Given Information:
Voltage of circuit A = Va = 208 Volts
Current of circuit A = Ia = 40 Amps
Voltage of circuit B = Vb = 120 Volts
Current of circuit B = Ib = 20 Amps
Required Information:
Ratio of power = Pa/Pb = ?
Answer:
Ratio of power = Pa/Pb = 52/15
Explanation:
Power can be calculated using Ohm's law
P = VI
Where V is the voltage and I is the current flowing in the circuit.
The power delivered by circuit A is
Pa = Va*Ia
Pa = 208*40
Pa = 8320 Watts
The power delivered by circuit B is
Pb = Vb*Ib
Pb = 120*20
Pb = 2400 Watts
Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is
Pa/Pb = 8320/2400
Pa/Pb = 52/15
Answer:
oh for real?
Explanation:
The solubility of glucose at 30°C is
125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water at 30°C. Explain your classification, and describe how you could increase the amount of glucose in the solution without adding more glucose.
Answer:
d = 13 miles
Explanation:
Lets say the position of court house is origin in this case
her office is located at 4 miles west and 4 miles south of court house
so here we have coordinate of the office with respect to court house is given as
now the position of her home is located at 1 miles east and 8 miles north of the court house
so the coordinates of her home is given as
now the change in the position is given as the distance between office and home
The electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
<h3>What is electric potential energy?</h3>
Electric potential energy can be defined as the energy needed to move a charge against an electric field.
It is calculated using the formula;
U = Kq1 q2 ÷ r
Where Q = electric potential energy
k = Coulombs constant
q1 and q2 = charges
r = distance of separation
Electric potential energy is inversely proportional to the distance of separation of the charges.
If the distance of the charges changes from 3mm to 6mm, then the electric potential energy of the charges is reduced because it decreases with increase in the distance of the charges.
Therefore, the electric potential energy of the charge is reduced because it decreases with increase in the distance between charges.
Learn more about electric potential energy here:
brainly.com/question/14812976
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I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
