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2 years ago

6

The frequency of light emitted from a source is changed. What visible evidence would indicate this?

2 answers:

Dvinal [7]2 years ago

7 0

Answer: If the light is visible, you would notice a change in the colour of the light. The frequency of a light ray is relationed with the wavelength in the next way λ = v/f.

where λ is the wavelength, v is the speed of the light ( in this case is c) and f is the frequency.

The visible wavelength are in the range of 380 nm to 750 nm

where 380 nm is a blue extreme, and the 750 nm is the "red" extreme.

Then if your frequency decreases, your wavelength increases and the light turns more reddish, if your frequency increases, your wavelength decreases and the light turns more blueish.

Llana [10]2 years ago

5 0

The change of color. Each color has a different wavelength therefore a different frequency.

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A Carnot heat engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a lar

kakasveta [241]

Answer:

$W=4037.36\ J$

Explanation:

Given:

mass of ice melted, $m=3.3\times10^{-2}\ kg$

time taken by the ice to melt, $t=5\ min=300\ s$

latent heat of the ice, $L=3.34\times 10^5\ J$

Now the heat rejected by the Carnot engine:

$Q_R=m.L$

$Q_R=0.033\times 3.34\times 10^5$

$Q_R=11022\ J$

Since we have boiling water as hot reservoir so:

$T_H=373\ K$

The cold reservoir is ice, so:

$T_L=273\ K$

Now the efficiency:

$\eta=1-\frac{T_L}{T_H}$

$\eta=1-\frac{273}{373}$

$\eta=26.81\%$

Now form the law of energy conservation:

Heat supplied:

$Q_S-W=Q_R$

where:

$Q_S=$ heat supplied to the engine

$Q_S-\eta\times Q_S=Q_R$

$Q_S(1-\eta)=Q_R$

$Q_S=\frac{11022}{1-0.2681}$

$Q_S=15059.36\ J$

Now the work done:

$W=Q_S-Q_R$

$W=15059.36-11022$

$W=4037.36\ J$

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2 years ago

Steam enters a turbine at 400K and exhausts at 360K. Calculate the maximum efficiency. =%

kozerog [31]

The maximum efficiency is 10%

Explanation:

The maximum efficiency of a machine is given by:

$\eta=1-\frac{T_C}{T_H}$

where

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the machine in this problem, we have:

$T_H=400K$ is the temperature of the hot reservoir

$T_C=360K$ is the temperature of the cold reservoir

substituting, we find the max efficiency:

$\eta=1-\frac{360}{400}=0.1$

So, the efficiency is 0.10 (10%).

Learn more about temperature:

brainly.com/question/1603430

brainly.com/question/4370740

#LearnwithBrainly

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2 years ago

Determine which factors will increase the reaction rate and which factors will decrease te reaction rate.

Elodia [21]

Answer: increase reaction rate side : increase surface area

Increase temperature

Add a catalyst

Decrease reaction rate side :

Decrease surface area

Decrease temperature

Decrease concentration

Explanation:

Got it right

8 0

3 years ago