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leslieperez67
03.02.2021 •
Physics
BRAINLIEST WILL BE GIVEN After communicating the information you requested above, he pushes the sled a total distance of 50 m and exerts a horizontal force of 800 N. During the sled push he pushes the sled 25 m, turns around and pushes it 25 m back to where he started. He argues that the work done is 40000 J, but you say the total work is actually 0 J. Why are you correct and he is wrong?
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Ответ:
W=F*D
40,000=800*50
With this you are wrong and he is right because work is not like displacement where if you go 10 ft to the right then 10 ft to the left you have traveled gone 0 ft from original spot. because your distance was 50 m and you exerted 800 N you multiply that together and come out with 40,000 so HE IS RIGHT YOU ARE WRONG.
PLZ mark as brainliest
Ответ:
The emissivity of the radiation shield is
Explanation:
From the question we are told that
The temperature of the first parallel plate is![T_1 = 650K](/tpl/images/0640/9277/f4ab3.png)
The temperature of the second parallel plate is![T_2 = 400K](/tpl/images/0640/9277/1556c.png)
The emissivity of first plate is![e_1 = 0.6](/tpl/images/0640/9277/aa656.png)
The emissivity of first plate is![e_2 = 0.9](/tpl/images/0640/9277/8001a.png)
Generally the total radiation heat that is been transferred without the shield is mathematically represented as
Where
is the Stefan-Boltzmann constant which has a value ![5.67 *10^{-8} \ W \cdot m^{-2} \cdot K^{-1}](/tpl/images/0640/9277/499c0.png)
Substituting values
From the question we are told the that using the radiation shield would reduce the radiation heat transfer by 15%
So the new heat transfer is
So![Q_2 = \frac{15}{100} * 4876.8](/tpl/images/0640/9277/45745.png)
Now this new radiation heat transfer can be mathematically represented as
Where
the emissivity of the radiation shield and n is the number of radiation shield
Substituting values