For a point charge, how does the potential vary with distance from the point charge, r?
a constant
b. r.
c. 1/r.
d. .
e. .
Answer:
The correct option is C
Explanation:
Generally for a point charge the electric potential is mathematically represented as
Here we can deduce that the electric potential varies inversely with the distance i.e
So
From the diagram The value of cos C × sin A =
First step : determine the values of cos C and sin A
cos C = adjacent / hypotenuse
= a / b
=
= √3/2
sin A = sin 60⁰
= √3/2
Therefore the numerical value of cos C * sin A =
In conclusion From the diagram The value of cos C × sin A =
Learn more about right angle : brainly.com/question/24323420
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Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F =
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²