Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, 
We need to find the y component of the normal force. We know that the y component of the normal force is given by :

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.
Answer: The new force is:
F1 = -(9/16)*F0
where F0 is the initial force.
Explanation:
The charges of the spheres is q1 = 8Q and q2 = -2Q
and as you know, the force between charged objects is:
F = k*q1*q2/r^2
where k is a constant and r is the distance
F0 = -(k16Q^2)/r^2
where the negative sign means that the force is attractive.
Now, when the spheres touch each other, the charge must be distributed equally in both spheres. So the total charge is q1 + q2 = 8Q - 2Q = 6Q
then each sphere has now a charge of 3Q.
The new force is:
F1 = k*3Q*3Q/r^2 = (k*9*Q^2)/r^2 = -(9/16)*F0
You can see that F1 is positive, this means that the force now is repulsive.
We need to the where the arrows are..
The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
<h3>What is angular velocity?</h3>
The velocity of a particle when moving in the circular path.
Let speed of the bug with respect to ground is u.
Speed of bag with respect to ring will be
v = u - (- Rω) =
Then, u = v- Rω...............(1)
Angular momentum of ring and bug will remain conserved.
Initial momentum: L ring + Lbug =0
Final momentum: -2m₁ R²ω + m₂uR =0...............(2)
Using equation (1) and (2), the angular velocity expression will be
ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction
Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].
Learn more about angular velocity.
brainly.com/question/17592191
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