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beks73 [17]
1 year ago
15

a crate is lifted vertically 1.5 m and then held at rest. the crate has weight 100 n (i.e., it is supported by an upward force o

f 100 n). how much work is being done to hold the crate 1.5 m above the ground in this way?
Physics
1 answer:
san4es73 [151]1 year ago
6 0

The required work to hold the crate above the ground is 150 joule.

We need to know about the work done to solve this problem. The work done by an object depends on the force applied and the distance. The work is proportional to force and displacement. It can be written as

W = F . s

where W is work done, F is the force and s is displacement

From the question above, the given parameters are

F = 100 N

s = 1.5

Thus, the required work to hold the crate above the ground can be calculated

W = F . s

W = 100 . 1.5

W = 150 joule

Find more on work at: brainly.com/question/25573309

#SPJ4

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<em>Ceteris Paribus</em>

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The process of examining a change in one variable in a model while assuming that all the other variables remain constant is called <em><u>Ceteris Paribus</u></em>.

<em>Ceteris Paribus</em> is a Latin phrase that means "all other things being equal" or "all other things held constant" in English. The phrase has found application in disciplines like Economics and Statistics. This phrase as being adopted as a process of examining a change in one variable in a model while assuming that all the other variables remain constant to ascertain the relationship between the variables or make deductions from an experimental study. An example of <em>Ceteris Paribus</em> application is the law of demand and supply in Economics. The law of demand states, <em>Ceteris Paribus</em>, the higher the price, the lower the quantity demanded and <em>vice versa. </em>Conversely, the law of supply states, <em>Ceteris Paribus</em>, the higher price, the higher the quantity supply and <em>vice versa</em>.

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How can stretching affect the range of motion of the neck? Hypothesis
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How do rechargeable batteries work?

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A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th
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Answer:E=75\ N/m

Explanation:

Given

First charge of q_1=15\ nC is placed at x=1.5\ m

Second charge  q_2=-20\ nC is placed at y=-2\ m

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E=\frac{kq}{r^2}

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