The horizontal displacement of a projectile launched at an angle
Answer:
DOUBLE CHECK BECUASE IM ONLY 68.030303039999999% SURE!!!
(ANSWER IS HERE) ( D) It lacked practical examples in supporting theory
Know it's not B becuase there was no scientific community back then.
Know it's not C becuase it actully had lots of evidence.
But I'm not sure about A
Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs
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.