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maks197457 [2]
3 years ago
5

Two plane mirrors are facing each other. They are parallel, 5.00 cm apart, and 30.0 cm in length, as the drawing indicates. A la

ser beam is directed at the top mirror from the left edge of the bottom mirror. What is the smallest angle of incidence with respect to the top mirror, such that the laser beam hits each mirror only twice.
Physics
1 answer:
ycow [4]3 years ago
6 0

Answer:

The minimum angle of incidence is, ∅  = 56° 18' 35'' to the normal

Explanation:

Given data,

The distance between the two plane mirror, d = 5 cm

The length of the plane mirror, L = 30 cm

According to the laws of reflection, the angle of incidence is equal to the angle of reflection.

If the laser beam touches each mirror only twice, then the base of the triangle formed is not less than L/2.

Let ∅ be the angle of incidence to the normal. The normal divides the triangle into two equal halves where base becomes L/4.

Therefore,

                       <em> tan ∅ = opp / adj</em>

                                  = (L/4) / d

                                   = (30/4) / 5

                        tan ∅  = 1.5

                               ∅  = tan ⁻¹ (1.5)

                                    = 56° 18' 35''

Hence, the minimum angle of incidence is, ∅  = 56° 18' 35'' to the normal.

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Lorico [155]

the answer is correct why.

EXPLANATION:

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Explanation:

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3 years ago
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

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From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

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v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

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