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brilliants [131]
2 years ago
8

If the surface temperature of that person's skin is 30∘C (that's a little lower than healthy internal body temperature becaus

e your skin is usually a little colder than your insides), what is the total power that person will radiate? For now, assume the person is a perfect blackbody (so ϵ=1).
Physics
1 answer:
anastassius [24]2 years ago
5 0

Answer:

E=477.92\ W.m^{-2}

Explanation:

Given that:

Absolute temperature of the body, T=273+30=303\ K

  • emissivity of the body, \epsilon=1

<u>Using Stefan Boltzmann Law of thermal radiation:</u>

E=\epsilon. \sigma.T^4

where:

\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}   (Stefan Boltzmann constant)

Now putting the respective values:

E=1\times 5.67\times 10^{-8}\times 303^4

E=477.92\ W.m^{-2}

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The motion of an object looks different to observers in different
Lubov Fominskaja [6]

Positions. Happy to help! Please mark as Brainliest!

7 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
The greatest number of thunderstorms occur in the?
Phantasy [73]

Answer:

On the other hand, Florida's Gulf Coast experiences the greatest number of thunderstorms out of any U.S. location. These types of storms occur on average 130 days per year in Florida.

6 0
2 years ago
Read 2 more answers
A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv
zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

    A = 1pu    B = 1.685∠60.8°Ω    C = 0 S    D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 KV_{LL}

(c) The sending-end voltage for 0.9 leading power factor is 33.40 KV_{LL}

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0
2 years ago
Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is
vaieri [72.5K]
1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle)  to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.
 
angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos 

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. :  sin(40) x '?' = 4905
-divide both sides by sin(40):  '?' = 4905/ sin(40)
--> Solve.

8 0
2 years ago
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