Answer:
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Answer:
a)
346.67 N/C, downward
b)
1.3 m
Explanation:
(a)
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
r = distance of location from the charged particle = 0.225 m
E = magnitude of electric field at the location
Magnitude of electric field at the location is given as

Inserting the values

E = 346.67 N/C
a negative charge produce electric field towards itself.
Direction : downward
(b)
E = magnitude of electric field at the location = 10.5 N/C
r = distance of location from the charged particle = ?
q = magnitude of charge on the particle = 1.95 x 10⁻⁹ C
Magnitude of electric field at the location is given as

Inserting the values

r = 1.3 m
Answer:
Displacement and acceleration
charge = current * time
q = 15 * 300 [5 min = 300 sec ]
q = 4500 C
Answer:
35.16 degrees
Explanation:
Knowing that the index of refraction of the guide is 1.33, calculate the resulting angle of refraction for a ray of light that falls on a pool with an angle of incidence of 50º
Refractive index, n = 1.33
The angle of incidence, i = 50°
We need to find the angle of refraction. let it is r. It can be calculated using Snells law as follows:

So, the angle of refraction is 35.16 degrees.