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kicyunya [14]
4 years ago
5

What is the angle of deviation in a plane mirror at normal incidence?​

Physics
1 answer:
Tems11 [23]4 years ago
7 0

Answer:

The deviation of a mirror is equal to twice the angle of incidence.The total angle between the straight-line path and the reflected ray is twice the angle of incidence. This is called the deviation of the light and measures the angle at which the light has strayed from its initial straight-line path.

HOPE IT HELPS :)

PLEASE MARK IT THE BRAINLIEST!

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Suppose two waves collide, and the temporary combined wave that results is smaller than the original waves. what term best descr
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The collision of the two waves would bring force and that would make the new wave smaller
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According to The Flash, The heroes top speed is Mach 3.3, or 2,500 miles per hour. How
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Explanation:

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Define linear expansivity
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Read 2 more answers
The moon Phobos orbits Mars
shepuryov [24]

27.9816 \times 10^{3} s is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

                  \frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant  6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}

Here, given data

M=6.23 \times 10^{23} \mathrm{kg}

R=9.38 \times 10^{6} \mathrm{m}

Substitute the given values, we get T as

      \frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}

      T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}

      T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}

Taking square root, we get

       T=27.9816 \times 10^{3} s

4 0
3 years ago
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
Lelechka [254]

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ T_1Cos30^{\circ}-T_2Cos60^{\circ}=0

or,

    T_1Cos 30^{\circ}=T_2Cos60^{\circ}

On substituting the values, we get

      T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}

      T_1\times \sqrt{3} =T_2....(equation 1)

On the Y-axis,

⇒ T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N

                        \frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N

                    T_1+\sqrt{3}T_2=139.2\times 2

From equation 1, we get

           T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N

                        T_1+3T_1=278.4 \ N

                                4T_1=278.4 \ N

                                  T_1=\frac{278.4}{4}

                                       =69.6 \ N  

6 0
3 years ago
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