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4 years ago

Find the kinetic energy of a bullet weighing 0.10 N if its velocity is 2.0 X 103 m/s?

Physics

1 answer:

irina1246 [14]4 years ago

5 0

The kinetic energy of the bullet is 20.4 kJ.

<u>Explanation:</u>

Kinetic energy of a bullet will be equal to the product of mass of the bullet with the square of velocity or speed of the bullet and then the half of that product value.

But here the mass of the bullet is not given, instead the weight of the bullet is given in terms of force. So from this, we have to first find the mass of the bullet.

We know that as per Newton's second law of motion, force is directly proportional to the product of mass and acceleration. So here the acceleration will be equal to the acceleration due to gravity as it is weight of the object.

So F = mg

0.10 N = m × 9.8

So ,the mass of the bullet is 0.0102 kg.

Now, we know the mass and velocity of the bullet is given as 2000 m/s.

So,

kinetic energy = $\frac{1}{2}$ × m × v²

kinetic energy = 0.5 × 0.0102 × 2000 × 2000 = 20.4 kJ

Thus, the kinetic energy of the bullet is 20.4 kJ.

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Answer:

a) The planet experiences a centripetal force towards its star

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Explanation:

This problem raise several claims, let's review some aspects of circular motion

F = m a

the centripetal acceleration is

a = v² / r

where v is the speed (modulus of velocity) that is constant and r is the radius

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Let's review the different claims

Part a) the orbital velocity is constant

The correct statement is: The planet experiences a centripetal force towards its star

Part b) what is the centripetal force

The correct statement: The universal attractive force (Gravitational force)

Part c) which statement is true

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3 years ago

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3 years ago

Consider a system made of a ramp that has a run of 8 m and a rise of 6 m. Two equal masses of 4-kg are connected to each other b

djyliett [7]

Answer:

The force of friction is 15.68 N.

Explanation:

Given:

Rise is 6 m and run is 8 m.

Mass of each body (m) = 4 kg

The whole system is in equilibrium.

Now, consider the diagram below representing the scenario given in the question.

The forces acting on the mass that is hanging are tension force in the upward direction and weight of the body acting vertically downward.

As the mass is in equilibrium, the total upward force equals total downward force. So,

$T= mg=4\times 9.8=39.2\ N$ -------- (1)

Now, the forces acting on the other mass along the ramp are:

1. Tension (T) up the ramp

2. $mg sin\theta$ and frictional force (f) down the ramp

Now, as per question:

Rise = 6 m and run = 8 m

So, from figure,

$\tan\theta=\frac{6}{8}=0.75\\\\\theta=\tan^{-1}(0.75)=37\°$

Now, $\sin\theta=\sin(37\°)=0.6$

Now, as the other mass is also at rest, net force acting on it is also 0. So,

$F_{net}=0\\\\T-(mg\sin\theta+f)=0\\\\mg\sin\theta+f=T\\\\f=T-mg\sin\theta$

Now, plug in the given values and solve for 'f'. This gives,

$f=39.2-(4\times 9.8\times 0.6)\\\\f=39.2-23.52=15.68\ N$

Therefore, the force of friction is 15.68 N

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4 years ago

An athlete runs four laps of a 400 m track. What is the athlete's total displacement?

sasho [114]

Answer:

○ 0 $\checkmark$

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