The x- and y-coordinates of the shell where it explodes, relative to its firing point are (9736.5 m, 762.3 m) respectively.
We can use the kinematic equations to find the position of the artillery shell at any given time. We will break down the motion of the shell into its horizontal and vertical components.
First, we can find the initial horizontal and vertical velocities of the shell as follows:
\begin{align} v_{0x} &= v_0 \cos(\theta) = 300 \cos(52.0^\circ) \approx 192.9\text{ m/s}\ v_{0y} &= v_0 \sin(\theta) = 300 \sin(52.0^\circ) \approx 245.4\text{ m/s} \end{align}
We can use the vertical motion of the shell to find the time it takes to reach its maximum height, using the following kinematic equation:
$$y = v_{0y}t - \frac{1}{2}gt^2$$
At maximum height, the vertical velocity will be zero, so we can solve for the time it takes to reach this point:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y}}{g} \approx 25.2\text{ s} \end{align}
Therefore, the time it takes for the shell to reach maximum height is 25.2 seconds. Using this time, we can find the maximum height, as follows:
\begin{align} y_\text{max} &= v_{0y}t - \frac{1}{2}gt^2\ &= 245.4\text{ m/s} \cdot 25.2\text{ s} - \frac{1}{2}(9.81\text{ m/s}^2)(25.2\text{ s})^2\ &\approx 762.3\text{ m} \end{align}
The time it takes for the shell to hit the mountainside can be found by solving for the time when y = 0:
\begin{align} 0 &= v_{0y}t - \frac{1}{2}gt^2\ t &= \frac{v_{0y} + \sqrt{(v_{0y})^2 + 2gy_\text{max}}}{g} \approx 50.5\text{ s} \end{align}
Therefore, the time it takes for the shell to hit the mountainside is 50.5 seconds. The x-coordinate of the explosion can be found by using the horizontal velocity and the time it takes for the shell to hit the mountainside:
\begin{align} x &= v_{0x}t\ &= 192.9\text{ m/s} \cdot 50.5\text{ s}\ &\approx 9736.5\text{ m} \end{align}
Therefore, the x-coordinate of the explosion is 9736.5 meters. The y-coordinate of the explosion is simply the height of the mountainside:
$$y = 0 + 762.3\text{ m} = 762.3\text{ m}$$
Therefore, the y-coordinate of the explosion is 762.3 meters.
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Plsss help Bumper car A (282 kg) moving +2.82 m/s
makes an elastic collision with bumper
car B (210 kg) moving +1.72 m/s. What is
the velocity of car A after the collision?
(Unit = m/s)
Remember: right is +, left is -
Answer:
Approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
Let [tex]m_{A}[/tex] and [tex]m_{B}[/tex] denote the mass of the two vehicles. Let [tex]u_{A}[/tex] and [tex]u_{B}[/tex] denote the velocity before the collision. Let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] denote the velocity after the collision.
Since the collision is elastic, both momentum and kinetic energy should be conserved.
For momentum to conserve:
[tex]m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B}[/tex].
For kinetic energy to conserve:
[tex]\displaystyle \frac{1}{2}\, m_{A} \, ({v_{A}}^{2}) + \frac{1}{2}\, m_{B} \, ({v_{B}}^{2}) = \frac{1}{2}\, m_{A}\, ({u_{A}}^{2}) + \frac{1}{2}\, m_{B}\, ({u_{B}}^{2})[/tex].
Simplify to obtain:
[tex]\displaystyle m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})[/tex].
It is given that [tex]m_{A} = 282\; {\rm kg}[/tex], [tex]m_{B} = 210\; {\rm kg}[/tex], [tex]u_{A} = 2.82\; {\rm m\cdot s^{-1}}[/tex], and [tex]u_{B} = 1.72\; {\rm m\cdot s^{-1}}[/tex]. The value (in [tex]{\rm m\cdot s^{-1}}[/tex]) of [tex]v_{A}[/tex] and [tex]v_{B}[/tex] can be found by solving this nonlinear system of two equations and two unknowns:
[tex]\left\lbrace \begin{aligned} & m_{A} \, v_{A} + m_{B} \, v_{B} = m_{A}\, u_{A} + m_{B}\, u_{B} \\ & m_{A} \, ({v_{A}}^{2}) + m_{B} \, ({v_{B}}^{2}) = m_{A}\, ({u_{A}}^{2}) + m_{B}\, ({u_{B}}^{2})\end{aligned}\right.[/tex].
[tex]\left\lbrace \begin{aligned} & 282 \, v_{A} + 210 \, v_{B} = 282\, (2.82) + 210\, (1.72) \\ & 282 \, ({v_{A}}^{2}) + 210 \, ({v_{B}}^{2}) = 282\, ({2.82}^{2}) + 210\, ({1.72}^{2})\end{aligned}\right.[/tex].
Solving this system gives two possible sets of solutions:
[tex]\left\lbrace\begin{aligned}v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].[tex]\left\lbrace\begin{aligned}v_{A} &\approx 2.82\; {\rm m\cdot s^{-1}} \\ v_{B} &\approx 1.72\; {\rm m\cdot s^{-1}}\end{aligned}\right.[/tex].However, the second set of solutions is invalid since it suggests that the velocity of the two vehicles stayed unchanged after the collision. Hence, only the first set of solutions ([tex]v_{A} &\approx 1.89\; {\rm m\cdot s^{-1}}[/tex], [tex]v_{B} &\approx 2.98\; {\rm m\cdot s^{-1}}[/tex]) is valid.
Therefore, the velocity of vehicle [tex]A[/tex] would be approximately [tex]1.89\; {\rm m\cdot s^{-1}}[/tex] after the collision.
A spring oriented vertically is attached to a hard horizontal surface as in the figure below. The spring has a force constant of 1.30 kN/m. How much is the spring compressed when a object of mass m = 2.70 kg is placed on top of the spring and the system is at rest? Answer should be in centimeters.
The spring is compressed by approximately 2.04 cm. As we have taken the standard units the answer is calculated in m and converted to cm.
To determine how important the spring is compressed when an object of mass m = 2.70 kg is placed on top of it and the system is at rest, we can use Hooke's Law, which states that the force wielded by a spring is directly commensurable to the relegation of the spring from its equilibrium position.
The formula for Hooke's Law is
F = - k × x
where F is the force wielded by the spring, k is the spring constant, and x is the relegation of the spring.
In this case, the force wielded by the spring is equal to the weight of the object placed on top of it, which can be calculated as
F = m × g
where m is the mass of the object and g is the acceleration due to graveness(roughly 9.8 m/ s²).
Given
Mass( m ) = 2.70 kg
Spring constant( k) = 1.30 kN/ m( Note 1 kN = 1000 N)
Converting the spring constant to Newtons
k = 1.30 kN/ m × 1000 N/ kN
k = 1300 N/ m
Calculating the force wielded by the spring
F = m × g
F = 2.70 kg × 9.8 m/ s²
F ≈26.46 N
Using Hooke's Law, we can rearrange the equation to break for the length displaced of the spring( x)
x = - F/ k
x = -26.46 N/ 1300 N/ m
x ≈-0.0204 m
The negative sign indicates that the spring is compressed. thus, when the object of mass m = 2.70 kg is placed on top of the spring and the system is at rest.
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a current of a 6 flows through a light bulb for 12 s, how many coulombs of charge pass through the light bulb during this time
A current of a 6 flows through a light bulb for 12 s. The total charge that passes through the light bulb during the given time is 72 coulombs.
To calculate the total charge that passes through the light bulb, we need to use the formula Q = I * t, where Q represents the charge in coulombs, I represents the current in amperes, and t represents the time in seconds.
Step 1: Identify the known values:
Current (I) = 6 amperes
Time (t) = 12 seconds
Step 2: Calculate the charge using the formula:
Q = I * t
Step 3: Substitute the known values into the formula:
Q = 6 amperes * 12 seconds
Q = 72 coulombs
Therefore, the total charge that passes through the light bulb during the given time is 72 coulombs.
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.An electron of charge 1.6 x 10-19is situated in a uniform electric filed strength of 120 vm-1 Calculate the force acting on it
The force acting on the electron is 1.92 x 10^-17 N.
The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.
We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.
The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.
F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.
Therefore, the force acting on the electron is 1.92 x 10^-17 N.
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One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.5 m/s. The first one is thrown at an angle of 58.0° with respect to the horizontal. Find a - At what angle should the second snowball be thrown to arrive at the same point as the first?, find b - How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?
The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.
the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.
To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.
Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.
The horizontal component of the velocity for the first snowball is given by:
V1x = V1 * cos(angle1)
= 26.5 m/s * cos(58.0°)
= 26.5 m/s * 0.530
= 14.045 m/s
Now, let's find the vertical component of the velocity for the first snowball:
V1y = V1 * sin(angle1)
= 26.5 m/s * sin(58.0°)
= 26.5 m/s * 0.848
= 22.472 m/s
Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.
The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:
t = (2 * V1y) / g
= (2 * 22.472 m/s) / 9.8 m/s²
≈ 4.582 s
Now, let's find the vertical displacement for the second snowball:
Δy = V1y * t - (0.5 * g * t²)
= 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)
≈ 103.049 m
To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:
tan(angle2) = Δy / Δx
= 103.049 m / (2 * 14.045 m/s * t)
= 103.049 m / (2 * 14.045 m/s * 4.582 s)
≈ 1.085
Now, we can find the angle2 by taking the arctan of both sides:
angle2 ≈ arctan(1.085)
angle2 ≈ 48.196°
Therefore,
To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.
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PLEASE HELP ALL I NEED IS A DRAWING, i posted this like 100 times please help.
Answer:
Hope this helps
Explanation:
A woman stands at the edge of a cliff and throws a pebble horizontally over the edge with a speed of v0 = 20.5 m/s. The pebble leaves her hand at a height of h = 55.0 m
above level ground at the bottom of the cliff, as shown in the figure. Note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the pebble leaves the hand. Answer parts a-f.
(a)The time taken for the pebble to reach the ground is approximately 2.01 seconds, and
(b) the horizontal distance traveled by the pebble is approximately 41.02 meters.
(c) The vertical distance traveled by the pebble is 55 meters.
(d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
(e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
(f) The negative sign indicates that the pebble is moving downward.
a) To find the time taken for the pebble to reach the ground, we can use the equation for vertical motion:
h = (1/2)gt^2, where h is the vertical distance and g is the acceleration due to gravity.
Rearranging the equation, we have:
t = √((2h) / g), where t is the time taken.
Substituting the given values, we get:
t = √((2 * 55) / 9.8) ≈ 2.01 seconds.
b) The horizontal speed of the pebble remains constant throughout its motion. Therefore, the horizontal distance traveled by the pebble can be found by multiplying the horizontal speed by the time taken:
d = v0 * t, where d is the horizontal distance and v0 is the initial horizontal speed.
Substituting the given values, we have:
d = 20.5 * 2.01 ≈ 41.02 meters.
c) The vertical distance traveled by the pebble is given as 55 meters.
d) The initial vertical velocity of the pebble is 0 m/s because it is thrown horizontally.
e) The vertical acceleration of the pebble is due to gravity and is approximately -9.8 m/s^2.
f) The final vertical velocity of the pebble when it reaches the ground can be found using the equation:
v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the initial vertical velocity is 0 m/s and the acceleration due to gravity is -9.8 m/s^2, we have:
v = 0 + (-9.8) * 2.01 ≈ -19.8 m/s.
The negative sign indicates that the pebble is moving downward.
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how can i write answers to get points
Use your knowledge of conjunction, disjunction, negation and truth tables to determine whether the argument is valid or invalid or unknown.
~( R · S )
~ R · P / ~ S
Using truth tables, we determined the validity of the argument ~(R · S) ~ R · P / ~ S. By examining the truth values of the expression ~ S · P, we found that it can be both true and false in different scenarios. Therefore, the argument is invalid.
To determine the validity of the argument ~(R · S) ~ R · P / ~ S, we can use truth tables. First, let's assign truth values to the variables:For more questions on truth tables
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Two blocks, 1 and 2, are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105
Find the magnitude 2 of the acceleration of block 2.
The magnitude of acceleration of block 2 is 4.67 m/s².
The diagram representing the blocks is shown below:It can be observed that the two blocks are connected by a massless string that passes over a massless pulley. 1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. 2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.Now let's derive the equation for acceleration, a2.
A key concept that must be understood to solve the problem is the difference in tension on either side of the string. Since the pulley is massless and frictionless, the tension must be the same on both sides. We can derive this concept using the following equations:Tension on block 1 side:T1 = m1(g)sin(1) - m1(g)cos(1) * f1Tension on block 2 side:T2 = m2(g)sin(2) + m2(g)cos(2) * f2Where g is acceleration due to gravity, which is equal to 9.8 m/s².Then:T1 = T2T1 + m1(g)cos(1) * f1 = m2(g)sin(2) + m2(g)cos(2) * f2Substitute the values into the above equation:2.25(9.8)cos(42.5) * 0.205 + 2.25(9.8)sin(42.5) = 5.55(9.8)sin(33.5) + 5.55(9.8)cos(33.5) * 0.105T2 = 25.836 N (correct to 3 significant figures)Now we can find the acceleration of block 2.
The acceleration of block 1 can be determined using the following equation:a1 = g(sin(1) - f1 cos(1))a1 = 9.8(sin(42.5) - 0.205cos(42.5))a1 = 5.748 m/s² (correct to 3 significant figures)Using the equation for acceleration of block 2:a2 = (T1 - T2) / m2a2 = (25.836 - 0) / 5.55a2 = 4.667 m/s² (correct to 3 significant figures).
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A ball is thrown straight upward and returns to the thrower's hand after 2.55 s in the air. A second ball thrown at an angle of 31.0° with the horizontal reaches the same maximum height as the first ball. Answer parts a-b.
(a) The speed at which the first ball was thrown is approximately 12.6735 m/s.
(b) The speed at which the second ball was thrown is approximately 22.84 m/s.
To solve this problem, we'll use the kinematic equations of motion and the fact that the maximum height reached by both balls is the same.
(a) Let's start with the first ball. We know that the time of flight (t) is 2.55 s, and since the ball is thrown straight upward, the time to reach the maximum height is half of the total time of flight (t/2).
Using the equation for vertical displacement:
Δy = Vyi * t - (1/2) * g * [tex]t^2[/tex]
At the maximum height, the vertical displacement (Δy) is zero. So we can rewrite the equation as:
0 = Vyi * (t/2) - (1/2) * g * [tex](t/2)^2[/tex]
Rearranging the equation, we can solve for the initial vertical velocity (Vyi):
Vyi = (1/2) * g * (t/2)
Plugging in the known values:
Vyi = (1/2) * 9.8 [tex]m/s^2[/tex] * (2.55 s / 2)
Simplifying the equation, we find:
Vyi = 12.6735 m/s
Since the ball was thrown straight upward, the initial vertical velocity is equal to the final vertical velocity when the ball reaches the thrower's hand. Therefore, the speed at which the first ball was thrown is approximately 12.6735 m/s.
(b) Now let's move on to the second ball thrown at an angle of 31.0° with the horizontal. We know that the maximum height reached by this ball is the same as the first ball.
The vertical component of the initial velocity (Vyi) for the second ball can be calculated using the equation:
Vyi = V * sin(θ)
To find the total initial velocity (V), we need to know the horizontal component of the initial velocity (Vxi), which can be calculated as:
Vxi = V * cos(θ)
Since the maximum height reached by the second ball is the same as the first ball, the time taken to reach the maximum height will also be the same. Therefore, we can use the same time of flight (t) value.
Using the equation for the maximum height (H):
H = [tex](Vyi)^2[/tex] / (2 * g)
We can rewrite this equation in terms of V:
V = √(2 * g * H) / sin(θ)
Plugging in the known values:
V = √(2 * 9.8 [tex]m/s^2[/tex] * 12.6735 m) / sin(31.0°)
Simplifying the equation, we find:
V ≈ 22.84 m/s
Therefore, the speed at which the second ball was thrown is approximately 22.84 m/s.
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1. Describe the path light takes as it travels through air and into glass.
When light travels from air into glass, it refracts towards the normal at the air-glass interface, propagates through the glass, and upon exiting, refracts away from the normal back into the air.
When light travels from air into glass, it undergoes several changes in its path due to the difference in optical properties between the two mediums. The path light takes can be described as follows:
1. Incident Ray: The journey begins with an incident ray, which is the incoming light ray traveling through the air towards the glass surface.
2. Refraction: As the incident ray reaches the interface between air and glass, it encounters a change in the refractive index. Refractive index is a measure of how much a medium can bend light. Glass has a higher refractive index than air, so the incident ray bends towards the normal, which is an imaginary line perpendicular to the surface of the glass.
3. Transmission: After refraction, the incident ray enters the glass and continues its path through the medium. Inside the glass, the light travels in a straight line until it encounters another interface or is affected by other optical phenomena.
4. Internal Reflection: If the incident ray encounters a glass-air interface at an angle greater than the critical angle, total internal reflection can occur. In this case, the light reflects back into the glass instead of transmitting out, effectively bouncing off the interface.
5. Refraction (again): If the incident ray does not undergo total internal reflection, it continues to propagate through the glass. At another glass-air interface, the light exits the glass and enters the air again. This transition involves refraction once more, but this time the light bends away from the normal, returning to its original path in air.
6. Transmitted Ray: Finally, the light ray continues to travel through the air, maintaining its original direction and path as it moves away from the glass surface.
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What are the six digit grid coordinates for the windtee?
The six digit grid coordinates for the windtee should be 3.
How do we we calculate?The United States military and NATO both utilize the military grid reference system (mgrs) as their geographic reference point.
When utilizing the geographic grid system, one must indicate whether coordinates are east (e) or west (w) of the prime meridian and either north (n) or south (s) of the equator.
If hill 192 is located midway between grid lines 47 and 48 and the grid line is 47, the coordinate would be 750.
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The six digit grid coordinates for the windtee is determined as 100049.
What is a coordinate point?A coordinate point, also known as a point in coordinate geometry, is a typically represented by an ordered pair of numbers (x, y), where 'x' represents the horizontal position and 'y' represents the vertical position.
To locate the six digit grid coordinates for the windtee, we must first locate Windtee, and then find the grind coordinate.
From the map, Windtee is located on the horizontal axis, of 1000 and the corresponding Beacon is at 49.
So the six digit grid coordinates = 100049.
Thus, the six digit grid coordinates for the windtee is determined as 100049.
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A playground is on the flat roof of a city school, hb = 5.90 m above the street below (see figure). The vertical wall of the building is h = 7.40 m high, to form a 1.5-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launching it at an angle of = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. Answer parts a-c.
(a) The speed at which the ball was launched is approximately 10.91 m/s.
(b) The ball clears the wall by approximately 1.50 m vertically.
(c) The horizontal distance from the wall to the point on the roof where the ball lands is approximately 24.02 m.
To solve this problem, we'll analyze the motion of the ball in two dimensions: horizontal and vertical.
(a) First, let's calculate the initial speed at which the ball was launched. We can use the time of flight and the horizontal distance traveled to find the initial horizontal velocity (Vx) of the ball.
The horizontal distance traveled by the ball (d) is given as 24.0 m, and the time of flight (t) is given as 2.20 s.
Using the equation for horizontal distance:
d = Vx * t
Rearranging the equation, we can solve for Vx:
Vx = d / t
Plugging in the known values:
Vx = 24.0 m / 2.20 s
Simplifying the equation, we find:
Vx ≈ 10.91 m/s
The initial horizontal velocity of the ball is approximately 10.91 m/s.
(b) Next, let's find the vertical distance by which the ball clears the wall. We can use the time of flight and the vertical motion of the ball to calculate this.
The vertical distance traveled by the ball is the difference between the height of the wall (h) and the height of the playground (hb).
Δy = h - hb
Plugging in the known values:
Δy = 7.40 m - 5.90 m
Simplifying the equation, we find:
Δy = 1.50 m
The ball clears the wall by approximately 1.50 m vertically.
(c) Finally, let's determine the horizontal distance from the wall to the point on the roof where the ball lands.
Since the time of flight and the horizontal distance traveled by the ball are given, we can calculate the horizontal distance (x) using the equation:
x = Vx * t
Plugging in the known values:
x = 10.91 m/s * 2.20 s
Simplifying the equation, we find:
x ≈ 24.02 m
The ball lands approximately 24.02 m horizontally from the wall on the roof.
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Select the correct answer.
Before a collision, the x-momentum of an object is 8.0 × 103 kilogram meters/second, and its y-momentum is 1.2 × 104 kilogram meters/second. What is the magnitude of its total momentum after the collision?
A.
1.4 × 104 kilogram meters/second
B.
2.0 × 104 kilogram meters/second
C.
3.2 × 104 kilogram meters/second
D.
5.7 × 104 kilogram meters/second
Particles q₁ = -66.3 μC, q2 = +108 μC, and
q3 = -43.2 μC are in a line. Particles q₁ and q2 are
separated by 0.550 m and particles q2 and q3 are
separated by 0.550 m. What is the net force on
particle q₂?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
About 1.75% of water on Earth is in Greenland and Antarctica's icecaps, and about 97.5% is in the oceans. Assume the icecaps have an average temperature of -28°C, and the oceans have an average temperature of 4.8°C. If all the icecaps slid into the ocean and melted, how much would the average temperature of the ocean decrease?
If all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
To calculate the decrease in the average temperature of the ocean when all the icecaps melt, we need to consider the heat exchange between the icecaps and the ocean.
Let's start by calculating the heat released by the icecaps when they melt. We can use the specific heat capacity formula:
Heat released = Mass of icecaps × Specific heat capacity of ice × Temperature change
Since the icecaps constitute 1.75% of the Earth's water, the mass of icecaps is 0.0175 times the total mass of water on Earth.
Assuming the icecaps have an average temperature of -28°C and melt into liquid water at 0°C, the temperature change is 0°C - (-28°C) = 28°C.
Next, we need to calculate the heat absorbed by the ocean when the icecaps melt. Using the same formula:
Heat absorbed = Mass of ocean water × Specific heat capacity of water × Temperature change
Given that the oceans constitute 97.5% of the Earth's water, the mass of the ocean water is 0.975 times the total mass of water on Earth.
Assuming the oceans have an average temperature of 4.8°C, the temperature change is 4.8°C - 0°C = 4.8°C.
Now we can calculate the change in temperature of the ocean:
Change in temperature = Heat released / (Mass of ocean water × Specific heat capacity of water)
Substituting the values, we get:
Change in temperature = (0.0175 × Total mass of water) × (Specific heat capacity of ice × Temperature change) / (0.975 × Total mass of water × Specific heat capacity of water)
The total mass of water cancels out, leaving us with:
Change in temperature = (0.0175 × Specific heat capacity of ice × Temperature change) / (0.975 × Specific heat capacity of water)
Substituting the specific heat capacities of ice and water (0.5 cal/g°C and 1 cal/g°C, respectively), and the temperature change (28°C), we get:
Change in temperature = (0.0175 × 0.5 cal/g°C × 28°C) / (0.975 × 1 cal/g°C)
Simplifying the equation, we find:
Change in temperature ≈ -0.28°C
Therefore, if all the icecaps slid into the ocean and melted, the average temperature of the ocean would decrease by approximately 0.28°C.
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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.2 m/s at an angle of 52.0° to the horizontal. Answer parts a-b.
(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).
(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.
To solve this problem, we'll analyze the vertical motion of the ball.
(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.
The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.
The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.
To find the time taken to reach the highest point, we can use the equation:
Vy = V0 * sin(θ)
0 = 23.2 m/s * sin(52.0°)
Solving for sin(52.0°), we find:
sin(52.0°) ≈ 0.7880
Dividing both sides by 23.2 m/s, we get:
0.7880 = sin(52.0°)
Taking the inverse sine, we find:
52.0° ≈ arcsin(0.7880)
Using a calculator, we find:
52.0° ≈ 56.43°
Now we can calculate the time (t) it takes to reach the highest point using the equation:
t = (2 * Vy) / g
Since Vy = 0, we have:
t = 0
This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.
To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.
Using the equation for vertical displacement, we have:
Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]
Plugging in the known values:
Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])
Since t = 0, the equation simplifies to:
Δy = 0
Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.
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Masses m and 2m are joined by a light inextensible string which runs without slipping over a uniform circular pulley of mass 2m and radius a. Using the angular position of the pulley as generalized coordinate, write down the Lagrangian function and Lagrange's equation. Find the acceleration of the masses.
The acceleration of the mass 2m is - (8/5) a θ´´.
We have two masses m and 2m connected by a string without slipping over a uniform circular pulley of mass 2m and radius a. We have to find the acceleration of the masses and write down the Lagrangian function and Lagrange's equation. The angular position of the pulley as generalized coordinate is used. Lagrangian function
L = T – VL = Kinetic energy - Potential energy
The kinetic energy is the sum of the kinetic energies of the two masses and the pulley. The potential energy is the sum of the potential energies of the two masses. The potential energy of the pulley can be ignored since it is fixed. Let θ be the angular position of the pulley, x be the distance fallen by the mass m and y be the distance fallen by the mass 2m.Kinetic energy of mass m (K1)K1 = ½ mv²where v = (dx/dt) is the velocity of mass mKinetic energy of mass 2m
(K2)K2 = ½ (2m) (dy/dt)²where (dy/dt) is the velocity of mass 2mKinetic energy of pulley (K3)The pulley is rolling without slipping, so the velocity of the point at the edge of the pulley is given byv = R(θ´)where R = a is the radius of the pulley. Hence, the kinetic energy of the pulley is
K3 = ½ I (θ´)²where I = (2/5) M R² = (2/5) (2m) a² is the moment of inertia of the pulleyPotential energy of mass m (V1)V1 = mgywhere g is the acceleration due to gravityPotential energy of mass 2m (V2)V2 = 2mgxThe Lagrangian function isL = K1 + K2 + K3 - V1 - V2L = ½ m(dx/dt)² + ½ (2m) (dy/dt)² + ½ (2/5) (2m) a² (θ´)² - mgy - 2mgxL = ½ m(dx/dt)² + ½ (2m) (dy/dt)² + ½ (4/5) ma² (θ´)² - mgy - 2mgxLagrange's
equationLet's find the equation of motion of the mass m using Lagrange's equation. The Lagrangian function depends on three variables, so we need three equations of motion.Lagrange's equation isd/dt (δL/δ(dx/dt)) - δL/δx = 0The first term gives usd/dt (δL/δ(dx/dt)) = m(dx/dt) + (4/5) ma² (θ´)(d/dt)(θ´) = m(dx²/dt²) + (4/5) ma² θ´´The second term gives usδL/δx = -d/dx (mgy) = 0The third term gives usδL/δ(θ) = d/dt (δL/δ(θ´))δL/δ(θ) = d/dt [(4/5) ma² (θ´)] = (4/5) ma² θ´´
The equation of motion ism(dx²/dt²) + (4/5) ma² θ´´ = 0We can solve this equation to find the acceleration of the mass m.The acceleration of the mass mThe acceleration of the mass m is given bya = dx²/dt² = - (4/5) a θ´´Therefore, the acceleration of the mass m is - (4/5) a θ´´.The equation of motion of the pulley is obtained in
the same way as above but we need to use the moment of inertia I of the pulley in the Lagrangian. We get(4/5) ma² θ´´ + 2mgRθ´² = 0Dividing by (4/5) ma², we getθ´´ + (5/8gR) θ´² = 0The acceleration of the mass 2m is given by the same expression as above but with m replaced by 2m.
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RHETORICAL ANALYSIS: How does Robinson use language in effective and engaging ways to develop his argument to his younger self-and, in the process, to young readers in the present? In your response, consider such techniques as metaphor, repetition, and sentence structure.
In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.
Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.
Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.
Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.
In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.
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In Bosnia, the ultimate test of a young man's courage used to be to step off a 400-year-old bridge (destroyed in 1993; rebuilt in 2004) into the River Neretva, 23 m below the bridge. Find a - How long did the drop last?, find b - How fast was the man traveling upon impact with the river?, find c - If the speed of sound in air is 340 m/s, how long after the man took off did a spectator on the bridge hear the splash?.
(a) the drop lasted approximately 2.17 seconds.
(b) the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
(a) To find the time it took for the drop, we can use the equation for free fall motion:
Δy = (1/2) * g * [tex]t^2[/tex]
Given:
Initial height, h = 23 m
Acceleration due to gravity, g = 9.8 [tex]m/s^2[/tex]
Rearranging the equation, we get:
t^2 = (2 * h) / g
Substituting the values:
t^2 = (2 * 23 m) / 9.8 [tex]m/s^2[/tex]
t^2 ≈ 4.6949 s^2
Taking the square root of both sides, we find:
t ≈ √(4.6949 [tex]s^2[/tex])
t ≈ 2.17 s
Therefore, the drop lasted approximately 2.17 seconds.
(b) To find the speed of the man upon impact with the river, we can use the equation for final velocity in free fall:
v = g * t
Substituting the values:
v = 9.8 [tex]m/s^2[/tex] * 2.17 s
v ≈ 21.26 m/s
Therefore, the man was traveling at approximately 21.26 m/s upon impact with the river.
(c) To find the time it takes for the sound of the splash to reach a spectator on the bridge, we can use the speed of sound:
Given:
Speed of sound, v_sound = 340 m/s
The time it takes for the sound to travel from the river to the spectator is the same as the time it took for the man to fall. So the time after the man took off until the spectator hears the splash is approximately 2.17 seconds.
Therefore, approximately 2.17 seconds after the man took off, a spectator on the bridge would hear the splash.
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A home run is hit in such a way that the baseball just clears a wall 18 m high, located 116 m from home plate. The ball is hit at an angle of 37° to the horizontal, and air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) Answer parts a-c.
(a) The initial speed of the ball is 60.4 m/s
(b) The time of motion of the ball is 1.92 seconds.
(c) The velocity component of the ball is , x - component = 48.24 m/s
and y - component = 36.35 m/s.
(d) The speed of the ball as it reaches the wall is 64.8 m/s.
What is the initial speed of the ball?(a) The initial speed of the ball is calculated as;
t = √ (2h/g)
where;
h is height g is gravityt = √ (2 x 18 / 9.8 )
t = 1.92 s
v = d / t
v = 116 m / 1.92 s
v = 60.4 m/s
(b) The time of motion of the ball is 1.92 seconds.
(c) The velocity component of the ball is calculated as;
x - component = 60.4 m/s x cos (37) = 48.24 m/s
y - component = 60.4 m/s x sin (37) = 36.35 m/s
(d) The speed of the ball as it reaches the wall is calculated as;
v = vi + gt
where;
vi is the initial speed of the ballthe time to travel 1 m high = √ (2 x 1 / 9.8 )
t = 0.45 s
v = 60.4 m/s + 0.45(9.8)
v = 64.8 m/s
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WHOEVER ANSWERS IS THE BRAINLIEST!!! PLS HELP!!
Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.
What do we see in the image?In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.
Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.
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A certain car is capable of accelerating at a rate of 0.65 m/s2. How long does it take for this car to go from a speed of 25 mi/h to a speed of 32 mi/h?
It takes about 4.85 seconds for the car to accelerate from a speed of 25 mi/h to a speed of 32 mi/h.
The given information includes the acceleration rate of a certain car which is 0.65 m/s², and the initial speed of the car which is 25 miles per hour. The question is asking about the time taken by the car to accelerate from the initial speed of 25 miles per hour to a speed of 32 miles per hour. This is a simple problem in kinematics that can be solved by using the formula of acceleration. Here’s how:
First, convert the initial and final speeds of the car into meters per second.
Given that:
Initial speed of the car, u = 25 miles/hour
Final speed of the car, v = 32 miles/hour
To convert miles/hour to meters/second, multiply it by 0.447:
u = 25 miles/hour × 0.447 = 11.175 meters/second
v = 32 miles/hour × 0.447 = 14.324 meters/second
Now, let’s use the formula of acceleration:
v = u + at
Where,
v = final speed = 14.324 m/s
u = initial speed = 11.175 m/s
a = acceleration = 0.65 m/s²
t = time taken
Substitute the given values in the formula:
14.324 = 11.175 + (0.65)t
Solve for t:
t = (14.324 - 11.175) / 0.65
t = 4.85 seconds
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Two particles move about each other in circular orbits under the influence of gravitational forces, with a period 7, Their motion is suddenly stopped at a given instant of time and they are then released and allowed to fall into each other......
Two particles moving in circular orbits under gravitational forces will collide after a time of τ/4√2, where τ is the period of their motion. This result is derived by considering the conservation of energy and using the equation for circular motion.
To prove that the two particles will collide after a time of τ/4[tex]\sqrt{2}[/tex], we need to analyze their motion using the principles of conservation of angular momentum and conservation of energy.
Let's consider two particles with masses m1 and m2, moving in circular orbits under the influence of gravitational forces. The period of their motion is given as τ.
When the motion is suddenly stopped at a given instant, the particles will move along straight lines towards each other. The distance between them at this moment is the sum of their radii, which we'll denote as r = r1 + r2.
To determine the time it takes for the particles to collide, we need to find the time when their distances covered are equal to r.
Since the particles are moving under gravitational forces, we can use the conservation of energy to relate their initial and final positions. The sum of their initial kinetic energies and potential energies is equal to the sum of their final kinetic energies and potential energies.
Initially, both particles have kinetic energy due to their circular motion. When the motion is stopped, their kinetic energies become zero. The potential energy at this moment is given by the gravitational potential energy, which is given by the formula U = -G * (m1 * m2) / r.
Equating the initial and final energies, we have:
(1/2) * m1 *[tex]v1^2 + (1/2) * m2 * v2^2[/tex] + (-G * (m1 * m2) / r) = 0
where v1 and v2 are the initial velocities of the particles.
Since the particles start from rest, their initial velocities are zero.
Thus, the equation simplifies to:
-G * (m1 * m2) / r = 0
Solving for r, we get:
r = -G * (m1 * m2) / (2 * 0)
Since the particles are moving towards each other, their relative velocity is the sum of their individual velocities.
[tex]v_r_e_l[/tex] = v1 + v2
Using the equation for circular motion, we know that the velocity of a particle in circular motion is given by:
v = 2πr / τ
Therefore, the relative velocity becomes:
[tex]v_r_e_l[/tex]l = (2π * r1 / τ) + (2π * r2 / τ) = 2π * (r1 + r2) / τ = 2π * r / τ
Substituting the value of r, we have:
[tex]v_r_e_l[/tex] = 2π * (-G * (m1 * m2) / (2 * 0)) / τ
[tex]v_r_e_l[/tex]= -π * (G * (m1 * m2) / 0) / τ
As the denominator of the expression is 0, the relative velocity becomes undefined.
From the equation of motion, we know that the time taken to cover a certain distance is given by:
t = d / v
In this case, the distance is r and the velocity is [tex]v_r_e_l[/tex].
Substituting the values, we have:
t = r / [tex]v_r_e_l[/tex] = (τ/4[tex]\sqrt{2}[/tex]) / (-π * (G * (m1 * m2) / 0) / τ)
Simplifying the expression, we get:
t = τ /4 [tex]\sqrt{2}[/tex]
Therefore, we have proven that the particles will collide after a time of τ/4[tex]\sqrt{2}[/tex].
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3. A cylindrical steel drum is tipped over and rolled along the floor of a ware house. If the drum has radius of 0.40m and makes on complete turns in every 8.0 s, how long does it take to roll the drum 36m?
It takes approximately 9.05 seconds to roll the drum a distance of 36 meters.
What is circumference of a circle?We can use the formula for the circumference of a circle:
Circumference = 2 * π * radius
Given:
Radius (r) = 0.40 m
Circumference (C) = 2 * π * 0.40 m
We must figure out how many full rotations the drum makes to go 36 meters in order to calculate how long it takes to roll the drum. Since we are aware of the circumference, we can determine the number of full turns as follows:
Number of turns = Distance / Circumference
Given:
Distance = 36 m
Number of turns = 36 m / (2 * π * 0.40 m)
Now that we know how many turns there are, we can calculate the time by multiplying that number by the length of a turn, which is given as 8.0 seconds:
Time = Number of turns * Time per turn
Time = (36 m / (2 * π * 0.40 m)) * 8.0 s
By substituting the values into the equation, we can calculate the time:
Time = (36 / (2 * 3.14159 * 0.40)) * 8.0 s
Time ≈ 9.05 s
So, it takes approximately 9.05 seconds to roll the drum a distance of 36 meters.
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Look at the diagram describing the
energy changes that occurred in a
collision between two identical cars.
Describe, in words, a possible
scenario for the collision. Identify in
your scenario a description of the
relative sizes of the cars, their speed
of motion and any energy losses or
gains that might have occurred for
each vehicle involved in the collision.
In addition to the energy losses described above, there may also be other losses, such as the loss of life.
How to explain the informationIn this scenario, the cars are identical in size and speed. However, in a real-world collision, the cars may not be identical. For example, one car may be heavier than the other. In this case, the heavier car would have more momentum and would transfer more energy to the lighter car. This could result in more damage to the lighter car.
The speed of the cars also plays a role in the severity of the collision. The faster the cars are traveling, the more kinetic energy they have. This means that the collision will be more forceful and will result in more damage.
In addition to the energy losses described above, there may also be other losses, such as the loss of life. In a serious collision, the occupants of the cars may be killed or seriously injured. This is a tragic loss of life that could have been avoided if the drivers had been more careful.
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Look at the velocity versus time graph below. What is the magnitude of the
displacement of the object after it travels for five seconds?
Velocity (m/s)
Time (s)
A. 30 m
OB. 20 m
OC. 25 m
OD. 35 m
The magnitude of displacement of the object after five seconds, calculated from the velocity-time graph, is 32.5 m. The correct answer is option E.
Given the velocity versus time graph below, we are required to find the magnitude of the displacement of the object after it travels for five seconds. Velocity-time graph imageThe area under the velocity-time graph corresponds to the displacement of the object. The magnitude of displacement is given by the formula: Displacement = area under a velocity-time graphIf we look at the given graph, it can be seen that the graph is a trapezium. Therefore, we need to split it into two parts: a rectangle and a triangle. The displacement is given by the sum of the area of both parts. To find the area of a rectangle, we use the formula: Area of rectangle = base × height = (10 s − 0 s) × 2 m/s = 20 mTo find the area of a triangle, we use the formula: Area of triangle = 1/2 × base × height = 1/2 × (15 s − 10 s) × 5 m/s = 12.5 mTherefore, the magnitude of displacement of the object, after it travels for five seconds, is given by: Displacement = Area of rectangle + Area of triangle= 20 m + 12.5 m= 32.5 mHence, the correct answer is option E. 32.5 m.For more questions on the magnitude of displacement
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what is a shargaff rule
According to Chargaff's rule, the amounts of adenine (A), thymine (T), and guanine (G) in the DNA molecule are equal to each other. The amounts of cytosine (C) and guanine (G) are also equal.
Who is Chargaff ?Erwin Chargaff was a biochemist, author, Bucovinian Jew who immigrated to America during the Nazi era, and professor of biochemistry at Columbia University's medical school.
Chargaff found patterns among the four bases, or chemical building blocks, of DNA, which are directly related to DNA's function as the genetic material of living things.
He was born in Austria-Hungary. Heraclitean Fire: Sketches from a Life Before Nature, an autobiography he penned, received positive reviews.
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Which statement best describes the refraction of light as it moves from air to glass?
A. Light bends due to the difference in the speed of light in air and glass.
B. Although the light bends, its speed remains the same as before.
C. Although the light changes speed, it continues in the same direction as before.
D. Light undergoes diffraction due to the difference in the speed of light in air and glass.