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vivado [14]
4 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]4 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]4 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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Answer:

756.88 Volts will be the potential difference across each capacitor.

Explanation:

Q=C\times V

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Capacitance of first capacitor = C_1=2.81 \mu F=2.81\times 10^{-6} F

Charge of first capacitor = Q_1

Voltage across first capacitor = V_1=1220 V

Q_1=C_1V_1

Q_1=2.81\times 10^{-6} F\times 1220 V=0.0034282 C

Capacitance of first capacitor = C_2=6.61\mu F=6.61\times 10^{-6} F

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Voltage across first capacitor = V_2=560 V

Q_2=C_2V_2

Q_1=6.61\times 10^{-6} F\times 560 V=0.0037016 C

Both the capacitors are disconnected and positive plates are now connected to each other and the negative plates are connected to each other. These capacitors are connected in parallel combination.

Total charge = Q

Q =Q_1+Q_2=0.0034282 C+ 0.0037016 C=0.0071298 C

Total capacitance in parallel combination:

C_p=C_1+C_2=2.81\times 10^{-6} F+6.61\times 10^{-6} F=9.42\times 10^{-6} F

Potential across both capacitors = V

V=\frac{Q}{C}=\frac{0.0071298 C}{9.42\times 10^{-6} F}=756.88 V

756.88 Volts will be the potential difference across each capacitor.

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Explanation:

The chloroplast in the leaf are used to make food for the plant. If the leaf loses its chloroplast, the leaf with not be able to use solar energy to manufacture their food.

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Explanation:

The volume of an average room is:

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Now let’s transform this V_{room} to units of cm^{3}, knowing 1 ft=30.48 cm:

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Now, the number n of Ping-Pong balls that can be packed into the room is:

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