gfx.wad
No. If the load requires 168 W, then an 84 W inverter is not going to maintain the load.
Yes, a W-8X18 will carry the weight of a basement load wall 2X6 over a 200 feet span.
The strength, S, of the beam is Mc/I where M = max moment to fail = PL/4 for load concentrated in the middle of the beam or WL/8 for uniformly distributed load. Here P is the concentrated load, W = distributed load, c = distance to outer fiber from neutral axis and I the area moment of inertia of the beam. L = length Solving for load maximum, P = 4IS/Lc for concentrated center load W = 8IS/Lc for distributed load
The answer is D. Why is it 1000W? How did you get that answer?
The equation you are looking for is I = W/E. Amps = Watts/Volts.
No. If the load requires 168 W, then an 84 W inverter is not going to maintain the load.
It depends on the load.
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Example: Attach a rope to the ceiling and pass it through a pulley wheel attached to a load W. Then pass the rope through a pulley wheel attached to the ceiling. Bring it down and pull on it. There are 2 ropes lifting the weight W so the tension in the rope is W/2. You are pulling on the rope so you feel a pull of W/2. So you have saved one half. Every centimetre that the load rises, the lengths of rope between the load and ceiling each reduce by 1cm so you have to pull on your end of rope through a distance 2cm. So the work done is W/2 x 2 which is the same as for lifting the load which is W x 1. So you don't save energy, only force. If you put extra pulleys on the roof and ceiling and take the rope round and round you can reduce the force to W/3, or W/4 or W/5 etc.
Yes, a W-8X18 will carry the weight of a basement load wall 2X6 over a 200 feet span.
Michael W. Reid has written: 'Ratebasing of utility conservation and load management programs'
Apparent power is VA. Real power is W reactive power is VAR. Under an inductive+resistive load the VA is higher than W
The strength, S, of the beam is Mc/I where M = max moment to fail = PL/4 for load concentrated in the middle of the beam or WL/8 for uniformly distributed load. Here P is the concentrated load, W = distributed load, c = distance to outer fiber from neutral axis and I the area moment of inertia of the beam. L = length Solving for load maximum, P = 4IS/Lc for concentrated center load W = 8IS/Lc for distributed load
w do u get gplayrt off so i can load adtrader
Depends on the load and the powder used.
w(l^2)/8 w = 38N l = 5m
The answer is D. Why is it 1000W? How did you get that answer?