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2 years ago

What is the answer to this question

Physics

1 answer:

deff fn [24]2 years ago

3 0

Answer:75%effecient

Explanation:

6000-1500=4500

6000=100%

4500=?

(4500×100)÷6000

=75%

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Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.

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2 years ago

Function do lappet which faced vultures serve in a temperate grassland

vfiekz [6]

Um this doesn't make since to me since you did not clearly state your awnser

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2 years ago

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In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$