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3 years ago

Will anyone please help me solve this physics Problem?

Physics

1 answer:

ArbitrLikvidat [17]3 years ago

5 0

Answer:

I. Clockwise moment = 150 Nm

II. Anticlockwise moment = 400 Nm

III. 200 N

III. The anticlockwise moment.

Explanation:

From the question given above, we can obtain the answers to the questions as follow:

I. Determination of the clockwise moment.

Distance (d) = 1.5 m

Force (F) = 100 N

Clockwise Moment =?

Moment = Force × distance.

Clockwise moment = 100 × 1.5

Clockwise moment = 150 Nm

II. Determination of the anticlockwise moment.

Distance (d) = 2 m

Force (F) = 200 N

Anticlockwise Moment =?

Moment = Force × distance.

Anticlockwise moment = 200 × 2

Anticlockwise moment = 400 Nm

III. Determination of who will turn the sea saw.

Clockwise moment = 150 Nm

Anticlockwise moment = 400 Nm

Since the anticlockwise moment is greater than the clockwise moment, it therefore means that the anticlockwise moment will turn the sea saw.

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Here $T_i$ is the inside temperature

while $T_o$ is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes $\frac{T_i}{T_o - T_i}$ heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

$Q \ \alpha \ (T_o - T_i)$

=> $Q= k (T_o - T_i)$

Here k is the constant of proportionality

since 1 unit of electricity removes $\frac{T_i}{T_o - T_i}$ amount of heat

E unit of electricity will remove $Q= k (T_o - T_i)$

$E = \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }$

=> $E = \frac{k}{T_i} (T_o - T_i)^2$

given that $\frac{k}{T_i}$ is constant

=> $E \ \alpha \ (T_o - T_i)^2$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

Considering the second question

Assuming that $T_i = 30 ^oC$

and $T_o = 40 ^oC$

Hence

$E = K (T_o - T_i)^2$

Here K stand for a constant

$E = K (40 - 30)^2$

=> $E = 100K$

Now if the $T_i = 20 ^oC$

Then

$E = K (40 - 20)^2$

=> $E = 400 \ K$

So from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low is much higher than the electricity required when the inside temperature is higher

Considering the third question

Now in the case where the heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

$Q = k (T_o - T_i )^{\frac{1}{2} }$

$E = \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }$

=> $E = \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }$

Assuming $\frac{k}{T_i}$ is a constant

Then

$E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }$

From this above equation we see that the electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root of the cube of the temperature difference.

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3 years ago

Will anyone please help me solve this physics Problem?​

Will anyone please help me solve this physics Problem?