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soldi70 [24.7K]
3 years ago
8

If a rock is dropped from the top of a tower at the front of and it takes 3.6 seconds to hit the ground. Calculate the final vel

ocity of the penny in m/s.
Physics
1 answer:
KengaRu [80]3 years ago
7 0

Answer:

36 m/s

Explanation:

t = 3.6s

u = 0m/s

a = +g = 10m/s²

v = ?

using,

v = u + at

v = 0 + 10(3.6)

v = 36 m/s

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madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

So, the force due to the sphere of mass m is

F = GMmx/[√(a² + x²)]³

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3 years ago
What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.
Softa [21]

<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

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Answer:

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b) 0.314159

c) 1.8326

Explanation:

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