The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
The angular momentum(L) of an electron moving in a circular path is given by the formula,
L = mvr ........(i)
We know that the radius of the path of an electron in a magnetic field is
r = mv/qB
Putting this value in equation (i),
L = mv x mv/qB
or L = (mv)^2/qB
Putting the given values in the above equation,
4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3
v comes out to be 8.88 x 10^7 m/s.
Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.
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Answer:
Final speed after 2 seconds = 34.6 m/s
Explanation:
Given:
Initial speed of coin (u) = 15 m/s
Time taken = 2 seconds
Find:
Final speed after 2 seconds
Computation:
Gravitational acceleration of earth = 9.8 m/s²
Using first equation of motion;
v = u + at
or
v = u + gt
where,
v = final velocity
u = initial velocity
g = Gravitational acceleration
t = time taken
v = 15 + 9.8(2)
v = 15 + 19.6
Final speed after 2 seconds = 34.6 m/s
Answer:
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
Explanation:
Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.
At maximum height , the speed is zero and then the projective comes back on the ground.
Use the third equation of motion
Now let the velocity at the time of strike is v'.
Use third equation of motion, here initial velocity is zero.
Thus, the velocity at the time of strike is same as the velocity at the time of projection.
Answer:
The capacity of doing work is called energy.
its forms are :
1. mechanical energy
2.potential energy
3. kinetics energy
4. heat energy
5. light energy
6.sound energy
7. wind energy .etc.
