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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.

Explanation:

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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The electromagnetic spectrum comprise a lot of waves length. Usually, different waves length are called as different lights, and a light source can emit in more than a different wave length, as the sun does, for example. The sun emit the visible light, UV light, infrared, etc.

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3 years ago

Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.

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Answer and Explanation:

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3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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3 years ago

5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?

GenaCL600 [577]

Force required is 100 N

Given that;

Rate of acceleration = 5 m/s²

Mass of object = 20kg

Find:

Force required

Computation:

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

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3 years ago

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Bas_tet [7]

Answer:

Explanation:

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3 years ago

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