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weqwewe [10]
4 years ago
13

A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10

^8 m. How fast does the moon move?
Physics
1 answer:
Alchen [17]4 years ago
7 0

Answer:

v = 895 m/s

Explanation:

Time period is given as 39 Earth Days

T = 39 days \times 24 hr \times 3600 s

T = 3369600 s

now the radius of the orbit is given as

r = 4.8 \times 10^8 m

so the total path length is given as

L = 2 \pi r

L = 2\pi (4.8 \times 10^8)

L = 3.015 \times 10^9

now the speed will be given as

v = \frac{L}{T}

v = \frac{3.015 \times 10^9}{3369600}

v = 895 m/s

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Four identical balls are thrown from the top of a cliff, each with the same speed. The
Jlenok [28]

Answer:

the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Explanation:

The kinetic energy, K.E. = (1/2) × m × v²

The velocity of the ball, v = u × sin(θ)

Where;

u = The initial velocity of the ball

θ = The reference angle

1) For the ball thrown straight up, we have;

θ = 90°

∴ v = u

The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh

Where;

h = The height of the cliff

∴ K.E. = (1/2) × m × (u² + 2gh)²

2) For the second ball thrown 30° to the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

3) For the third ball thrown at 30° below the horizontal, we have;

K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²

4)  For the fourth ball thrown straight down, we have;

K.E. = (1/2) × m × (u² + 2gh)²

Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal

Learn more about object kinetic energy of object if free fall here;

brainly.com/question/14872097

6 0
3 years ago
By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles
Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r²   -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

<u>F' = (4/9)F</u>

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3 years ago
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Answer:

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Explanation:

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