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trasher [3.6K]
3 years ago
13

What is the temperature of a 3.72 mm cube (e=0.288) that radiates 56.6 W?

Physics
1 answer:
blsea [12.9K]3 years ago
7 0

Answer:

The temperature is 2541.799 K

Explanation:

The formula for black body radiation is given by the relation;

Q = eσAT⁴

Where:

Q = Rate of heat transfer 56.6

σ = Stefan-Boltzman constant = 5.67 × 10⁻⁸ W/(m²·k⁴)

A = Surface area of the cube = 6×(3.72 mm)² = 8.3 × 10⁻⁵ m²

e = emissivity = 0.288

T = Temperature

Therefore, we have;

T⁴ = Q/(e×σ×A) = 56.6/(5.67 × 10⁻⁸ × 8.3 × 10⁻⁵ × 0.288) = 4.174 × 10¹⁴ K⁴

T  =  2541.799 K

The temperature = 2541.799 K.

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En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)
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The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
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In a ballistics test, a 52g bullet hits a sand bag and stops after moving 1.34 m. If the initial bullat
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Answer:

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Explanation:

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after travelling into the sand bag by distance d = 1.34 m it comes to rest

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v_f = 0

now we can use kinematics top find the acceleration of the bullet

v_f^2 - v_i^2 = 2 a d

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3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN
timurjin [86]

Answer:

f_{o} = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency (f_{o}).

The For a vibrating string, the fundamental frequency (f_{o}) can be determined by:

f_{o} = \frac{v}{2L}

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f_{o} = 391.6667 Hz

The fundamental frequency of the string is 391.67 Hz.

3 0
2 years ago
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