C)Weight
A)An attraction between two objects that have mass
B)Force decreases as distance increases.
grav force of massive planet increases your weight
Centripetal acceleration is the motion inwards towards the center of a circle. It is given by the square of velocity, divided by the radius of the circular path.
ac = v²/r, where ac is the centripetal acceleration in m/s²
Therefore,
Centripetal acceleration = 3²/20m
= 9/20m
= 0.45 m/s²
Thus, the centripetal acceleration is 0.45 m/s²
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Explanation:
Given that,
Potential difference, V = 412 V
Magnitude of magnetic field, B = 188 mT
(a) The potential energy of electron is balanced by its kinetic energy as :

v is speed of the electron

(b) When the charged particle moves in magnetic field, it will move in circular path. The radius of the circular path is given by :

Hence, this is the required solution.
Answer:
Explanation:
Given that,
Initial angular velocity is 0
ωo=0rad/s
It has angular velocity of 11rev/sec
ωi=11rev/sec
1rev=2πrad
Then, wi=11rev/sec ×2πrad
wi=22πrad/sec
And after 30 revolution
θ=30revolution
θ=30×2πrad
θ=60πrad
Final angular velocity is
ωf=18rev/sec
ωf=18×2πrad/sec
ωf=36πrad/sec
a. Angular acceleration(α)
Then, angular acceleration is given as
wf²=wi²+2αθ
(36π)²=(22π)²+2α×60π
(36π)²-(22π)²=120πα
Then, 120πα = 8014.119
α=8014.119/120π
α=21.26 rad/s²
Let. convert to revolution /sec²
α=21.26/2π
α=3.38rev/sec
b. Time Taken to complete 30revolution
θ=60πrad
∆θ= ½(wf+wi)•t
60π=½(36π+22π)t
60π×2=58πt
Then, t=120π/58π
t=2.07seconds
c. Time to reach 11rev/sec
wf=wo+αt
22π=0+21.26t
22π=21.26t
Then, t=22π/21.26
t=3.251seconds
d. Number of revolution to get to 11rev/s
∆θ= ½(wf+wo)•t
∆θ= ½(0+11)•3.251
∆θ= ½(11)•3.251
∆θ= 17.88rev.