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
Picture one prediction there would be more clouds picture two precipitation picture 3 95%
Answer:
The current is proportional to the voltage and inversely proportional to the resistance. This mean if the resistance decreased and the voltage stayed constant, the current will increase.
Explanation:
Let's star with a closed circuit with a constant voltage of 48V, and a variable resistor selector of 6, 4 and 2 Ohms.
Using Ohm's Law equation
With the variable resistor selector on 6 Ohms
Changing the selector to a 4 Ohms
Last, changing the selector to a 2 Ohms
Answer: k = ma + uk×mgcosθ/ xf
Explanation: The body is placed on a frictionless inclined ramp.
The weight of the object has 2 components, horizontal component (mgsinθ) and vertical component (mgcosθ).
The horizontal component of weight is responsible for making tje object slide down the plane even with no applied force.
So from newton's second law of motion
mgsinθ - uk×R = ma
Where uk = coefficient of kinetic friction.
R = normal reaction = mgcosθ
mgsinθ - uk×mgcosθ = ma
mgsinθ = ma + uk×mgcosθ
mgsinθ is the applied force in this case. This applied force compresses a spring.
According to hooke's law,
F =ke
Where F = ma + uk×mgcosθ, e =xf
F = applied force , e = extension and k = spring constant.
k = F/e
k = ma + uk×mgcosθ/ xf