Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as
the initial height of the block. Will A make it completely around the loop without losing contact with the track?
1 answer:
Answer:No
Explanation:
Given
Block A is at height h and released from rest
Initial Energy possessed by block A is equal to Potential energy of Block which is given by
where m=mass of block
After releasing the block, block first reaches to the bottom of the circle and then uses this energy to reach a top point of the loop.
But as soon as block reaches the top point of the loop it acquires the energy which is equal to Initial energy i.e. all the energy is stored in the form of potential energy and there is no kinetic energy so the block will not able to move further and fall from the top point.
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Answer:
Resolving Power=625000
Explanation:
See attached picture.
Answer:
The value of v2 in each case is:
A) V2=3v for only Vs1
B) V2=2v for only Vs2
C) V2=5v for both Vs1 and Vs2
Explanation:
In the attached graphic we draw the currents in the circuit. If we consider only one of the batteries, we can consider the other shorted.
Also, what the problem asks is the value V2 in each case, where:
If we use superposition, we passivate a battery and consider the circuit affected only by the other battery.
In the first case we can use an equivalent resistance between R2 and R3:
And
In the second case we can use an equivalent resistance between R2 and (R1+R4):
And
If we consider both batteries:
