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3 years ago

6

How is energy transfer related to the ozone layer and its temperature?

1 answer:

Leno4ka [110]3 years ago

6 0

Energy is transferred<span> between the earth's surface and the atmosphere via conduction, convection, and radiation. Conduction is the process by which heat</span>energy is<span> transmitted through contact with neighboring molecules. ... During the day, solar radiation heats the ground, which heats the air next to it by conduction.</span>

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Which statement about atmospheric pressure near Earth’s surface is true

notka56 [123]

What are the statements?? I need to know the statements so i can help you.
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3 years ago

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You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler

scZoUnD [109]

Answer:

q=1.4*10^{-9}C

Explanation:

Given data:

charge on ruler = -14μC

Mass of tissue is 5 g

To Know the minimum charge, equate electrostatic force to weight

we have F = W

so $\frac{KQq}{r^2} =mg$

putting all value in equation,

$=\frac{9*10^9*(14*10^{-6})*q}{0.06^2} = 5* 10^{-3}*9.8$

solving for q

$q =\frac{5* 10^{-3}*9.8 *0.06^2}{9*10^9*(14*10^{-6})}$

or q=1.4*10^{-9}C

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3 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

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3 years ago

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

$5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s$

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

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3 years ago

What is liquidity preferences?

ZanzabumX [31]

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