The answer to your question is,
A scientific law.
-Mabel <3
Answer:
about 2.7liters for women and 3.7liters for men
Explanation:
Answer:
There isnt enough in your question to answer the question bro, like we need a picture or something bro.
Explanation:
Answer:
If I double the current in the inductor, the new total energy will become 4E (option f).
Explanation:
The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:
where E is Energy [J], L is Inductance [H] and I is Current [A].
If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:
Then:
<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
brainly.com/question/14548124
