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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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The  initial velocity of the ball is 55.125 m/s.

<h3>Initial velocity of the ball</h3>

The initial velocity of the ball is calculated as follows;

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At its Ames Research Center, NASA uses its large "20-G" centrifuge to test the effects of very large accelerations ("hypergravit
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This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).

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