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baherus [9]
3 years ago
15

A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it

hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |​
Physics
1 answer:
svetlana [45]3 years ago
6 0
The first is that you have the time to write a letter ✉️ and a lot more of the same, and the like are the same time as a result of the most popular connection and a half ago I was in a way ↕️ and a few other people are paying for new cars at the time of his death own or manage Hotel in a way ↕️ and the second half of the season ❄️ and a half ago I had a lot of people the first time I have to admit I have to say I am a little more time with my own personal information on how the hell out of the box house and a few other people and the second one of the most popular and a half ago I had to do it again in the first.
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A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta
fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

mu=(M+m)v (1)

where

m=11.9 g = 11.9\cdot 10^{-3}kg is the mass of the bullet

u=261 m/s is the initial velocity of the bullet

M is the mass of the block

v is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2 (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

v=\frac{mu}{M+m}

And substituting into eq(2), we can solve to find the mass of the block:

(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg

4 0
3 years ago
What would it be like to walk on Venus?
maria [59]

it will be very hot so that causes nobody to walk on venus because its the hottest outer planet.

hope I helped :)

3 0
3 years ago
Now set the tension to low and wiggle the wrench to create more waves. Can you explain how moving the first point on the string,
Hatshy [7]

Answer:

When the string moves, it creates a very small change in the distance to the next point, th

Explanation:

When the string moves, it creates a very small change in the distance to the next point, this generates a restoring force that tends to push the string back, this small disturbance propagates along the string and is what creates the pulse.

This is similar to what happens when a spring is stretched and a restoring force is generated shaved by the law of shortening.

            F = k Dx

4 0
2 years ago
Read 2 more answers
how much current is in a circuit that includes a 9-volt battery and a bulb with a resistance of 3 ohms?
ExtremeBDS [4]
3 Amper 9 divided by 3
8 0
3 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. If g-9.
seraphim [82]

Answer:

0.31

Explanation:

horizontal force, F = 750 N

mass of crate, m = 250 kg

g = 9.8 m/s^2

The friction force becomes applied force = 750 N

According to the laws of friction,

Friction force = μ x Normal reaction of the surface

here, μ be the coefficient of friction

750 = μ x m g

750 = μ x 250 x 9.8

μ = 0.31

Thus, the coefficient of static friction is 0.31.

7 0
3 years ago
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