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3 years ago

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A person is in a room whose walls are maintained at a temperature of 17 °C. The temperature of the person’s skin is 32 °C. The s

urface area of the person’s skin is 1.4 m2. Determine the netheat transfer rate leaving the person.

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

sned me short notes of Physics of all branches (post graduation level) thanks

+923466867221

Explanation:

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Certain gases in the atmosphere trap heat on Earth.

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The answer is <span>A.)the greenhouse effect
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Two 1-kg objects, C and D, increase in temperature by the same amount, but the

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The object D is made up of material Lead. The correct option is D.

<h3>What is specific heat?</h3>

The specific heat is the amount of heat required to change the temperature by 1°C. It is denoted by C.

Two 1-kg objects, C and D, increase in temperature by the same amount, but the thermal energy transfer of object C is greater than the thermal energy transfer of object D. The object C has a specific heat of 235 J/kg-K.

Q = m C ΔT

Qc > Qd

The energy transfer is proportional to specific heat.

Specific heat of D must be less. The possible material with specific heat less than the given value is for Lead material.

Thus, the correct option is D.

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1 year ago

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws

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Explanation:

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2 years ago

17 copper wires of length l and diameter d are connected in parallel to form a single composite conductor of resistance R. What

Lubov Fominskaja [6]

Answer:

$\frac{D}{d} = 4.12$

Explanation:

As we know that resistance of one copper wire is given as

$r = \rho \frac{L}{a}$

here we know that

$a = \pi (\frac{d}{2})^2$

now we have

$r = \rho \frac{L}{\pi (\frac{d^2}{4})}$

$r = \rho \frac{4L}{\pi d^2}$

now we know that such 17 resistors are connected in parallel so we have

$R = \frac{r}{17}$

$R = \rho \frac{4L}{17 \pi d^2}$

Now if a single copper wire has same resistance then its diameter is D and it is given as

$R = \rho \frac{4L}{\pi D^2}$

now from above two equations we have

$\rho \frac{4L}{\pi D^2} = \rho \frac{4L}{17 \pi d^2}$

$D^2 = 17 d^2$

now we have

$\frac{D}{d} = 4.12$

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