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2 years ago

What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?

Physics

1 answer:

ikadub [295]2 years ago

8 0

Answer:

A convex mirror is sometimes referred to as a diverging mirror due to the fact that incident light originating from the same point and will reflect off the mirror surface and diverge. ... After reflection, the light rays diverge; subsequently they will never intersect on the object side of the mirror.

Explanation:

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What is the label for coefficient of friction

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Force]/[force] = Newon/Newton = 1

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It takes 180 kj of work to accelerate a car from 21.0 m/s to 27.0 m/s. what is the car's mass?

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3 years ago

A stack of 55 business cards is 1.85 cm tall. Use this information to determine the thickness of one business

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3 years ago

Read 2 more answers

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

3 years ago

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

dangina [55]

1) In the first case, the correct answer is
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.

2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
$f'= \frac{c}{c+v_s} f_0$
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.

6 0

2 years ago