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Airida [17]
3 years ago
11

A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

en that Stefan Boltzmann constant = 5.67 × 10^8W^-8m^-8K^-4
Physics
1 answer:
cestrela7 [59]3 years ago
3 0
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

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3 years ago
A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W
lianna [129]

Answer:

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Explanation:

We have given mass of the block m = 0.5 kg

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Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

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\mu =0.4081

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Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso
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Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

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Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

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There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

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Answer:

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If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.

Also work depends on the sum of the velocities

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W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2

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Answer:

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