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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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If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
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Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

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2 years ago
With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c
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Answer:

Then the difference of weight between the two cars are:

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An object's weigh due to the gravitational attraction force of the earth is:

w = mg

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                         g is the  gravitational acceleration in the surface earth

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