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Otrada [13]
3 years ago
13

The resource speaks of the atmosphere as if it were living, in the sense that as the earth's surface is heated by the light from

the sun, it is heated unequally and as a result of this unequal heating of the surface, air and water begin to move around in "an effort to accomplish something."
Physics
1 answer:
Shtirlitz [24]3 years ago
6 0

Explanation:

The uneven heating causes temperature differences, which in turn cause air currents (wind) to develop, which then move heat from where there is more heat (higher temperatures) to where there is less heat (lower temperatures). The atmosphere thus becomes a giant "heat engine", continuously driven by the sun.

Wind is the result of pressure changes in the atmosphere due to temperature.

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A sliver cylindrical rod has a length of 0.5 m and a radius of 0.4 m, find the density of the rod if it's mass is 2.640 kg
Serga [27]

Answer:

Density=10.50kg/m³

Explanation:

Solution,\\Height(h)=0.5m\\Radius(r)=0.4m\\Now, \\Volume=\pi r^{2} h\\\\Volume=\pi *(0.4m)^{2} *0.5m\\\\Volume=0.251327m^{3} \\\\Again,\\\\Density=\frac{mass}{volume} \\\\Density=\frac{2.640kg}{0.251327} \\\\Density=10.50kg/m^{3}

7 0
2 years ago
A(n) _______ studies physical components and characteristics of celestial objects.
inna [77]

Answer:

Astronomers?

Explanation:

5 0
2 years ago
What would happen if your stomach wasint working properly
adelina 88 [10]
Your stomach, as in JUST your stomach?
Well the role of your stomach is to break down large clumps of food. Without that it would be very hard to impossible to digest food.<span />
7 0
3 years ago
Um móvel realiza movimento retilíneo uniformemente
Alona [7]
B is the correct answer hope that helped
8 0
3 years ago
You then measure Polly's internal temperature to be 13oC, which is quite a drop from the normal human body temperature of 37oC.
ankoles [38]

Answer:

The specific heat is 3.47222 J/kg°C.

Explanation:

Given that,

Temperature = 13°C

Temperature = 37°C

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We need to calculate the specific heat

Using formula of energy

Q= mc\Delta T

c =\dfrac{Q}{m\Delta T}

Put the value into the formula

c=\dfrac{5000}{60\times(37-13)}

c=3.47222\ J/kg^{\circ}C

Hence, The specific heat is 3.47222 J/kg°C.

5 0
3 years ago
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