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
Answer:
Explanation:
wave length of light λ = 623 x 10⁻⁹ m .
Distance of screen D = 76.5 x 10⁻² m
width of slit = d
Distance on the screen between the second order minimum and the central maximum = 2 λ D / d
1.11 x 10⁻² = (2 x 623 x 10⁻⁹ x 76.5 x 10⁻² )/ d
d = ( 2 x 623 x 10⁻⁹ x 76.5 x 10⁻²) / 1.11 x 10⁻²
= 85872.97 x 10⁻⁹
= 85.87297 x 10⁻⁶
= 85.87 μm
width a of the slit is = 85.87 μm
Answer:
Explanation:
Given
Object is thrown with a velocity of 
Acceleration due to gravity is -g (i.e. acting downward)
Vertical distance traveled by object is given by
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
at maximum height final velocity is zero
time taken to reach maximum height
using
v=u+at
0=9-gt
The time described above is known as the waves Period.
The time which it takes for a particle to complete one full cycle is known as the period. Period is normally measured in seconds. Frequency on the other hand is the number of cycles which are completed in a given period of time e.g a second. periodic time T is given by reciprocal of frequency (1/f).
