Answer:
ω = 0.467 rad/s
Explanation:
given,
tangential force exerted by the person = 37.7 N
radius of merry-go-round = 2.75 m
mass of merry-go-round = 144 Kg
angle = 33.2°
moment of inertia
I = 544.5 kg.m²
torque = force x radius
τ = 37.7 x 2.75
τ = 103.675 N.m
angular acceleration
α = 0.190 rad/s²
now ,
distance =
d = 0.579 rad
we know,
using equation of rotational motion
t = 2.46 s
angular speed
ω = α x t
ω = 0.19 x 2.46
ω = 0.467 rad/s
Answer:
a)N = 3.125 * 10¹¹
b) I(avg) = 2.5 × 10⁻⁵A
c)P(avg) = 1250W
d)P = 2.5 × 10⁷W
Explanation:
Given that,
pulse current is 0.50 A
duration of pulse Δt = 0.1 × 10⁻⁶s
a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles
N = Δq/e
charge is given by Δq = IΔt
so,
N = IΔt / e
N = 3.125 * 10¹¹
b) Q = nqt
where q is the charge of 1puse
n = number of pulse
the average current is given as I(avg) = Q/t
I(avg) = nq
I(avg) = nIΔt
= (500)(0.5)(0.1 × 10⁻⁶)
= 2.5 × 10⁻⁵A
C) If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,
eV = K
V = K/e
the power is given by
P = IV
P(avg) = I(avg)K / e
= 1250W
d) Final peak=
P= Ik/e
=
P = 2.5 × 10⁷W