Answer:
Part a)

Part b)

Part c)

Part d)

Explanation:
Part a)
While bucket is falling downwards we have force equation of the bucket given as

for uniform cylinder we will have

so we have


now we have




now we have


Part b)
speed of the bucket can be found using kinematics
so we have



Part c)
now in order to find the time of fall we can use another equation



Part d)
as we know that cylinder is at rest and not moving downwards
so here we can use force balance



The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction
You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.
That's <em>false</em>. It's just the opposite. As you become more fit, your heart becomes able to accomplish more with each beat, so your resting heart rate DEcreases.
Here, as the charge is uniformly distributed in the sphere, we will consider s as an area of the sphere which is, s=4πr2 and r is radius of the gaussian surface shown in the figure above. From this, it can be seen that
Answer:
<em>The velocity of the carts after the event is 1 m/s</em>
Explanation:
<u>Law Of Conservation Of Linear Momentum
</u>
The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is
P=mv.
If we have a system of bodies, then the total momentum is the sum of the individual momentums:

If a collision occurs and the velocities change to v', the final momentum is:

Since the total momentum is conserved, then:
P = P'
In a system of two masses, the equation simplifies to:

If both masses stick together after the collision at a common speed v', then:

The common velocity after this situation is:

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

The velocity of the carts after the event is 1 m/s