The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C ,
which delivers 3 kW of power to the shaft while it is turning at 50rev/s. Gears A and B remove 1 kW and 2 kW, respectively. The shaft is free to turn in its support bearings D and E.
1 answer:
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Answer:
21870.3156 N
Explanation:
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 1.6 m/s²
Equation of motion
The acceleration of the craft should be 1.02234 m/s²
Weight of the craft
Thrust
The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N
Answer:
The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.
Explanation:
It is given that,
A high jumper jumps over a bar that is 2 m above the mat, h = 2 m
We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :
g is acceleration due to gravity
v = 6.26 m/s
So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.