Answer:
Hello some part of your question is missing below is the missing part
2. What is the force on the charged particle if it is now located at the 0V potential difference line? (mN) (hint: The electric field can be obtained as above using the 0V and -10V equipotential lines.)
answer :
1) 0.8 mN
2) 0.8 mN
Explanation:
Given data:
1) Calculate the force on the charged particle
q = 80 μC , Va = 30v , Vb = 40v, ∝ = 1 m
E = ( Δv ) / ∝
= ( Vb - Va ) / ∝
F = qE
= 80 μC * ( 40 - 30 ) / 1 m
= 800 μC
F = 0.8 mN
<u>2) Calculate the force on the charged particle when it is located at 0V</u>
Va = -10V , Vb = 0V, q = 80 μC, ∝ = 1 m
F = qE
where E = ( 0 - ( -10 ) / 1
F = 80 μC * ( 0 - ( -10 ) / 1
= 800 μC = 0.8 mN
Answer:
option B
Explanation:
Given,
Refractive index of the glass, n₂ = 1.50
Refractive index of plastic, n₁ = 1.30
critical angle = ?
![\theta_{critical}=\sin^{-1} (\dfrac{n_1}{n_2})](https://tex.z-dn.net/?f=%5Ctheta_%7Bcritical%7D%3D%5Csin%5E%7B-1%7D%20%28%5Cdfrac%7Bn_1%7D%7Bn_2%7D%29)
n₁ is the refractive index of the rare medium
n₂ is the refractive index of the denser medium.
![\theta_{critical}=\sin^{-1} (\dfrac{1.30}{1.50})](https://tex.z-dn.net/?f=%5Ctheta_%7Bcritical%7D%3D%5Csin%5E%7B-1%7D%20%28%5Cdfrac%7B1.30%7D%7B1.50%7D%29)
![\theta_{critical}=\sin^{-1} (0.8667)](https://tex.z-dn.net/?f=%5Ctheta_%7Bcritical%7D%3D%5Csin%5E%7B-1%7D%20%280.8667%29)
![\theta_{critical}=60.1^\circ](https://tex.z-dn.net/?f=%5Ctheta_%7Bcritical%7D%3D60.1%5E%5Ccirc)
The correct answer is option B
You mean like a box sitting on a table.
One force is the force of gravity, pulling downward on the box.
Now, you know that the forces acting on the box must be balanced, because
if they're not, then the box would be accelerating. But it's just sitting there, so
there must be some other force, just exactly the right strength and direction to
exactly cancel the force of gravity on the box, so that the net force on it is zero.
The other force is the force of the table pushing upward on the box. It's called
the "normal force".
It is the Pacific Ocean.
There's not really a way to give evidence, so hope it helps!