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3 years ago

7

A glowing electric light bulb placed 15 cm from a concave spherical mirror produces a real image 8.9 from the mirror. the light

bulb is moved to a position 20cm from the mirror
a) what is the image position? (answer with -1000 if the image does not exist)

1 answer:

marta [7]3 years ago

5 0

Answer:

The image will form at distance d = 7.75 cm from the mirror

Explanation:

As we know that the bulb is placed at 15 cm distance from the concave mirror

Then the image will form at distance d = 8.9 cm

now we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

so we have

$\frac{1}{8.9} + \frac{1}{15} = \frac{1}{f}$

$f = 5.58 cm$

Now when bulb is placed at distance 20 cm from the mirror

then again we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$\frac{1}{d_i} + \frac{1}{20} = \frac{1}{5.58}$

now we have

$d_i = 7.75 cm$

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Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

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As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

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where the suffix i and f means initial and final respectively.

The initial momentum will be:

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So, this means:

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$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

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