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denis-greek [22]

3 years ago

6

If a star is moving away from you at a constant speed, how do the wavelengths of the absorption lines change as the star gets fa

rther and farther?

1 answer:

Minchanka [31]3 years ago

8 0

Answer:

they stay shifted the same amount to the red

Explanation:

Redshift is given by

$z=\dfrac{\lambda_o-\lambda_e}{\lambda_e}$

Where,

$\lambda_o$ = Wavelength observed

$\lambda_e$ = Wavelength emitted

Also

Transverse redshift is given by

$1+z=\dfrac{1}{\sqrt{1-v^2/c^2}}$

v = Velocity of object

c = Speed of light = $3\times 10^8\ m/s$

So, if the velocity is constant the redshift remains the same

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A constant 20 N force is applied to a 7 kg box to push it along the ground. How

Elis [28]

Answer:

40 joules

Explanation:

Work Done=Force*Distance

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3 years ago

Describe the motion of a swing that requires 6 seconds to complete one cycle. What is its period and the frequency? Round to the

shutvik [7]

Period = 6 seconds and $frequency = 0.167Hz$ .

<u>Explanation:</u>

We have , the motion of a swing that requires 6 seconds to complete one cycle. Period is the amount of time needed to complete one oscillation . And in question it's given that 6 seconds is needed to complete one cycle. Hence ,Period of the motion of a swing is 6 seconds . Frequency is the number of vibrations produced per second and is calculated with the formula of $\frac{1}{t}$ . SI unit of frequency is Hertz or Hz. We know that time period is 6 seconds so frequency = $\frac{1}{t}$

⇒ $frequency = \frac{1}{time}$

⇒ $frequency = \frac{1}{6}$

⇒ $frequency = 0.167Hz$

Therefore , Period = 6 seconds and $frequency = 0.167Hz$ .

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3 years ago

What happens to beaches over time?

Rama09 [41]

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3 years ago

Read 2 more answers

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

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