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lakkis [162]
2 years ago
12

The Doppler shift can be observed when ____________. Check all that apply. Group of answer choices the source of sound moves tow

ards an observer the source of sound moves away from an observer an observer moves towards a stationary sound source an observer moves away from a stationary sound source the observer and sound source move at the same speed in the same direction the observer and sound source move at the same speed towards one another none of these
Physics
1 answer:
castortr0y [4]2 years ago
6 0

Answer:

the source of sound moves towards an observe

Explanation:

The Doppler effect is related to waves such as sound or light. the effect causes an increase or decrease in the frequency of sound light or other waves when the souces either move towards or away from the observer. For example the siren of the train to a person on the platform, the redshift seen by astronomers.

Therefore, The Doppler shift can be observed when the source of sound moves towards an observer From a place closer to the observer than the last wave's crest, each consecutive wave crest is sent. Each wave therefore, takes a little less time than the preceding wave to reach the observer.

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Debris from the solar system strikes the Earth as which of the following​
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Answer: Meteors

Explanation:

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A physics student uses a spring loaded launcher to fire a 58 gram projectile horizontally. The projectile leaves the launcher at
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Answer:

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Explanation:

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3 0
3 years ago
Which describes the changes in visible light moving from red to violet?
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2 years ago
5) [Honors]A seagull, ascending straight upward at 5.2 m/s, drops a shell when it is 12.5m above the ground. (A)
jolli1 [7]

Answer:

(B) 13.9 m

(C) 1.06 s

Explanation:

Given:

v₀ = 5.2 m/s

y₀ = 12.5 m

(A) The acceleration in free fall is -9.8 m/s².

(B) At maximum height, v = 0 m/s.

v² = v₀² + 2aΔy

(0 m/s)² = (5.2 m/s)² + 2 (-9.8 m/s²) (y − 12.5 m)

y = 13.9 m

(C) When the shell returns to a height of 12.5 m, the final velocity v is -5.2 m/s.

v = at + v₀

-5.2 m/s = (-9.8 m/s²) t + 5.2 m/s

t = 1.06 s

3 0
3 years ago
To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

W = 11.5* \frac{\pi}{3}

W = 12.09J

4 0
3 years ago
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