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Fantom [35]
3 years ago
11

A 15.0 cm tall object is placed 20.0 cm from a converging lens with a 10.0 cm focal length. Where is the image located?

Physics
1 answer:
Ksju [112]3 years ago
4 0

Answer: 20 cm

Explanation:

Given that

F = 10 cm

U = 20 cm

1/F = 1/V - 1/U

1/10 = 1/V - 1/20

1/V = 1/10 - 1/20

1/V = (2 - 1)/20

1/V = 1/20

V = 20 cm

The image will be located at 20cm away from the lens.

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storchak [24]

Answer:

65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

Explanation:

Given :

Frequencies of the sinusoids,

$f_{m_1}= 65 \ Hz$ ,  and

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Sampling rate f_s = \ 245 \ Hz

The positive frequencies at the output of the sampling system are :

$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $

When n = 0,

$f_{o_1}= f_{m_1} = 65 \ Hz,\ \  f_{o_2}= f_{m_2} = 95 \ Hz $

when n  = 1,

$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm  f_s $

$f_{o_1}= \pm 65 \pm 245,\ \  f_{o_2}=\pm 95 \pm 245$

$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \  f_{o_2}= 150 \ Hz,340 \ Hz$

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$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \  f_{o_2}= 395 \ Hz,585 \ Hz$

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65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz

8 0
3 years ago
The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earth’s surface.
aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

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Weight of probe on Mars = \frac{1}{3}  m * g₁ = \frac{1}{3} w₁

⇒Weight of probe on Mars =\frac{1}{3} Weight of probe on earth

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