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Marta_Voda [28]
3 years ago
9

A 5.20-N force is applied to a 1.05-kg object to accelerate it rightwards. The object encounters 3.29-N of friction. Determine t

he acceleration of the object.
Physics
1 answer:
tatiyna3 years ago
8 0

Answer:

Explanation:

We will use the equation F - f = ma, which is just a fancy way of stating Newton's 2nd Law. For us:

F = 5.20 to the right (+)

f = 3.29 to the left (-)

m = 1.05 kg. Therefore,

5.20 - 3.29 = 1.05a and

1.91 = 1.05a so

a = 1.82 m/s/s to the right

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A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa
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7 0
3 years ago
A rock is dropped from the edge of a cliff into a pool of water. Assume free-fall acceleration is 10 m/s per second, and air res
il63 [147K]

Answer:

20 m

Explanation:

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S = ut+1/2gt²................................. Equation 1

Where S = Height, u = initial velocity, t = time, g = acceleration due to gravity.

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Substituting into equation 1

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S = 5(4)

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