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3 years ago

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A 1500 kg car is parked at the top of a hill 5.2 m high. At the bottom of the hill, what is the kinetic energy, in Joules, of th

e car?

1 answer:

irga5000 [103]3 years ago

4 0

Answer:

10.1

Explanation:

I think it's right?

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A hammer taps on the end of a 5.0-mm-long metal bar at room temperature. A microphone at the other end of the bar picks up two p

katrin2010 [14]

Answer:

810.37 m/s

Explanation:

s = Displacement = 5 m

$v_a$ = Speed of sound in air = 343 m/s

Time taken to travel the distance

$t'=\dfrac{s}{v_a}\\\Rightarrow t'=\dfrac{5}{343}\\\Rightarrow t'=0.01457\ s$

Time for the sound in metal

$t=0.01457-8.4\times 10^{-3}=0.00617\ s$

Speed of sound is given by

$v=\dfrac{s}{t}\\\Rightarrow v=\dfrac{5}{0.00617}\\\Rightarrow v=810.37277\ m/s$

The speed of sound in the metal is 810.37 m/s

6 0

3 years ago

The largest watermelon ever grown had a mass of 118 kg. Suppose this watermelon were exhibited on a platform 5.00 m above the gr

WINSTONCH [101]

Answer: height = 3.98m

Explanation: by placing the watermelon at a height above the ground, it has a potential energy of the formulae

p = mgh

p = potential energy = 4.61kJ = 4610J

m = mass of watermelon = 118 kg

g = acceleration due gravity = 9.8 m/s²

4610 = 118 * 9.8 * h

h = 4610/ 118 * 9.8

h = 4610/ 1156.4

h = 3.98m

6 0

3 years ago

When work is done on a spring to stretch it, elastic potential energy is stored in the spring.

aniked [119]

Explanation:

I think its true

hope it helps

6 0

3 years ago

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0

3 years ago

a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?

julsineya [31]

Answer:

Approximately $7.0\; \rm m \cdot s^{-1}$ .

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant $g = 9.81 \; \rm m \cdot s^{-2}$ near the surface of the earth.

For an object that is accelerating constantly,

$v^2 - u^2 = 2\, a \, x$ ,

where

$u$ is the initial velocity of the object,
$v$ is the final velocity of the object.
$a$ is its acceleration, and
$x$ is its displacement.

In this case, $x$ is the same as the change in the ball's height: $x = 2.5\; \rm m$ . By assumption, this ball was dropped with no initial velocity. As a result, $u = 0$ . Since the ball is accelerating due to gravity, $a = 9.81\; \rm m \cdot s^{-2}$ .

$v^2 - u^2 = 2\, g \cdot h$ .

In this case, $v$ would be the velocity of the ball just before it hits the ground. Solve for

$v^2 = 2\, a\, x + u^2$ .

$\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}$ .

3 0

3 years ago