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rosijanka [135]

3 years ago

13

How many joules are expended by a 70 kg man climbing up 6m of stairway? How much work is done if the climbing takes place on the

surface of the moon? (Assume that the acceleration due to gravity on the moon’s surface is 1.6 m/s2

1 answer:

GuDViN [60]3 years ago

7 0

Answer:

On the earth surface ,W= 4200 J

On the moon surface ,W= 672 J

Explanation:

Given that

mass of the man ,m = 70 kg

Height of the stairway ,h= 6 m

On the earth surface :

We know that acceleration due to gravity at earth ,g= 10 m/s²

The work done by man is given as

W= m g h

Now by putting the values in the above equation we get

W= 70 x 10 x 6 J

W=4200 J

On the moon surface :

The gravity on the moon ,g= 1.6 m/s²

The work done W

W= 70 x 1.6 x 6 J

W=672 J

On the earth surface ,W= 4200 J

On the moon surface ,W= 672 J

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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

So

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

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

So

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

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

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

So

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

=> $E = 100K$

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

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

So

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

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

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.

4 0

3 years ago

a ball is thrown downward with an initial speed of 7 m/s. the ball's velocity after 3 seconds is ____ m/s (g= -9.8m/s^{2}

Karo-lina-s [1.5K]

The speed downwards is 7 + (3*9.8)
7+ 29.4 = 36.4 m/s

5 0

3 years ago

An 80.0-gram sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its

Sidana [21]

Internal energy, U, is equal to the work done or by the system, plus the heat of the system: <span>ΔU=q+w
</span>in the question they tell you the work done by the system, and the internal energy:
8185 J= -346 J + q work is negative because it was done BY the system.
substitute in: <span>q=m∗Cp∗ΔT</span> and solve for <span>Cp</span><span>.
</span>
-------------------------------------
remember that <span>ΔT=<span>Tf</span>−<span>Ti
</span></span>
so the equation, really, is: <span>q=m∗Cp∗(<span>Tf</span>−<span>Ti</span>)</span><span>
------------------------------------------
</span>
<span>185J=−346J+[m∗Cp∗(<span>Tf</span>−<span>Ti</span>)]

</span>plug in the rest of your values and solve for <span><span>Cp</span></span>

4 0

3 years ago

Benny wants to estimate the mean lifetime of Energizer batteries, with a confidence level of 97%, and with a margin of error not

VikaD [51]

Answer:

143 batteries does Benny need to sample

Explanation:

Given data

confidence level = 97%

error = ±10 hours

standard deviation SD = 55 hours

to find out

how many batteries does Benny need to sample

solution

confidence level is 97%

so a will be 1 - 0.97 = 0.03

the value of Z will be for a 0.03 is 2.17 from standard table

so now we calculate no of sample i.e

no of sample = (Z× SD/ error)²

no of sample = (2.16 × 55 / 10)²

no of sample = 142.44

so 143 batteries does Benny need to sample

7 0

3 years ago

Please help me i need help only way i’ll pass

pickupchik [31]

Answer:

they require a medium

Explanation:

your welcome.............

3 0

3 years ago