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MrRissso [65]
3 years ago
6

A ray of light that strikes a surface is known as?

Physics
2 answers:
Lena [83]3 years ago
7 0

A ray of light that strikes a surface is known as the <em>incident ray.  (d)</em>

lisov135 [29]3 years ago
6 0

In the physics of optics, it has been defined that the incident rays are the rays which strikes the surface.

Answer : Option D

<u>Explanation:</u>

When the ray of light strikes the surface it is know as ''incident ray''. The angle which it makes with the normal or the perpendicular imagined or drawn to the surface is the ''incident angle''.

According to the ''Law of Reflection'', the incident ray after striking the surface is reflected back, the ray which leaves the surface is called the ''Reflected Ray''. The angle which is made between the same normal and reflected ray is the reflected angle.

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3 years ago
Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at
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Answer:

1. the electric potential energy of the electron when it is  at the midpoint is - 2.9 x 10^{-17} J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x  10^{-17} J

Explanation:

given information:

q_{1} =  3 nC = 3 x 10^{-9} C

q_{2} =  2 nC = 2 x 10^{-9} C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is  at the midpoint

potential energy of the charge, F

F = k \frac{q_{e}q}{r}

where

k = constant (8.99 x 10^{9} Nm^{2} /C^{2})

electron charge, q_{e} = - 1.6 x 10^{-19} C

since it is measured at the midpoint,

r = \frac{0.5}{2}

  = 0.25 m

thus,

F = F_{1}+ F_{2}

  = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = \frac{kq_{e} }{r} (q_{1} +q_{2})

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} +2 x 10^{-9})/0.25

  = - 2.9 x 10^{-17} J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

r_{1} = 10 cm = 0.1 m

r_{2} = 0.5 - 0.1 = 0.4 m

F = k\frac{q_{e} q_{1} }{r} + k\frac{q_{e} q_{2} }{r}

  = kq_{e}(\frac{q_{1} }{r_{1} }+\frac{q_{2} }{r_{2} })

  = (8.99 x 10^{9})( - 1.6 x 10^{-19} )(3 x 10^{-9} /0.1+2 x 10^{-9}/0.4)

  = - 5.04 x  10^{-17} J

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