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Sveta_85 [38]
3 years ago
5

The energy released from condensation in thunderstorms can be very large. Calculate the energy (in J) released into the atmosphe

re for a small storm of radius 2.1 km, assuming that 2.3 cm of rain is precipitated uniformly over this area. (Assume that the rain is precipitated over a circular area and that the density of water is 1,000 kg/m3.)

Physics
2 answers:
Wewaii [24]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

makkiz [27]3 years ago
7 0

Answer:

7.19 * 10^14J

Explanation:

Given that

Density of water Pwater= 1000kg/m3

R=2.1km = 2.1*10^3m

H= 2.3cm. = 2.3*10^-2m

Lv water= 2256 * 10^3J/kg

First, mass of water need to be calculated, using an imaginary cylinder

Density= Mass /Volume

Mass= Density* Volume

Volume of a cylinder= πR2h

Therefore mass= PπR2H

= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)

= 3.18 *10^8

Heat Released Qv = mLV

= 3.18*10^8 * 2236*10^3

= 7.19 * 10^14J

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Answer:

Explanation:

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r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

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Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

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Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

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B_{max}=2.901\times10^{-13}\ T

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