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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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Answer:

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Four charges of equal magnitude q = 2.16 µC are situated as shown in the diagram below. If d = 0.88 m, find the electric potenti
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Answer:

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<u>Electric Potential </u>

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Where k is the Coulomb's constant, q is the charge, and r is the distance from the charge. Let's find the potential of the rightmost charge:

\displaystyle V_1=\frac{9\cdot 10^{9}\times -2.16\cdot 10^{-6}}{0.88}=-22090.91\ V

The potential of the leftmost charge is exactly the same as the above because the charges and distances are identical

V_2=-22090.91\ V

The potential of the topmost charge is almost equal to the above computed, is only different in the sign:

V_3=+22090.91\ V

The bottom charge has double distance and the same charge, thus the potential's magnitude is half the others':

\displaystyle V_4=\frac{9\cdot 10^{9}\times 2.16\cdot 10^{-6}}{1.76}=+11045.45 \ V

The total electric potential in A is

V=-22090.91\ V-22090.91\ V+22090.91\ V+11045.45 \ V

V=-11045.45 \ V

The total potential (magnitude only) is 11045.45 V

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Answer:

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See attachment below.

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