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Marrrta [24]
4 years ago
14

If the Earth and distant stars were stationary (motionless) in space, what would we observe about the wavelength from these star

s?
A.Wavelengths measured would match the actual wavelengths emitted.
B.Wavelengths measured would be shorter than the actual wavelengths emitted.
C.Wavelengths measured would be longer than the actual wavelengths emitted.

If a star is moving toward us in space, what would we observe about the wavelength from these stars?

A.Wavelengths measured would be shorter than the actual wavelengths emitted.

B.Wavelengths measured would match the actual wavelengths emitted.

C.Wavelengths measured would be longer than the actual wavelengths emitted.
Physics
1 answer:
torisob [31]4 years ago
7 0
There's no such thing as "stationary in space".  But if the distance
between the Earth and some stars is not changing, then (A) w<span>avelengths
measured here would match the actual wavelengths emitted from these
stars. </span><span>

</span><span>If a star is moving toward us in space, then (A) Wavelengths measured
would be shorter than the actual wavelengths emitted from that star.

</span>In order to decide what's actually happening, and how that star is moving, 
the trick is:  How do we know the actual wavelengths the star emitted ?


 
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A student put her ear on one end of a long piece of lead. She asked her friend to
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1) The length of the lead is 1700 m

2) The frequency of the wave is 400 Hz

Explanation:

1)

The wave in this problem is a sound wave, which is produced by the vibrations of the particles of air. It is a type of longitudinal wave (which means that the vibrations occur back-and-forth along the direction of propagation of the wave).

The speed of a sound wave in air is approximately

v = 340 m/s

In this problem, the sound wave is heard after a time of

t = 5 s

after the shot. Therefore, the distance travelled by the wave in this time (and therefore, the length of the lead) is

d=vt=(340)(5)=1700 m

2)

The frequency of a wave is the number of complete cycles made by the wave in one second.

For the wave in this problem, we are told that the wave has 2000 whole wavelengths in 5 seconds: this means that the wave completes 2000 cycles in 5 seconds. Therefore, we can find the frequency by setting up the following proportion:

\frac{2000}{5}=\frac{x}{1}

where x is the number of cycles made in 1 second. Solving for x,

x=\frac{2000\cdot 1}{5}=400 s^{-1}

Therefore, the frequency of the wave is

f=400 Hz

Learn more about waves and frequency:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

7 0
4 years ago
4. Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is tra
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55*3-40*5=x*95
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4 years ago
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At locations A and B, the electric potential has the values VA=1.51 VVA=1.51 V and VB=5.81 V,VB=5.81 V, respectively. A proton r
Oksana_A [137]

Answer:

<u>For proton:</u>

A. The proton is released from Vb (highest potential)

B. v = 2.9x10⁴ m/s

<u>For electron:</u>

A. The electron is released from Va (lowest potential)

B. v = 1.2x10⁶ m/s    

Explanation:

<u>For a proton we have</u>:

A. To find the origin from which the proton was released we need to remember that in a potential difference, a proton moves from the highest potential to the lowest potential.                

Having that:

Va = 1.51 V and Vb = 5.81 V

We can see that the proton moves from Vb to Va, hence the proton was released from Vb.

B. We now that the work done by an electric field is given by:

W = \Delta Vq    (1)                                        

Where:

q: is the proton's charge = 1.6x10⁻¹⁹ C    

V: is the potential    

Also, the work is equal to:

W = \Delta K = (K_{a} - K_{b}) = \frac{1}{2}mv_{a}^{2} - \frac{1}{2}mv_{b}^{2}     (2)      

Where:

K: is the kinetic energy

m: is the proton's mass = 1.67x10⁻²⁷ kg

v_{a}: is the velocity in the point a

v_{b}: is the velocity in the point b = 0 (starts from rest)

Matching equation (1) with (2) we have:

\Delta Vq = \frac{1}{2}mv_{a}^{2}

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}1.67 \cdot 10^{-27} kg*v_{a}^{2}

v_{a} = 2.9 \cdot 10^{4} m/s

<u>For an electron we have</u>:

A. For an electron we know that it moves from the lowest potential (Va) to the highest potential (Vb), so it is released from Va.

B. The speed is:

\Delta Vq = \frac{1}{2}mv_{b}^{2} - \frac{1}{2}mv_{a}^{2}

Since v_{a} = 0 (starts from rest) and m_{e} = 9.1x10⁻³¹ kg (electron's mass), we have:

(5.81 V - 1.51 V)*1.6 \cdot 10^{-19} C = \frac{1}{2}9.1 \cdot 10^{-31} kg*v_{b}^{2}    

v_{b} = 1.2 \cdot 10^{6} m/s

I hope it helps you!

6 0
4 years ago
Kathy had a heavy metal ball and Rich had a light wooden ball they drop the balls from the same height. they were surprised to d
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Answer: I scored a 9/10

Explanation:

I agree with Kathy gravity must be acting the same on both balls just because one ball is heavier doesn’t mean they will hit different times if the wooden ball was drop closer to the ground then the wooden ball would hit the ground first

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The answer is Ai just took it
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