Answer:
The magnitude of the new electric field is <u>35820 N/C</u>.
Explanation:
Given:
Original magnitude of electric field (E₀) = 2388 N/C
Original voltage = 'V' (Assume)
Original separation between the plates = 'd' (Assume)
Now, new voltage is three times original voltage. So, 
New distance is 1/5 the original distance. So, 
Now, electric field between the parallel plates originally is given as:
Let us find the new electric field based on the above formula.
Now, 
. So,
Therefore, the magnitude of the new electric field is 35820 N/C.
Answer:
B.
Explanation:
K is the SI unit for Kelvin which measures temperature.
kg is NOT an SI unit, g is the SI unit for mass.
L is NOT an SI unit, m^3 is the SI unit for volume.
Therefore the answer is B.
Answer:
the tangential velocity of the student is 4.89 m/s.
Explanation:
Given;
the radius of the circular path, r = 3.5 m
duration of the motion, t = 4.5 s
let the student's tangential velocity = v
The tangential velocity of the student is calculated as follows;
Therefore, the tangential velocity of the student is 4.89 m/s.
Answer:
a) 46.5º b) 64.4º
Explanation:
To solve this problem we will use the laws of geometric optics
a) For this part we will use the law of reflection that states that the reflected and incident angle are equal
θ = 43.5º
This angle measured from the surface is
θ_r = 90 -43.5
θ_s = 46.5º
b) In this part the law of refraction must be used
n₁ sin θ₁ = n₂. Sin θ₂
sin θ₂ = n₁ / n₂ sin θ₁
The index of air refraction is n₁ = 1
The angle is this equation is measured between the vertical line called normal, if the angles are measured with respect to the surface
θ_s = 90 - θ
θ_s = 90- 43.5
θ_s = 46.5º
sin θ₂ = 1 / 1.68 sin 46.5
sin θ₂ = 0.4318
θ₂ = 25.6º
The angle with respect to the surface is
θ₂_s = 90 - 25.6
θ₂_s = 64.4º
measured in the fourth quadrant
