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kap26 [50]
4 years ago
10

why doesn't light bend (refract) when it initially enters a glass? what actually happens to the light inside the semi-circle?

Physics
1 answer:
tiny-mole [99]4 years ago
7 0
The light does not bend when it initially enters the glass because it travels from a less dense medium such as air to a dense medium such as glass. The refraction of light happens only on the flat side of the semi-circle, while on the circular side, it strikes on the interface.
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What elements are solid liquid and gas at room temperature?
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What is the gravitational force acting on a 556 kg object standing on the Earth's surface?
Vitek1552 [10]

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gravity

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2 When a cube of hot metal is placed in a beaker of cold water, the temperature of the water -
jek_recluse [69]
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3 0
3 years ago
Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
JulijaS [17]

Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

4 0
3 years ago
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