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vivado [14]
3 years ago
14

Two adjoining plane mirrors are separated by an angle of 100°. Light is incident on the first mirror at an angle of 70°. The lig

ht reflects toward the second mirror and is then reflected from it. At what angle is the light reflected from the second mirror?

Physics
2 answers:
Ludmilka [50]3 years ago
8 0

Answer:

The angle is the light reflected from the second mirror is 30°

Explanation:

Given that,

Angle between mirrors = 100°

Angle of incident = 70°

We know that,

The incident angle is equal to the reflected angle.

We need to calculate the angle BCA

According to figure,

In \Delta ABC,

\angle ABC+\angle BCA+\angle C=180

Put the value into the formula

20+100+\angle BCA=180

\angle BCA = 60

We need to calculate the angle is the light reflected from the second mirror

Using relation of incident and reflection

i+\angle BCA=90

i=90-60=30^{\circ}

We know that, i = r

So, i=r=30°

Hence, The angle is the light reflected from the second mirror is 30°

vladimir1956 [14]3 years ago
4 0

Answer:

The angle is the light reflected from the second mirror is 30°

Explanation:

Given that,

Angle between mirrors = 100°

Angle of incident = 70°

We know that,

The incident angle is equal to the reflected angle.

We need to calculate the angle BCA

According to figure,

In ,

Put the value into the formula

We need to calculate the angle is the light reflected from the second mirror

Using relation of incident and reflection

We know that, i = r

So, i=r=30°

Hence, The angle is the light reflected from the second mirror is 30°

Read more on Brainly.com - brainly.com/question/13152660#readmore

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mel-nik [20]

Answer:

(a) 0.17 m

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(c) 6.38 × 10^{-26} N

(d) 7.37 ×10^{-29} N

Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

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3 years ago
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Answer:

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The cylinder with piston locked in place is immersed in a mixture of ice and water and allowed to come to thermal equilibrium wi
lukranit [14]

Answer:

a. volume of gas:  (decreases)

b. temperature of gas:  (same)

c. internal energy of gas: (same)

d. pressure of gas: (increases)

Explanation:

We have a gas (let's suppose that is ideal) in a piston with a fixed volume V.

Then we put in a reservoir at 0°C (the mixture of water and ice)

remember that the state equation for an ideal gas is:

P*V = n*R*T

and:

U = c*n*R*T

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Now, we have equilibrium at T = 0°C, then we can assume that T is also a constant.

Then in the equation:

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all the terms in the left side are constants.

P*V = constant

And knowing that:

U = c*n*R*T

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c) Internal energy of the gas:

we have:

P*V = n*R*T = constant

and:

P*V = U/c = constant.

Then:

U = c*Constant

This means that the internal energy does not change.

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P = (constant)/V

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Answer:

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