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Reika [66]
3 years ago
11

When you fire a shot gun, you feel that you are pushed back by the gun. Can you explain why this happens? is the recoil velocity

of the gun equal to the velocity of the bullet? If not is there any other quantity that is equal for the bullet and the gun?
Physics
2 answers:
yarga [219]3 years ago
8 0

This is a prime example of newtons 3rd law, for every action there is an equal and opposite reaction. This means the shotgun hits your shoulder with the same energy as the bullet hits the target. The stock has enough mass though that the energy is distributed evenly to the surface area of your shoulder touching the gun.

Vilka [71]3 years ago
3 0
The recoils is caused by the gun trying to balance out the momentum of the bullet and the gas being discharged. The gas helps balance out the momentum of the bullet a little, but the recoil is still needed for the complete balance. This need for balance is proved by Newtons third law. 

The formula for finding the velocity of recoil is as follows:
(Weight of Bullet * Velocity of Bullet + Weight of Gases * Velocity of Gases) / (7000 * Weight of Gun)

<span>Hope this helped!</span>
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which measurment describes the amount of heat needed to raise the temperature of one gram of a material by one degree celsius?
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<u>Answer:</u>

Specific Heat

<u>Explanation:</u>

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7 0
4 years ago
A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.
Daniel [21]

Answer:

a) a_{center} = 7.38 ~m/s^2

b) a_{end} = 14.77 ~m/s^2

c) v_{center} = 2.43~m/s

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

\tau =I\alpha

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2

So, the Newton's Second Law yields

\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

a = \alpha r = 7.38~m/s^2

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

a = \alpha L = 14.77~m/s^2

c) The conservation of energy can be used to find the velocity when the rod is vertical.

K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s

The linear velocity is v = ωr, so

v = 2.43 m/s.

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