This is an obstetrician. Also called for short OB/GYN. They are doctors that specialize in delivering babies and the female reproductive system.
Answer:
I am calculating the total area of a solar panel for a particular load demand by ... designing according to energy demand for a day then how will it affect total solar area? ... Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency ... would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.
Explanation:
Answer:
So to increase current of the circuit what you can do is :
1. Use conductor of low resistivity, ¶.
2. Use conductor of small length.
3. Use thick wire.
4. Decrease the temperature of the circuit.
5. If operating temprature is high than use semiconductor, because it have negative temprature coefficient.
6. Minimise the circuit losses.
Answer:
Explanation:
The Carnot cycle is a special case of a thermodynamic cycle that produces an ideal gas and consists of two isothermal processes and two adiabatic processes. This cycle is a theoretical solution given by Sadi Karnot to refine heat engines for their efficient use.
The formula for the coefficient of efficiency is:
η = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁
Where Q₁ is is the amount of heat of the heater supplied to the working body and Q₂ is the amount of heat that the working body transfers to the refrigerator according to this T₁ is the temperature of the heater T₂ is the temperature of the refrigerator.
This formula provides a theoretical limit for the maximum value of the coefficient of efficiency of heat engines.
God is with you!!!
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90