According to the inverse square law of light, how will the apparent brightness of an object change if its distance to us triples
1 answer:
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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
Missing question: "What is the spring's constant?"
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
And by using Hook's law, we can find the constant of the spring:
Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
And by using Hook's law, we can find the constant of the spring: