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3 years ago

For 1983 through 1989, the per capita consumption

Physics

1 answer:

Tresset [83]3 years ago

8 0

Answer:

y = 43.55 + 2.15t

Explanation:

We were told that in 1983, the per capita consumption was 37.1 pounds, and in 1989 it was 50 pounds.

If we assume t = 0 corresponds to year 1980. Then, for 1983 it will be t = 3 and for 1989,it will be t = 9.

Thus, expressing the information as ordered pairs, we have; (3,37. 1) and (9,50).

Let us now find slope of the linear function:

m1 = (y2 - y1)/(t2 - t1)

m1 = (50 - 37.1)/(9 - 3)

m1 = 2.15

So, we can find the linear equation as;

y - 37.1 = 2.15(t - 3)

y = 37.1 + 2.15t - 6.45

y = 43.55 + 2.15t

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When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

Andre45 [30]

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A$

(b) Again the power is given by :

$P=\dfrac{V^2}{R}$

R is resistance of the dryer

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega$

(c) The magnetic field produced by the dryer at the user's hand is given by :

$B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T$

Hence, this is the required solution.

4 0

2 years ago

The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between

erma4kov [3.2K]

Answer:

Considering first question

Generally the coefficient of performance of the air condition is mathematically represented as

$COP = \frac{T_i}{T_o - T_i}$

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.