A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o
1 answer:
Answer:
2940 kg m/s
Explanation:
First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:
where
v0 =0 is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration due to gravity
t = 3 s is the time of the fall
Substituting into the formula, we find
And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:
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Answer:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s
Explanation:
The momentum of the particle is related to force by the following equation:
Δp = F · Δt
Where:
Δp = change in momentum = final momentum - initial momentum
F = constant force.
Δt = time interval.
Let´s calculate the x-component of the momentum after the 0.13 s:
final momentum - 8 kg m/s = -7 N · 0.13 s
final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s
final momentum = 7.09 kg m/s
Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:
final momentum = 5 kg m/s² · 0.13 s
final momentum = 0.65 kg m/s
Then, the mometum vector will be as follows:
p = (7.09 kg m/s, 0.65 kg m/s)
The magnitude of this vector is calculated as follows:
The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s
Answer
given,
I = 0.140 kg ·m²
decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.
a)
τ = -1.467 N m
b) angle at which fly wheel will turn
θ = 20.35 rad
c) work done on the wheel
W = τ x θ
W = -1.467 x 20.35 rad
W = -29.86 J
d) average power of wheel