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3 years ago

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You will begin with a relatively standard calculation.Consider a concave spherical mirror with a radius of curvature equal to 60

.0 centimeters. An object 6.00 centimeters tall is placed along the axis of the mirror, 45.0 centimeters from the mirror. You are to find the location and height of the image.a) What is the focal length Image for this mirror?b) Now use the spherical mirror equation to find the imagedistance s'.c) Find the magnification m,using sand s'.d) Finally, use the magnification to find the height of theimage y'.e) Solve the spherical mirror equation for s'.f)What is the magnification m? Use your answer from Part E

1 answer:

Strike441 [17]3 years ago

6 0

a) 30.0 cm

For any mirror, the radius of curvature is twice the focal length:

r = 2f

where

r is the radius of curvature

f is the focal length

For the mirror in this problem, we have

r = 60.0 cm is the radius of curvature

Therefore, solving the equation above for f, we find its focal length:

$f=\frac{r}{2}=\frac{60.0 cm}{2}=30.0 cm$

b) 90 cm

The mirror equation is:

$\frac{1}{s'}=\frac{1}{f}-\frac{1}{s}$

where

s' is the distance of the image from the mirror

f is the focal length

s is the distance of the object from the mirror

For the situation in the problem, we have

f = +30.0 cm is the focal length (positive for a concave mirror)

s = 45.0 cm is the object distance from the mirror

Solving the formula for s', we find

$\frac{1}{s'}=\frac{1}{30.0 cm}-\frac{1}{45.0 cm}=0.011 cm^{-1}$

$s'=\frac{1}{0.011 cm^{-1}}=90 cm$

c) -2

The magnification of the mirror is given by

$M=-\frac{s'}{s}$

where in this problem we have

s' = 90 cm is the image distance

s = 45.0 cm is the object distance

Solving the equation, we find:

$M=-\frac{90 cm}{45 cm}=-2$

So, the magnification is -2.

d) -12.0 cm

The magnification can also be rewritten as

$M=\frac{y'}{y}$

where

y' is the height of the image

y is the heigth of the object

In this problem, we know

y = 6.0 cm is the height of the object

M = -2 is the magnification

Solving the equation for y', we find

$y'=My=(-2)(6.0 cm)=-12.0 cm$

and the negative sign means that the image is inverted.

Part e and f are exactly identical as part b) and c).

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Answer:

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Explanation:

The period of oscillation is given by

$T=2\pi \sqrt{\frac {m}{k}}$

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Also, net force is given by

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Substituting the equation of $\frac {k}{m}$ into the above we obtain

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3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

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Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

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The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

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Answer:

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Explanation:

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