An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with
the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?
1 answer:
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Answer:
The weight of measuring stick is 9.8 N
Explanation:
given information:
the mass of the rock, = 1 kg
measuring stick, x =1 m
d = 0.25 m
to find the weight of measuring stick, we can use the following equation:
τ = Fd
τ = 0
-
= 0
F_{r} = the force of the rock
F_{s} = the force of measuring stick
= m g
= 1 kg x 9.8 m/s
= 9.8 N
thus, the weight of measuring stick is 9.8 N
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
, where 
is the coefficient of friction and 
is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
And so, this is the force that the worker must apply to the crate.
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:
And so, this is the force that the worker must apply to the crate.
Answer:
3.97305 m
Explanation:
a = Acceleration due to gravity = 9.81 m/s²
If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.
u = Initial velocity = 0
Equation of motion
So, height of the jump is 3.97305 m.