Answer:231.16 N/C
Explanation:
Given
Electric Flux
Area(A)
Given Field point above 
Therefore angle between Area vector Electric Field =90-31.6=
We know that Flux is given by



E=231.16 N/C
Answer:
Explanation:
v² = u² + 2as
v = 0
u = 96 / 3.6 = 26.7 m/s
0² = 26.7² + 2a100
a = -3.5555555... ≈ -3.6 m/s²
the negative sign indicated the acceleration vector opposes the (assumed positive) initial velocity vector direction.
Answer: (1) 3.83x10^3 J
Explanation:
(1) Fx=(255N)cos60°
dx=30.0m
w=Fx dx =(255)(cos60°)(30.0m)
Answer:
Frequency, 
Explanation:
Visible red light has a wavelength of 680 nanometers (6.8 x 10⁻⁷ m). The speed of light is 3.0 x 10 ⁸ m / s. What is the frequency of visible red light?
It is given that,
Wavelength of a visible red light is, 
Speed of light is, 
We need to find the frequency of visible red light. It can be calculated using below relation.

So, the frequency of visible red light is
.