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Gre4nikov [31]
3 years ago
15

What is the magnitude of the gravitational force of attraction between two 0.425-kilogram soccer balls when the distance between

their centers is 0.500 meter?2.41 Ă— 10’11 N4.82 Ă— 10’11 N5.67 Ă— 10’11 N1.13 Ă— 10’10 N
Physics
1 answer:
Lapatulllka [165]3 years ago
7 0

<u>Answer</u>

4.8212×10⁻¹¹ N


<u>Explanation</u>

The gravitational force F, between 2 masses m₁ and m₂ is given by:

F = (Gm₁m₂)/d²

Where G =  6.673 x 10⁻¹¹ N m²/kg² and d is the distance between the 2 masses.

F = (Gm₁m₂)/d²

   =  (6.673 x 10⁻¹¹ × 0.425 × 0.425)/0.500²

   = 1.2053×10⁻¹¹/0.25

   = 4.8212×10⁻¹¹ N

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Assuming typical speeds of 8.5 km/s and 5.5 km/s for P and S waves, respectively, how far away did the earthquake occur if a par
taurus [48]

Answer:

The earthquake occurred at a distance of 1122 km

Explanation:

Given;

speed of the P wave, v₁ = 8.5 km/s

speed of the S wave, v₂ =  5.5 km/s

The distance traveled by both waves is the same and it is given as;

Δx = v₁t₁ = v₂t₂

let the time taken by the wave with greater speed = t₁

then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.

v₁t₁ = v₂t₂

v₁t₁ = v₂(t₁ + 1.2 min)

v₁t₁ = v₂(t₁ + 72 s)

v₁t₁ = v₂t₁ + 72v₂

v₁t₁ - v₂t₁ = 72v₂

t₁(v₁ - v₂) = 72v₂

t_1 = \frac{72v_2}{v_1-v_2}\\\\t_1 =   \frac{72*5.5}{8.5-5.5}\\\\t_1 = 132 \ s

The distance traveled is given by;

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Δx = (8.5)(132)

Δx = 1122 km

Therefore, the earthquake occurred at a distance of 1122 km

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A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00
ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

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