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adelina 88 [10]
3 years ago
15

A sled is being pulled along a horizontal surface by a horizontal force F of magnitude 600 N. Starting from rest, the sled speed

s up with acceleration 0.08 m/s^2 for 1 minute. Find the average power P created by force F.
Physics
1 answer:
Simora [160]3 years ago
6 0

Answer:

1440 W

Explanation:

First of all, we can find the total displacement of the sled, which is given by

d=\frac{1}{2}at^2

where

a = 0.08 m/s^2 is the acceleration

t = 1 min = 60 s is the time

Substituting,

d=\frac{1}{2}(0.08 m/s^2)(60 s)^2=144 m

Now we can find the wotk done on the sled, equal to the product between force and displacement:

W=Fd=(600 N)(144 m)=86,400 J

And finally we can fidn the average power, which is the ratio between the work done and the time taken:

P=\frac{W}{t}=\frac{86,400 J}{60 s}=1440 W

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Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

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h= \frac{1}{2}g{t}^{2}

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

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By substitution we obtain,

16= \frac{1}{2}\times 9.8{t}^{2}

\implies32=9.8{t}^{2}

Diving through by 9.8

\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}

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square root both sides, we obtain

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How does the Coriolis effect influence the direction of the Trade Winds in the Northern Hemisphere? Does it have the same effect
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Answer:

Part A

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Part B

In the Northern Hemispheres, the winds are known as northeasterly trade winds and in the Southern Hemisphere, they are known as the southeasterly trade wind. Therefore, Coriolis effect has the same effect on the direction of the Trade Winds in the Southern Hemisphere as it does in the Northern Hemisphere

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Consider a 4-m-long, 4-m-wide, and 1.5-m-high above-the-ground swimming pool that is filled with water to the rim. (a) Determine
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Answer:

44100 N

Explanation:

Each wall will have dimension of 4 m x 1.5 m

Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m

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