A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented above, and the other chamber is
1 answer:
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Answer:
1. False
2. True
3. True
Explanation:
1- False —> The relation between electric potential and electric field is given such that
Therefore, for a uniform E field, electric potential is linearly proportional to the distance.
2- True —> The electric field lines always cross the equipotential lines perpendicularly.
3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:
There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.
The angle is 53.5 degrees.
What is a flat mirror?
A plane mirror is a mirror with a flat (planar) reflective surface. For light rays striking a plane mirror, the angle of reflection equals the angle of incidence.
This ques have been solved in two steps:
Step 1:
given that;
Θ=107 degrees
from figure,
∠BAC=180-ψ-(90-φ)
∠BAC= 90-ψ-φ
∠AOC=180-2(ψ-φ)-2φ
∠AOC=180-2φ
Step 2: part(a)
writing an expression for the psi at point o:
Θ+180-2ψ=180
2ψ=Θ
ψ=Θ/2
above is the expression for ψ
Step 3: part(b)
The angle of ψ is,
ψ=Θ/2
ψ=107/2
ψ=53.5 degrees
Hence, the angle is 53.5 degrees.
To more know about flat mirrors and phi and psi angles the link is given below:
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Answer:
The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.
Explanation:
Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively
and the mass of big rock be 'M'
Initial momentum of the system equals
Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'
Thus the final momentum of the system is
Equating initial and the final momenta we get
Now since the surface is frictionless thus the energy is also conserved thus
Similarly the final energy becomes
\
Equating initial and final energies we get
Solving i and ii we get
Using this in equation i we get
Thus putting v = -v' in equation i we get V' = 0
This implies Smaller stone rebounds while as larger stone remains stationary.