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slamgirl [31]
3 years ago
15

A classroom that normally contains 40 people is to be airconditioned with window air-conditioning units of 5-kW cooling capacity

. A person at rest may be assumed to dissipate heat at a rate of about 360 kJ/h. There are 10 lightbulbs in the room, each with a rating of 100 W. The rate of heat transfer to the classroom through the walls and the windows is estimated to be 15,000 kJ/h. If the room air is to be maintained at a constant temperature of 21°C, determine the number of window air-conditioning units required.
Physics
1 answer:
oksano4ka [1.4K]3 years ago
7 0

Answer:

About  2 units.

Explanation:

We assume that there is no heat dissipating instrument in the room

Total cooling load of the room is defined from the given below equation

Q_{cooling}=Q_{light}+Q_{people}+Q_{heat gain}

where

Q_{light}= 10*100 W =1 KW

Q_{people} = 40*360 KJ/h= 4 KW

Q_{heat gain}=15000KJ/h= 4.17 KW

Q_{cooling}= 1+4+4.17=9.17 KW

the number of air conditioner unit is =\frac{9.17}{5}=1.83

 which is approximately 2 units

   

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Answer:

I=1.5\times10^{-28}W/m^2

Explanation:

The intensity is related to the power and surface area by I=\frac{P}{A}=\frac{P}{4\pi r^2}. We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m

So we have:

I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2

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If the kinetic and potential energy in a system are equal, then the potential energy increases. What happens as a result?
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Which of the following sentences shows the correct use of a transition word? Even though I was sick, I went to the doctor. My sh
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A man's higher initial acceleration means that a man can outrun a horse over a very short race. A simple - but plausible - model
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Answer:

 t_man = 10.16 s,   t_horse = 10.73 s,  the winner is the man

Explanation:

To solve this problem we are going to find the time of each one separately.

Man we look for distance and time during acceleration

         x₁ =  v₀ t₁ + ½ a₁ t₁²

as it comes out of rest its initial velocity is zero

        x₁ = ½ a₁ t₁²

        x₁ = ½ 6.0 1.8²

        x₁ = 9.72 m

at this point its speed is

        v₁ = v₀ + a t

        v₁ = 0 + 6  1.8

        v₁ = 10.8 m / s

From here on it continues at constant speed, the distance that the man needs to travel from the point where the man leaves at 100m is

        x₂ = 100 - x₁

        x₂ = 100- 9.72

        x₂ = 90.28 m

the time for this part is

        v₁ = x₂ / t₂

         t₂ = x₂ / v₂

         t₂ = 90.28 / 10.8

         t₂ = 8.36 s

the total time for the man is

        t_man = t₁ + t₂

        t _man = 1.8 + 8.36

        t_man = 10.16 s

We repeat the calculation for the horse

distance traveled during the acceleration period

         x₃ = v₀ t + ½ a₂ t₃²

as part of rest its initial velocity is zero

        x₃ = ½  a₂ t₃²

        x₃ = ½  5.0  4.8²

        x₃ = 57.6 m

the velocity at this point is

         v₃ = v₀ + a₂ t₃

         v₃ = 0 + 5  4.8

         v₃ = 24 m / s

the rest of the route is at constant speed, the remaining distance

         x₄ = 200 - x₃

         x₄ = 200 - 57.6

         x₄ = 142.4 m

the time to go through it is

         t₄ = x₄ / v₃

         t₄ = 142.4 / 24

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the total time for the horse is

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         t_horse = 4.8 + 5.93

         t_horse = 10.73 s

when we compare the times we see that the man arrives a little before the horse, the winner is the man

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