(b) Calculate the momentum when a mass of 40kg is moving with a velocity of 20m/s in the forward
2 answers:
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Answer:
x = 0.974 L
Explanation:
given,
length of inclination of log = 30°
mass of log = 200 Kg
rock is located at = 0.6 L
L is the length of the log
mass of engineer = 53.5 Kg
let x be the distance from left at which log is horizontal.
For log to be horizontal system should be in equilibrium
∑ M = 0
mass of the log will be concentrated at the center
distance of rock from CM of log = 0.1 L
now,
∑ M = 0
x = 0.974 L
hence, distance of the engineer from the left side is equal to x = 0.974 L
Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
