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fiasKO [112]
3 years ago
9

What is the period of a wave if the wavelength is 100 m and the speed is 200 m/s?

Physics
2 answers:
Stells [14]3 years ago
8 0
This is simple question and easy to understand.
so first you use a formula to calculate frequency of the wave.
v _ stands for velocity a.k.a speed
f_ stands for frequency which you have to find.
(pronounced lemda)_ stands for wavelength.
Then you gonna place all your values according to you have in the formula as shown then solve for f.
Finally, you use the time formula
where
T_ is period a.k.a time which equal to the reciprocal of frequency or equal to 1/f.
Simply input your values according to the formula and you your answer as 0.5s

dsp733 years ago
5 0

(C)   0.5 would be your ansewr, i took the quiz

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Due to the shape of the lens , parallel rays will be deviated
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3 years ago
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An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

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3 years ago
As temperature increase what energy increases
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4 years ago
How much energy is required to raise the temperature of 5g of air by 10°C?
Alex777 [14]
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3 years ago
A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I
scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens F_1=20cm

Focal length of the diverging lens F_2=30cm

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}

\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

M=\frac{F}{F-d}=\frac{60}{60-40}=3

So magnification will be equal to 3

3 0
3 years ago
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