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Crazy boy [7]
3 years ago
5

The full moon rises in the east at the time of sunset. After rising, it continuously moves to the west and is seen high in the s

ky close to midnight. Which of these is the reason for the change of the moon’s position in the night sky?
A) The moon rotates on its axis.
B) The continuous change in the phases of the Moon.
C) The moon orbits counterclockwise around the earth.
D) The earth rotates from west to east on its own axis.
Physics
1 answer:
Sever21 [200]3 years ago
6 0

Answer:

Earth rotates from West to East on its own axis. As a result of this, the Moon changes its position during the night and the Moon appears to move from East to West in the sky.

Explanation:

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A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red
Orlov [11]

The physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

<h3>How fast was the physicist traveling, according to his own testimony?</h3>

Using the formula for doppler shift for light,

λ' = λ√[(1 + v/c)/(1 - v/c)] where

  • λ = wavelength of source,
  • λ' = wavelength of observer,
  • v = speed of source and
  • c = speed of light

Given that the driver is moving away from the stop light, we take the driver as the source. Since, the Doppler shift made the red light of wavelength 650nm appear green to him, with a wavelength of 520nm. we have

  • λ' = wavelength of source = 650 nm,
  • λ' = wavelength of observer = 520 nm

So, substituting the values of the variables into the equation, we have

λ' = λ√[(1 + v/c)/(1 - v/c)]

520 nm = 650 nm√[(1 + v/c)/(1 - v/c)]

520/650 = √[(1 + v/c)/(1 - v/c)]

0.8 = √[(1 + v/c)/(1 - v/c)]

Squaring both sides, we have

0.8² = (1 + v/c)/(1 - v/c)

0.64 = (1 + v/c)/(1 - v/c)

0.64(1 - v/c) = (1 + v/c)

0.64 - 0.64v/c = 1 + v/c

0.64 - 1 = v/c + 0.64v/c

-0.36 = 1.64v/c

-0.2195 = v/c

v = -0.22c

v = -0.22 × 3 × 10⁸ m/s

v = -0.66 × 10⁸ m/s

v = -6.6 × 10⁷ m/s

So, the physicist traveling, according to his own testimony at -6.6 × 10⁷ m/s.

Learn more about doppler shift for light here:

brainly.com/question/28499579

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5 0
1 year ago
A 1-lb block and a 100-lb block are placed side by side at the top of a frictionless hill. Each is given a very light tap to beg
qwelly [4]

Answer:

(c). The two blocks end in a tie

Explanation:

the reason being the absence of any resistance offered to both of the blocks.

if the slope of the hill is for instance 60 deg.

then the acceleration in absence of any resistance is a= 9.81sin(60)

since the acceleration is same then both of the blocks will reach at the same instant

4 0
3 years ago
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. If the plane is traveling hori
11111nata11111 [884]

Answer:

481 m

Explanation:

To fall 235 m, the time required is

t = √(2H/g)

t= √(2\times235/9.8)

t=6.92 seconds.

The supplies will travel forward

6.92 \times 69.4 ≈ 481 m

Therefore, the goods must be dropped 481  m in advance of the recipients.

5 0
3 years ago
A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0
3 years ago
Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!
krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

r_2 = 47 cm

so now the distance from friend must be 47 cm

8 0
2 years ago
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