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Vadim26 [7]
3 years ago
11

If a burning log is a black object with a surface area of 0.25 m2 and a temperature of 800 8c, how much power does it emit as th

ermal radiation?
Physics
1 answer:
tresset_1 [31]3 years ago
8 0

The correct answer of this question: 18789.79 watt.

EXPLANATION:

As per the question, the surface area of the log is given as A = 0.25\ m^2

The temperature of the of the surface of the log t = 800 ^0C

We know that t^0C\ =(273+t)\ K

                      ⇒ 800 degree celsius= 800+273 K

                                                           = 1073 K.

Here, K stands for kelvin scale.

We are asked to calculate the thermal power radiation .

The power is defined as the amount of energy released per unit time.

Hence, power is calculated as-

                                Power P = \sigma AT^4

                                               = 5.67\times 10^{-8} \times 0.25\times (1073)^4\ watt

                                               = 18789.79 watt.      [ans]

Here, \sigma is the Stefan-Boltzmann constant and T is the absolute temperature of the log.

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D. 51 N. The minimum applied force that will cause the television slide is 51 N.

In order to solve this problem we have to use the force of static friction equation Fs = μs*n, where μs is the coefficient of static friction, and n is the normal force m*g.

With μs = 0.35, and n = 15kg*9.8m/s² = 147 N

Fs = (0.35)(147 N)

Fs = 51.45 N

Fs ≅ 51 N

3 0
3 years ago
Read 2 more answers
A thin rod of length 0.79 m and mass 130 g is suspended freely from one end. It is pulled to one side and then allowed to swing
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Answer:

a) 0.3965 j

b) 0.3112 m

Explanation:

The picture attached explains it all. Thank you

3 0
3 years ago
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o
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The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

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8 0
2 years ago
The half-life of cobalt-60 is 5.26 years. If 50 g are left after 15.8 years, how many
PtichkaEL [24]

Answer:

400 g

Explanation:

The computation of the number of grams in the original sample is shown below:

Given that

half-life = 5.26 years

total time of decay = 15.8 years

final amount = 50.0 g

Now based on the above information  

number of half-lives past is

=  15.8 ÷ 5.26

= 3 half-lives

Now

3 half-lives = 1 ÷ 8 remains = 50.0 g

So, the number of grams would be

= 50.0 g × 8

= 400 g

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Snow tends to sublime outside during cold, clear winter days when the atmospheric pressure is low. During this proces
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