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Xelga [282]
3 years ago
12

Which sets of data show a wave with the shortest wavelength? A. Speed=100 million m/s and frequency = 50 million Hz. B. Speed=15

0 million m/s and frequency=1500 Hz C.Speed=300 million m/s and frequency= 100 Hz
Physics
1 answer:
SashulF [63]3 years ago
3 0
The answer is <span>A. Speed=100 million m/s and frequency = 50 million Hz.</span>

Let's calculate for each choice the wavelength using the equation:
v = f × λ   ⇒ λ = v ÷ f<span>

where:
v - the speed,
f - the frequency,
</span>λ - the wavelength.

A:
v = 100 000 000 m/s
f = 50 000 000 Hz = 50 000 000 1/s (Since f = 1/T, so units are Hz = 1/s)
⇒ λ = 100 000 000 ÷ 50 000 000 = 2 m

B:
v = 150 000 000 m/s
f = 1 500 Hz = 1 500 1/s
⇒ λ = 150 000 000 m/s ÷ 1 500 = 100 000 m

B:
v = 300 000 000 m/s
f = 100 Hz = 100 1/s
⇒ λ = 300 000 000 m/s ÷ 100 = 3 000 000 m

According to these calculations, the shortest wavelength is needed for choice A.
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r = 4.24x10⁴ km.  

     

Explanation:

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\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

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