Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the 
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Answer:
Explanation:
1 )
We shall apply conservation of momentum law to solve the problem.
mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .
.5 x .2 = (1.5 + .5)V
V = .05 m /s
2 ) We shall use formula for velocity of object after elastic collision as follows
v₁ = 
m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.
Putting the values
= 
= - .66 m /s
Since the sign is negative so it will be in opposite direction .
The answer is true because A current carrying wire is surrounded by magnetic field
Answer:
option B
−1.92 m/s2
Explanation:
Given in the question,
time took by truck to slow down = 3.56 sec
initial speed of truck = 112 km/h
final speed of truck = 87.4 km/h
1 km/h = 0.277778 m/s
112 = 31.1 m/s
87.4 = 24.28 m/s
Formula use to calculate the acceleration
v - u = at
where v is final speed
u is initial speed
a is acceleration
t is time
plug values in the equation
24.28 - 31.1 = a(3.56)
-6.8 = a(3.56)
a = -6.8 / 3.56
a = -1.9 m/s²
