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Katena32 [7]
3 years ago
7

The speed of the proton when it reaches point

Physics
1 answer:
romanna [79]3 years ago
8 0
1) An electron is moving east in a uniform electric field of 1.53N/C directed to the west. At point A, the velocity of the electron is 4.51�105m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.400m east of point A? 2) A proton is moving in the uniform electric field of part A. At point A, the velocity
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A ray of light exits from a material with a refractive index of 1.75, traveling into air. The angle of refraction is 25°. What w
Viktor [21]

Answer:

θi = 47.7°

Explanation:

The formula for the refractive index is as follows:

n = \frac{Sin\theta_i}{Sin\theta_r}

where,

n = refractive index = 1.75

θi = angle of incidence = ?

θr = angle of refraction = 25°

Therefore,

1.75 = \frac{Sin\ \theta_i}{Sin\ 25^o} \\\\(1.75)(Sin\ \ 25^o) = Sin\ \theta_i\\\\\theta_i = Sin^{-1}(0.739)

<u>θi = 47.7°</u>

3 0
2 years ago
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Sphinxa [80]
Because of the parameters that we are given in the question, the right current formula that should be used for this physics problem is:
I = P/V
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I = Current
P = Power  and 
V = Voltage
From the question, we are told that:
P = 16 W
V = 9 V
Thus, current = P/V = 16/9 = 1.78
Therefore, The current used by the camcorder is 1.8 Amphere. 
8 0
2 years ago
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