assume that the plane is initially at rest.
Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0
Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:
Force (F) = 5000 N
Distance (s) = 500 m
Energy (E) =?
E = F × s
E = 5000 × 500
E = 2500000 J
Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J
Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:
Initial kinetic energy (KE₁) = 0
Final kinetic energy (KE₂) = 2500000 J
Change in kinetic energy (ΔKE) =?
ΔKE = KE₂ – KE₁
ΔKE = 2500000 – 0
ΔKE = 2500000 J
Hence, the change in the kinetic energy of the plane is 2500000 J.
A force of 100 newtons Is applled to a box at an angle of 36° with the horizontal. If the mass of the box Is 25 kilograms, what Is the horizontal
How should the mass of a harmonic oscillator be changed to double the frequency? Can the frequency be tripled by a suitable adjustment of the mass?
Answer:
a. [tex]m' = \frac{m}{4}[/tex]
b. [tex]m' = \frac{m}{9}[/tex]
Explanation:
The frequency of a harmonic oscillator is given by the following formula:
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] ----------------- equation (1)
a.
In order to double the frequency of this oscillator:
ω' = 2ω
m' = ?
Therefore,
[tex]\omega ' = 2\omega = \sqrt{\frac{k}{m'}}[/tex]
using equation (1):
[tex]2 \sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{4}{m} = \frac{1}{m'}[/tex]
[tex]m' = \frac{m}{4}[/tex]
a.
In order to triple the frequency of this oscillator:
ω' = 3ω
m' = ?
Therefore,
[tex]\omega ' = 3\omega = \sqrt{\frac{k}{m'}}[/tex]
using equation (1):
[tex]3\sqrt{\frac{k}{m}} = \sqrt{\frac{k}{m'}}\\\\ \frac{9}{m} = \frac{1}{m'}[/tex]
[tex]m' = \frac{m}{9}[/tex]
A) To double the Frequency of a harmonic oscillator ;
Divide the mass by four i.e. m₁ = m / 4
B) To triple the frequency of a harmonic oscillator :
Divide the mass by nine (9) i.e. m₂ = m / 9
Given that The frequency of a harmonic oscillator is expressed as
[tex]w = \sqrt{\frac{k}{m} }[/tex] -- ( 1 )
A) Doubling the frequency
[tex]2w = \sqrt{\frac{k}{m_{1} } }[/tex] ---- ( 2 )
Applying equation ( 1 ) and ( 2 )
[tex]2\sqrt{\frac{k}{m} } = \sqrt{\frac{k}{m_{1} } }[/tex]
squaring both sides
( 4 / m ) = 1 / m₁
∴ m₁ ( new mass ) = m / 4
B) Tripling the frequency
3w = [tex]\sqrt{\frac{k}{m_{2} } }[/tex] ---- ( 3 )
applying equation ( 1 ) and ( 3 )
[tex]3 \sqrt{\frac{k}{m} } = \sqrt{\frac{k}{m_{2} } }[/tex]
squaring both sides
( 9 / m ) = 1 / m₂
∴ m₂ = m / 9
Hence we can conclude that To double the Frequency of a harmonic oscillator m₁ = m / 4 and To triple the frequency of a harmonic oscillator : m₂ = m / 9
Learn more : https://brainly.com/question/20050933
Choose the best answer. A car traveling at constant speed has a net work of zero done on it.
A. True
B. False
Answer:
Explanation:
we know that,
force = mass × acceleration
∴ since speed/velocity is constant, acceleration should be zero.
∴ f = m × 0
f = 0 N
∴ If we apply this to ,
work = force × displacement
we get ,
w = 0 × s
∴ we can say that the net work is zero.
and hence the answer is true!!!
What is the speed that is measured in speedometer to track speed violation?
Answer:
The officer's unit detects this 135-mile-per-hour speed and should subtract the patrol car's 70-mile -per-hour ground speed to get your true speed of 65 miles per hour. Instead, the officer's ground-speed beam fixes on the truck ahead and measures a false 50-mile-per-hour ground speed.
Explanation:
A speedometer or speed meter is a gauge that measures and displays the instantaneous speed of a vehicle. Now universally fitted to motor vehicles, they started to be available as options in the early 20th century, and as standard equipment from about 1910 onwards.
Are volcanoes fed by highly viscous magma a greater threat to life and property than volcanoes supplied with very fluid magma?
Answer:
A volcano fed by highly viscous magma is likely to be a greater threat to life and property than a volcano supplied with very fluid magma because with high viscous magma gas is trapped more in the magma so the gas will build up and then eventually explode, whereas with fluid magma the gas can escape allowing the magma.
HOPE THIS HELPS!!!
Explanation:
Less fluid magma done great damaged to the property and life as compared to highly viscus magma.
Highly viscus magma VS less viscous magmaNo, volcanoes that are fed by highly viscous magma are not a greater threat to life and property than volcanoes supplied with very fluid magma because the highly viscous magma can't move to a large distance due to its large viscosity.
While on the other hand, those volcanoes that supplied with very fluid magma do great damaged to the property due to its easily flowing on the surface of earth so we can conclude that less fluid magma done great damaged to the property and life as compared to highly viscus magma.
Learn more about magma here: https://brainly.com/question/23661578
How does the spring constant of the smaller springs relate to that of the original spring? Choose the correct explanation.
A) The spring constant of each half will be half the spring constant of the original long spring since it will stretch twice as much under the same tension
B) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch twice as much under the same tension.
C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.
D) The spring constant of each half will be half the spring constant of the original long spring since it will stretch only half as much under the same tension.
Answer:
C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.
Explanation:
Hooke's law states that the force needed to extend or compress a spring by a distance is proportional to that distance. If is given as:
F = ke, where F is the force applied, k is spring constant and e is the extension.
If a force f is applied to a spring with a spring constant k and by a distance stretched (x) then:
k = F / x
For half the spring, if the same force F is applied, the stretch would be half (x/2), hence the spring constant C is:
C = F / (x/2)
C = 2 (F / x) = 2 * spring constant of original spring
define alpha and beta
alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.
beta is a measure of volatility relative to a benchmark ,such as the S&P 500.
Explanation:
alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answer: The red at the bottom represents the base of the tube, not the red liquid.
Explanation:
The densest materials have more weight per unit of volume.
This means that those elements will always flow to the bottom of the containers, like the one in the image.
The red liquid being the least dense one, can not go to the bottom by its own means.
There could be some cases, like:
The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.
Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.
The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.
An incandescent lightbulb is rated at 120 Watts when plugged into a 200 V-rms household outlet. Calculate the resistance of the filament and the rms current. g
Answer:
The resistance of the filament is 333.33 ohmsThe rms current is 0.6 AExplanation:
Given;
output power of the incandescent lightbulb , P = 120 W
input voltage, V = 200 V
The resistance of the filament is calculated as;
[tex]P = \frac{V^2}{R}[/tex]
Where;
R is the resistance of the filament
[tex]R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms[/tex]
The rms current is given as;
[tex]P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A[/tex]
Two stones, one with twice the mass of the other, are thrown straight up and rise to the same height h. Compare their changes in gravitational potential energy.
a. They rise to the same height, so the stone with twice the mass has twice the change in gravitational potential energy.
b. They rise to the same height, so they have the same change in gravitational potential energy.
c. The answer depends on their speeds at height h.
Answer:
the correct one is the a, U₂ = 2 U₁
Explanation:
The gravitational potential energy is
U = m g h
if the stones reach the same height, the energy of the first stone is
U₁ = m₁ g h
The second stone is twice the mass and reaches the same height
m₂ = 2 m₁
potential energy is
U₂ = m₂ g h
U₂ = 2 m₁ g h
U₂ = 2 U₁
therefore the energy is double.
When reviewing the statements, the correct one is the a
What happens when a population exceeds its carrying capacity?
Answer:
If a population exceeds carrying capacity, the ecosystem may become unsuitable for the species to survive. If the population exceeds the carrying capacity for a long period of time, resources may be completely depleted.