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Ierofanga [76]
3 years ago
6

Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After

reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
Physics
1 answer:
Setler [38]3 years ago
3 0

A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.

A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror.  This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.

In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.

Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.

For reference: brainly.com/question/20380620

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A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the p
frez [133]

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

a(t)=(5-10t)\ m/s^2

(a) Since, a=\dfrac{dv}{dt}

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v=\int(a.dt)

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(b) v=\dfrac{dx}{dt}

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(c) Velocity function is given by :

v=5t-5t^2

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7 0
3 years ago
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

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Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
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