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guapka [62]
3 years ago
11

A 3.1-kilogram gun initially at rest is free to

Physics
2 answers:
34kurt3 years ago
8 0
The conservation of momentum states that the total momentum in a system is constant if there is no external force acting on the system. The total momentum in the gun bullet system is 0 so it must stay that way.

The momentum of the bullet is mv = 0.015*500=7.5
The momentum of the gun must be the same to keep the total momentum of the system equal to zero, so we know that p = 7.5 for the gun.
Substituting this in we get:

7.5=3.1x
x=7.5/3.1
x=2.42

So the speed of the gun is 2.4m/s.
Svetradugi [14.3K]3 years ago
8 0
     Let us consider the gun with the index 1 and bullet with the index 2. Using the Momentum consevation Equation, we have:

\Delta Q= 0 \\ m_{1}*v_{1}-m_{2}*v_{2}=0 \\ m_{1}*v_{1}=m_{2}*v_{2}
 
     Entering the unknown, comes:

m_{1}*v_{1}=m_{2}*v_{2} \\ 3.1*v_{1}=0.015*500 \\ v_{1}= \frac{7.5}{3.1} \\ \boxed {v_{1}=2.4m/s}

Number 2

Obs: approximate results.

If you notice any mistake in my english, please let me know, because i am not native.
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Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
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where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
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Point 2:
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R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
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