The speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.
<h3>Relative velocity of the car</h3>
The speed of the car passing you is determined by applying relative velocity principle as shown below;
Vr = Va - Vb
Vr = 26 m/s - 32 m/s
Vr = -6 m/s
Thus, the speed of the car passing you is 6 m/s while car is moving 6 m/s behind the car.
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weight = mg acts
downwards <span>
normal force = N acts upwards.
and force F acts at an angle θ below the horizontal.
(Let us assume that the woman pushes from the left, so F is
acted towards the right, which is below the horizontal)
so that, Frictional force, f=us*N acts towards the left
Now we balance the forces along x and y directions:
y direction: N = mg + F sinΘ
x direction: us * N = F cosΘ
We let the value of µs be equal to a value such that any F
will not be able to move the crate. Then, if we increase F by an amount F',
then the force pushing the crate towards the right also increases by F' cosΘ. Additionally,
the frictional force f must raise by exactly this amount.
Since f can’t exceed us*N, so the normal force must increase
by F' cosΘ/us.
Also, from the y direction equation, the normal force exceeds
by F' sin Θ.
<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Answer:
found in the nucleus, has mass of one amu
Explanation: