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Leona [35]
2 years ago
11

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

at this instant? (b) How much net work is required to double his speed?
Physics
1 answer:
vladimir2022 [97]2 years ago
8 0

Answer:

a)KE=878.8 J

b)W=2636.4 J      

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

KE=\dfrac{1}{2}mu^2

m=mass

u=velocity

Now by putting the values in the above equation we get

KE=\dfrac{1}{2}\times 65\times 5.2^2\ J

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Now by putting the values in the above equation we get

W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

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Answer:

15\%.

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The question states that 34\; \rm J out of that 40\; \rm J of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the (40\; \rm J - 34\; \rm J) = 6\; \rm J of energy supplied to this lamp would be turned into useful energy output.

Thus, every second, this lamp would receive 40\; \rm J of energy input and would outputs 6\; \rm J of useful work. The efficiency of this lamp would be:

\begin{aligned}& \text{Efficiency} \\ =\; & \frac{\text{Useful energy out}}{\text{Total energy in}} \times 100\% \\ =\; & \frac{6\; \rm J}{40\; \rm J} \times 100\%\\ =\; &15\% \end{aligned}.

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If you want to make a strong battery, should you pair two metals with high electron affinities, low electron affinities, or a mi
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A mix

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The earth's radius is about 4000 miles. Kampala, the capital of Uganda, and Singapore are both nearly on the equator. The distan
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Answer:

a) the required angle in both radian and degree is  1.25 rad and 71.6°

b) the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

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we know that The expression for the angle subtended by an arc of circle at the center of the circle is,

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given that Length is 5000 miles and radius is 4000 miles

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b)

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given that θ is 1.25 rads and time t is 9 hours or ( 9 × 3600 sec ) = 32400 sec

we substitute

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Therefore, the plane's angular speed relative to the earth is 3.86 × 10⁻⁵ rad/sec

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