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rusak2 [61]
3 years ago
10

Two 20kg spheres are placed with their

Physics
1 answer:
Maslowich3 years ago
5 0

Answer:

F = 1.07 x 10⁻⁷ N

Explanation:

The gravitational force of attraction between two objects can be found by the use of Newton's Gravitational Law:

F = \frac{Gm_{1}m_{2}}{r^2}\\\\

where,

F = Gravitational Force of attraction = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = m₂ = mass of spheres = 20 kg

r = distance between the objects = 50 cm = 0.5 m

Therefore,

F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(20\ kg)(20\ kg)}{(0.5\ m)^2}\\\\

<u>F = 1.07 x 10⁻⁷ N</u>

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4 years ago
hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How
timurjin [86]

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

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t² = 84/4.9

t = 4.14 s

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4 0
4 years ago
Read 2 more answers
A small asteroid with a mass of 1500 kg moves near the earth. At a particular instant the asteroid’s velocity is ⟨3.5 × 104, −1.
zalisa [80]

Answer:

P_{f} =(5.7 x 10^{7 i - 2.24 x 10^{7 j) kgm/s

Explanation:

Due to earths gravity, force on asteroid is given by:

F= \frac{Gm_{1}m_{2} }{r^{2} } r^

Plugging in the values, we have

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F= 2736 i^ + 3078 j^

In order find the final momentum of the Asteroid, apply impulse momentum theorem

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P_{f} = 1500(3.5 x 10^{4 i - 1.8x10^{4 j) + (2736i + 3078j)(1.5x10^{3)

P_{f} =(5.7 x 10^{7  i- 2.24 x 10^{7 j)kgm/s

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