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3 years ago

What is the mass of a bullet moving at 970m/s if the bullet’s KE is 3.9x 10^3J

Physics

2 answers:

xxMikexx [17]3 years ago

7 0

I think the Answer is 0.08kg

motikmotik3 years ago

4 0

Answer:

0.08kg

Explanation:

K.E = 1/2 mv^2

v = 970m/s

K.E = 3.9x 10^3J= 3900J

K.E = 1/2 mv^2

3900 = 1/2 m x 970x 970

3900 = 1/2 ×940900m

3900 = 470450m

m = 3900/470450 = 0.00828993516 = 0.008kg

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2 years ago

Two forces act on an object. One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N direct

xeze [42]

Answer:

8 N North.

Explanation:

Given that,

One force has a magnitude of 10 N directed north, and the other force has a magnitude of 2 N directed south.

We need to find the magnitude of net force acting on the object.

Let North is positive and South is negative.

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2 years ago

A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9

Andreyy89

Answer:

$d = 0.0306 m$

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

$U + K = constant$

$\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2$

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

$\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0$