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BARSIC [14]
3 years ago
7

A ball is thrown forward at 5 m/s how would the path of the ball differ on earth than on the moon

Physics
2 answers:
Levart [38]3 years ago
6 0
In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases. 
Degger [83]3 years ago
6 0

Answer:

The ball on the moon would have a more straight path than on the Earth, and it would have a longer range

Explanation:

In both cases, the motion of the ball is the motion of a projectile, which has a parabolic trajectory. In fact, the path is the resultant of two indipendent motions:

- A horizontal uniform motion, with constant horizontal velocity 5 m/s

- An accelerated vertical motion, with constant acceleration downward equal to the gravitational acceleration of the planet

On the Moon, the value of the gravitational acceleration is about 1/6 that of the Earth. This means that the downward acceleration of the ball on the Moon is smaller than on the Earth: so, the ball will take more time to reach the ground on the Moon, and so it will have a longer range, and a more straight trajectory.

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3 years ago
What is the mass of a cannonball if the force a force of 2500 N gives the cannonball an acceleration of 200 m/s^2??
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A forklift raises a 1,020 N crate 3.50 m up to a shelf. How much work is done by the forklift on the crate?
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3 years ago
Read 2 more answers
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
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