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:
C. Quadruple the intensity
Explanation:
The intensity of the sound is proportional to square of amplitude of the sound.
I ∝ A²

When the given sound is twice loud as the initial value, then the new amplitude is twice the former.
A₂ = 2A₁

Thus, to make a given sound seem twice as loud, the musician should Quadruple the intensity
Answer:
P= 454.11 N
Explanation:
Since P is the only horizontal force acting on the system, it can be defined as the product of the acceleration by the total mass of the system (both cubes).

The friction force between both cubes (F) is defined as the normal force acting on the smaller cube multiplied by the coefficient of static friction. Since both cubes are subject to the same acceleration:

In order for the small cube to not slide down, the friction force must equal the weight of the small cube:

The smallest magnitude that P can have in order to keep the small cube from sliding downward is 454.11 N