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mixer [17]
3 years ago
13

A car accelerates uniformly from rest to 20 m/sec in 5.6 sec along a level stretch of road. Ignoring friction, determine the ave

rage power required to accelerate the car if (a) the weight of the car is 9,000 N, and (b) the weight of the car is 14,000 N.
Physics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

(a) P=33000W

(b) P=51000W

Explanation:

The average power is defined as the amount of work done during a time interval:

P=\frac{W}{t}(1)

According to work-energy theorem, the work done is equal to the change in kinetic energy. So, we have:

W=\Delta K\\W=K_f-K_0\\W=\frac{mv_f^2}{2}-\frac{mv_0^2}{2}\\(2)

Recall that the weight is given by:

w=mg\\m=\frac{w}{g}(3)

The car accelerates uniformly from rest (v_0=0). Replacing (3) in (2), we have:

W=\frac{wv_f^2}{2g}

(a) Finally, we replace this in (1):

P=\frac{wv_f^2}{2gt}\\P=\frac{9000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=33000W

(b)

P=\frac{14000N(20\frac{m}{s})^2}{2(9.8\frac{m}{s^2})(5.6s)}\\P=51000W

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Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30  m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let \theta be the angle .

Therefore,

30 cos\theta=12.6

cos\theta=\frac{12.6}{30}=0.42

\theta=cos^{-1}(0.42)=65.165^{\circ}

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v^2-v^2_0=2aS

S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}

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A 1.2 KG rubber ball is being thrown in the air if the ball is traveling at 2.0 M/S when it is 3.0 M off the ground what is the
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We see a full moon by reflected sunlight. How much earlier did the light that enters our eye leave the sun? the earth-moon and e
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The time taken by the light reflected from sun to reach on earth will be 8.4 minutes.

To find the answer, we need to know about the distance travelled by light.

<h3>How to find the time taken by the light reflected from sun to reach on earth?</h3>
  • So, in order to solve this problem, we must first know how far the moon is from Earth and how far the Sun is from the moon.
  • These distances are given as 3.8×10^5 km (Earth-Moon) and 1.5×10^8 km (Sun- Earth).
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                      t=\frac{1.51*10^{11}}{3*10^8}=503.33 s\\t=\frac{503.33}{60}=8.4min

Thus, the time it takes for the light from the Sun to reach Earth and be recognized as 8.4 minutes.

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