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trasher [3.6K]
3 years ago
12

9) Manypeoplearefamiliarwiththefactthatariflerecoilswhenfired.Thisrecoilistheresultof actionreactionforcepairs.Agunpowderexplosi

oncreateshotgaseswhichexpandoutward allowingtherifletopushforwardonthebullet.ConsistentwithNewton'sthirdlawofmotion, thebulletpushesbackwardsupontherifle.Theaccelerationoftherecoilingrifleis... A. Greaterthantheaccelerationofthebullet. B. Smallerthantheaccelerationofthebullet. C. Thesamesizeastheaccelerationofthebullet.
Physics
1 answer:
ycow [4]3 years ago
6 0

Answer:

B. Smaller than the acceleration of the bullet.

Explanation:

According to the law of  conservation of momentum; the total momentum of the gun and bullet after firing is equal to the total momentum of the gun and bullet before firing.

However, the mass of gun is much larger than that of the bullet hence it follows that the velocity and acceleration during the recoil of the rifle is much smaller in comparison to the velocity and acceleration of the of bullet.

So; the acceleration of the rifle is given by the force exerted on the rife by the bullet divided by the mass of the rifle. Given that the mass of the rifle is much greater than the mass of the bullet, the acceleration of the rifle is much less than that of the bullet.

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3 years ago
A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 15 m/sm/s. A passenger accidentally drops his camera
jeka94

To solve this problem we will apply the linear motion kinematic equations. With the information provided we will calculate the time it takes for the object to fall. From that time, considering that the ascent rate is constant, we will take the reference distance and calculate the distance traveled while the object hit the ground, that is,

h = v_0 t -\frac{1}{2} gt^2

-18 = 15*t + \frac{1}{2} 9.8*t^2

t = 3.98s

Then the total distance traveled would be

h = h_0 +v_0t

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h = 77.7m

Therefore the railing will be at a height of 77.7m when it has touched the ground

5 0
3 years ago
Block 1, of mass m1 = 1.30 kg , moves along a frictionless air track with speed v1 = 27.0 m/s . It collides with block 2, of mas
Dennis_Churaev [7]

(A) The total initial momentum of the system is

(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s

(B) Momentum is conserved, so that the total momentum of the system after the collision is

35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>

where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives

<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s

(C) The kinetic energy of the system after the collision is

1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J

and before the collision, it is

1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J

so that the change in kinetic energy is

∆<em>K</em> = 25.4 J - 474 J ≈ -449 J

6 0
3 years ago
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