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3 years ago

If the Doppler radar is measuring an echo of 41 dBZ, what is the corresponding rainfall rate?

Physics

2 answers:

Rashid [163]3 years ago

3 0

Answer:

R = 0.52 in/h

Explanation:

The rainfall rate (R) can be calculated using the Marshall-Palmer equation:

$R = (\frac{10^{\frac{dBZ}{10}}}{200})^{\frac{5}{8}}$ (1)

The dBZ given is equal to 41, so introducing this valor into the equation (1) we can find the rainfall rate:

$R = (\frac{10^{\frac{41}{10}}}{200})^{\frac{5}{8}} = 13.32 mm/h = 0.52 in/h$

Therefore, the rainfall rate corresponding to an echo of 41 dBZ is 0.52 in/h.

I hope it helps you!

Fantom [35]3 years ago

3 0

Answer:R = 0.52 in/h

Explanation:

The rainfall rate (R) can be calculated using the Marshall-Palmer equation:

(1)

The dBZ given is equal to 41, so introducing this valor into the equation (1) we can find the rainfall rate:

Therefore, the rainfall rate corresponding to an echo of 41 dBZ is 0.52 in/h

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A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

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Answer:

Part a)

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Explanation:

Part a)

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$75 = 10 Log(\frac{I}{10^{-12}})$

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Part b)

Intensity of sound wave is given as

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here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

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A 22.0 kg bucket of concrete is connected over a very light frictionless pulley to a 375 N box on the roof of a building as show

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Answer:

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Explanation:

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a = 3.58 [m/s^2]

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