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.
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
Since you are looking for the speed, you need to rearrange the formula which is f = speed / wavelength. That should give you speed = f (wavelength.) All you need to do next is to substitute the value to the following equation. speed = 250 Hz (6.0m) that should leave you with 1500 m/s which is very fast.
Answer: Say the big boy is 40kg and the small boy is 20kg... ... ...then by the rule that moments around a fulcrum or pivot should be equal for equilibrium, the distance of big boy from pivot=half of dist of small boy from pivot
Thicker wure enables higher AMP's, assuming your supply is capable, so indirectly thicker wire may help. Thicker wire takes up more space. So does more turns.
