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3 years ago

When the mirror is rotated, the normal will turn as well, but will the incident Ray and reflected ray turn?

1 answer:

6 0

When a mirror is rotated . . .

-- The incident ray doesn't turn. It's just the line from the source to the mirror.
It would be there, in the same place, even if there was no mirror.

-- The normal turns. It's the line perpendicular to the mirror, so it must turn
with the mirror.

-- Since the normal tuns and the incident ray doesn't, the angle between them
must change. And since the angle of the reflected ray is equal to the angle of
the incident ray, the reflected ray must also turn.

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andreev551 [17]

Answer:

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3 years ago

An electrical heater 100 mm long and 5 mm in diameter is inserted into a hole drilled normal to the surface of a large block of

slega [8]

Answer:

$T_{1}=94.9^{o}C$

Explanation:

Given data

length=100mm

Diameter=5mm

Thermal conductivity=5 W/m.K

Power=50 W

Temperature=25°C

The temperature of heater surface follows from the rate equation written as:

$T_{1}=T_{2}+\frac{q}{kS}$

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

$S=\frac{2\pi L}{ln(\frac{4L}{D} )} \\$

Substitute the given values

$S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m$

The temperature of heater is then:

$T_{1}=25^{o}C+\frac{50W}{5W/m.K*0.143m} \\T_{1}=94.9^{o}C$

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.

$T_{1}=94.9^{o}C$

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3 years ago

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

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Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

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$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

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3 years ago

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4 years ago

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