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steposvetlana [31]
3 years ago
6

When light (or another wave) bounces back from a surface ?

Physics
1 answer:
topjm [15]3 years ago
3 0
Light bounces back due to the phenomenon called reflection. When light incidents on a plane shiny surface like mirror, it travels again in the same medium(medium from which light was incident) with no loss in energy(energy before and after reflection will remain same iff incident surface completely reflects light)
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Incident rays parallel to the principle axis of a concave mirror will reflect _____. parallel to the principle axis through the
Svetradugi [14.3K]
I’m not sure if its correct but I think it’s focal Ray point


For concave mirrors, some generalizations can be made to simplify ray construction. They are: An incident ray traveling parallel to the principal axis will reflect and pass through the focal point. An incident ray traveling through the focal point will reflect and travel parallel to the principal axis.
8 0
3 years ago
Read 2 more answers
A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t
Sergio [31]

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

7 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
Ir
NemiM [27]

Answer:

what is your question because I seem to not see a question in this question....

3 0
2 years ago
"18. Milliliter is equivalent to 1 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Firlakuza [10]
18 mililiter is equivilent to 0.0047551
3 0
3 years ago
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