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Jet001 [13]
3 years ago
9

PLEASE HELP - In running a ballistics test at the police department, Officer Rios fires a 6.1g bullet at 350 m/s into a containe

r that stops it in 0.32 min. What average force stops the bullet?​
Physics
2 answers:
Arlecino [84]3 years ago
8 0

Answer:

–0.11 N.

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 6.1 g

Initial velocity (u) = 350 m/s

Time (t) = 0.32 mins

Final velocity (v) = 0 m/s

Force (F) =?

Next, we shall convert 0.32 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

0.32 mins = 0.32 × 60 = 19.2 s

Next, we shall determine the acceleration of the the bullet. This can be obtained as follow:

Initial velocity (u) = 350 m/s

Time (t) = 19.2 s

Final velocity (v) = 0 m/s

v = u + at

0 = 350 + (a × 19.2)

0 = 350 + 19.2a

Collect like terms

0 – 350 = 19.2a

–350 = 19.2a

Divide both side by 19.2

a = –350 / 19.2

a = –18.23 m/s²

Next, we shall convert 6.1 g to Kg.

1000 g = 1 Kg

Therefore,

6.1 g = 6.1 g × 1 Kg / 1000 g

6.1 g = 0.0061 Kg

Finally, we shall determine the force. This can be obtained as follow:

Mass (m) = 0.0061 Kg

Acceleration (a) = –18.23 m/s²

Force (F) =?

F = ma

F = 0.0061 × –18.23

F = –0.11 N

NOTE: The negative sign indicates that the force is in opposite direction.

vfiekz [6]3 years ago
7 0

Answer:

–0.11 N.

Explanation:

Hope this helps

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Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apar
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The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.

We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.

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<h3>What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?</h3>

The potential energy of the system of two charges separated by a distance is given by -

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In order to solve this question, it is important to remember the work - energy theorem which states -

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Hence, using this work -energy theorem in the question given to us we get -

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In our case -

U_{f}  = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} =   \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}

W = 0.018 joules

Hence, the total work done should be 0.018 joules.

To solve more question on potential energy, visit the link below -

brainly.com/question/15014856

#SPJ4

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