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umka2103 [35]
3 years ago
6

The outer surface of a grill hood is at 87 °c and the emissivity is 0.93. the heat transfer coefficient for convection between t

he hood and surroundings at 27 °c is 10 w/m2 /k. determine the net rate of heat transfer between the grill hood and the surroundings by convection and radiation, in kw per m2 of surface area.
Physics
1 answer:
Tju [1.3M]3 years ago
4 0
The net flux, or rate of heat transfer per area, for convection and radiation are the following:

Convection:
Q/A = hΔT
Radiation:
Q/A = [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
where
h is the heat transfer coefficient: 10 W/m²·K or 0.01 kW/m²·K
ε is the emissivity: 0.93
σ is the Stephen-Boltzmann constant: 5.67×10⁻¹¹ kW/m²·K⁴
A₁ = A₂ because they have the same surface area
T is absolute temperature in Kelvin: K = °C + 273

Thus,

Q/A = hΔT + [εσ(T₁⁴ - T₂⁴)]/{1 - [A₁/A₂(1-ε)²]}
Q/A = (0.01 kW/m²·K)(360 K - 300 K) + [0.93*5.67×10⁻¹¹ kW/m²·K⁴*(360⁴ - 300⁴)]/{1 - [1*(1-0.93)²]}
Q/A =  0.6 + 0.461
Q/A = 1.061 kW/m²
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Answer:

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Answer:

λ = 437 m.

Explanation:

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  • Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

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2 years ago
A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
Colt1911 [192]

Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

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ElenaW [278]

Answer:

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Explanation:

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We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.

The velocity just before the ball hits the ground can be found by the equation:

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Let's substitute h = 10 m and h = 40 m into this formula.

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We can see that the velocity increases by a factor of 4 (10 m → 40 m).

Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.

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Answer:

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