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DedPeter [7]
3 years ago
8

A cheetah of mass 68 kg can accelerate from rest to a speed of 20.1 m/s in 2.0 s. Assuming constant acceleration, what is the ne

t force that causes this acceleration?
272 N
78 N
137 N
680 N
Physics
1 answer:
Art [367]3 years ago
8 0

Answer:

680 N

Explanation:

We start by calculating the cheetah's acceleration knowing that goes from rest (0 speed) to 20.1 m/s in 2 seconds:

a=\frac{v_f-v_1}{t} \\a=\frac{20.1-0}{2} =10.05 \frac{m}{s^2}

Now we use the definition of force (F):

F=m*a\\F=68 kg * 10.05 \frac{m}{s^2} =683.4 N

Therefore the force that accelerates the cheetah is 683.4 Newtons which we can round to 680 N to match the fourth choice shown

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Let the Blaise runs for time "t" to complete the race

so the total distance he moved is given by

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Now Issac runs for time t = "t - 2*60"

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d_1 - d_2 = 10

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A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
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Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

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a. The maximum altitude occurs at the point during which v_y=0:

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At this point, the rocket has an altitude of

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b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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