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seraphim [82]
3 years ago
9

Ice-skater slides toward a sled sitting on the ice and hits it. The skater exerts a 12.6 N force on the sled at an angle of 15.3

° below the horizontal. The sled then moves 15.4 m forward. How much work did the skater do on the sled? Assume friction is negligible.
Knowns Unknown

F = ____________ d = _____________ W = ?

θ = ____________

Solve for the Unknown

a. Write the equation for work
b. Insert known quantities
Physics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 187.2 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 12.6 N

d = 15.4 m

\theta = 15.3 degree

so we will have

W = 12.6 \times 15.4 cos15.3

W = 187.2 J

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Compounds

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If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?
Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

<u>Given the following the details;</u>

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Mathematically, the quantity of charge with respect to electric potential is given by the formula;

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Substituting the values into the formula, we have;

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Therefore, the quantity of charge must be <em>5 Coulombs.</em>

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