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snow_tiger [21]
3 years ago
6

If a projectile hits a stationary target, and the projectile continues to travel in the same direction, the mass of the projecti

le is less than the mass of the target. the mass of the projectile is equal to the mass of the target. the mass of the projectile is greater than the mass of the target. nothing can be said about the masses of the projectile and target without further information. this is an unphysical situation and will not actually happen.
Physics
1 answer:
nika2105 [10]3 years ago
8 0

The correct arrangement of the question is;

If a projectile hits a stationary target, and the projectile continues to travel in the same direction,

A) the mass of the projectile is less than the mass of the target.

B) the mass of the projectile is equal to the mass of the target.

C) the mass of the projectile is greater than the mass of the target.

D) nothing can be said about the masses of the projectile and target without further information.

E) this is an unphysical situation and will not actually happen.

Answer:

Option C: The mass of the projectile is greater than the mass of the target.

Explanation:

We want to find what will happen when a projectile continues in motion after it hits a target.

Now, for the projectile to keep moving in that direction after it hits the target, it means it had a force bigger than the force of the target to overpower it and force it to move with it.

Now, from law of inertia, Force = ma.

But in this case acceleration is 0 because the speed of the projectile is constant.

Thus, the force depends on the mass. So for a higher force, the mass of the projectile has to be more than that of the stationary object.

Thus, option C is correct

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2 years ago
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
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Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
3 years ago
a glass bottle of soda is sealed with a screw cap the absolute pressure of the gas inside the bottle is 1.80*10^5 pa. assuming t
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The force exerted by a pressure of any gas over a surface its given by the formula P=F/S (where P is pressure, F force and S surface).

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Solve for P=1.80*10^5 Pa and S=4.10*10^-4 m^2 ([Pa] =[N/m^s])

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73.8 N =F

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The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n
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T - F = ma
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5 0
3 years ago
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