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
This is a problem of conservation of momentum
Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s
A) man throws the rock forward
=>
rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2
=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s
B) man throws the rock backward
this changes the sign of the velocity, v2 = -14.5 m/s
46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2
v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s
Answer:
The speed of q₂ is 
Explanation:
Given that,
Distance = 0.4 m apart
Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.
We need to calculate the speed of q₂
Using conservation of energy
Put the value into the formula
Hence, The speed of q₂ is
Angular velocity = (75x2pie)/60
=2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
v= 0.5 x 2.5pie
v=3.93 ms^-1
tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
F= (0.2 x 3.93^2)/0.5
F=6.18 N
(sorry if wrong)
Higher voltage means faster rotation. Changing the voltage polarity changes the direction of the motor.
