Answer:
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
Explanation:
Knowing that the density of lead is 
and the volume, we can calculate the true weight of the piece of lead:
When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.
When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.
Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.
Now that we have the two measurements, we can calculate the difference:
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
To solve this exercise it is necessary to apply the kinematic equations of angular motion.
By definition we know that the displacement when there is constant angular velocity is
From our given data we know that,
Moreover we know that
Therefore for time t=8.1s we have,
That number in revolution is:
Here, we see that there are 15 complete revolutions
And 0.108 revolutions i not complete, so the tunable rotation is
Therefore the angle of the speck at a time 8.1s is 
This is very good conceptual question and can clear your doubts regarding work-energy theorem.
Whenever force is perpendicular to the direction of the motion, work done by that force is zero.
According to work-energy theorem,
Work done by all the force = change in kinetic energy.
here, work done = 0.
Therefore,
0=change in kinetic energy
This means kinetic energy remains constant.
Hope this helps
Answer:
Explanation:
When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.
When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other .So both the shells lose their charges .The positive half shell also loses all its charges
When we separate the half shells , there will be no deflection in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.
