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lubasha [3.4K]
3 years ago
11

Arunner has a speed of 23 m/s. They see the finish line and speed up to 27 m/s. This happens in 5 seconds. If the runner has a m

ass of 78 kg, with what force did the
runner cross the finish line? Show all work to receive full credit.
Type your answer.
Physics
1 answer:
sp2606 [1]3 years ago
6 0
The answer is 100

Have a great day
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The label has been scratched off a tuning fork and you need to know its frequency. From its size, you suspect that it is somewhe
bonufazy [111]

Answer:

255 Hz

Explanation:

With 5 beats per second with the 250 Hz fork, we know the unknown fork is either 250 - 5 = 245Hz or 250 + 5 = 255 Hz

With 15 beats per second with the 270 Hz fork, we know the unknown fork is either 270 - 15 = 255Hz or 270 + 15 = 285 Hz (most people would have a hard time discerning 15 beats per second... 5 per second is hard enough)

As 255 is the common frequency, it is the one selected.

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What is an example of potential energy to kinetic energy?
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Potential energy is the store she energy from an object this could include rubber bands. Kinetic energy is the energy that deals with motion a good example is a person running

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2 years ago
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You find a micrometer (a tool used to
MakcuM [25]

Explanation:

A micrometer is a measuring device or an instrument which is used to measure very minute measurements very accurately and precisely. It is mathematical tool used to provide accurate measurement for any mechanical components.

Now, if the micrometer that I have found is badly bent, it would provide faulty or wrong measurements both in terms of  precision and accuracy when compared to a high quality meter stick.

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3 years ago
What is power?
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B) Power is the rate at which work is done

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The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it
Lilit [14]

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, m=0.02 kg

Kinetic energy imparted, K=1200 J

Length of rifle barrel, d=1 m

(a)

Let the speed of bullet when it leaves the barrel is v.

Kinetic energy, K=\frac{1}{2} mv^{2}

v=\sqrt{\frac{2K}{m} }

=\sqrt{\frac{2\times1200}{0.02} }

=346.4m/s

Initial speed of bullet, u=0

The average speed in the barrel, v_a_v_g=\frac{u+v}{2}

=\frac{0+346.4}{2} \\=173.2 m/s

Time taken by bullet to cross the barrel, t=\frac{d}{v}

=\frac{1}{173.2}\\ =0.00577 second

Power, P_a_v_g=\frac{W}{t}

=\frac{1200}{0.00577} \\=207.97kW

(b)

In projectile motion,

Maximum height, H_m=\frac{v^2\sin^2\theta}{2g} \\

Range, R=\frac{v^2\sin2\theta}{g}

given that, H_m=R

then, \frac{v^2\sin^2\theta}{2g}=\frac{v^2\sin2\theta}{g}\\\sin^2\theta=2\sin\theta\cos\theta\\\\\tan\theta=4\\\theta=\tan^-^14\\\theta=75.96^0\\R=\frac{v^2\sin2\theta}{g}\\=\frac{346.4^2\times\sin(2\times75.96)}{9.8}\\5768.6 meter

5 0
2 years ago
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