An object on the moon will weigh 1/6th of what it does on earth but have the same mass. Why is that?
1 answer:
Answer:
Because the weight depends of the gravity
Explanation:
This is because weight and mass are different, in order to better understand this problem we will apply an example with real values, which will help us to determine a person's weight.
A man has a mass of 80 [kg] on Earth when measuring his weight he realizes that it is 784.9 [N] and on the moon it is 130.8 [N]
<u>On Earth</u>
<u />
Where:
g = gravity
<u>Weight on the moon</u>
<u />
Wm = 80 * 1.635
Wm = 130.8[N]
<u>Weight on the earth</u>
<u />
We = 80 * 9.81
We = 784.8[N]
<u />
In this way we can see that the weight depends on the gravity of where the person is located.
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Refer to the figure shown below.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball: