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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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An airplane is flying in a horizontal circle at a speed of 480 km/h (). If its wings are tilted at angle =40° to the horizontal
german

Answer:

R = 2162 m

Explanation:

When wings of the airplane makes an angle of 40 degree with the horizontal so here we can say that force due to air is having two components

F_y = mg

F_x = \frac{mv^2}{R}

now we know that

F_y = F cos40

F_x = F sin40

also we know that

v = 480 km/h

v = 133.3 m/s

now plug in all data in above equations

tan 40 = \frac{v^2}{Rg}

R = \frac{v^2}{g tan40}

R = \frac{133.3^2}{9.8 tan40}

R = 2162 m

6 0
3 years ago
A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

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3 years ago
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