Answer:
I = 27.65A < 40.59°
PowerFactor = 0.76
Explanation:
Current on the heating load is:
I1 = 30KW / 4KV = 7.5A < 0°
Current on the inductive load:
I2 = (150KVA*0.6) /4KV = 22.5A with an angle of acos(0.6)=53.1°
The sum of both currents is:
It = I1 + I2 = 7.5<0° + 22.5<53.1° = 27.65<40.59°
Now, the power factor will be:
pf = cos (40.59°) = 0.76
Answer:
Explanation:
Samantha walks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the horizontal part is for me what is the magnitude of displacement
Displacement is given by the straight line distance between P and Q. Displacement will be length of straight line joining P and Q
a semi circle with a radius of 2 m
Length of this straight line=4+diameter
=4+(2*2)
=8 m
Answer:
Q = 5267J
Explanation:
Specific heat capacity of copper (S) = 0.377 J/g·°C.
Q = MSΔT
ΔT = T2 - T1
ΔT=49.8 - 22.3 = 27.5C
Q = change in energy = ?
M = mass of substance =508g
Q = (508g) * (0.377 J/g·°C) * (27.5C)
Q= 5266.69J
Approximately, Q = 5267J
Answer:
by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2