Answers are:
(1) KE = 1 kg m^2/s^2
(2) KE = 2 kg m^2/s^2
(3) KE = 3 kg m^2/s^2
(4) KE = 4 kg m^2/s^2
Explanation:
(1) Given mass = 0.125 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.125 * (16)
KE = 1 kg m^2/s^2
(2) Given mass = 0.250 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.250 * (16)
KE = 2 kg m^2/s^2
(3) Given mass = 0.375 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.375 * (16)
KE = 3 kg m^2/s^2
(4) Given mass = 0.500 kg
speed = 4 m/s
Since Kinetic energy = (1/2)*m*(v^2)
Plug in the values:
Hence:
KE = (1/2) * 0.5 * (16)
KE = 4 kg m^2/s^2
        
             
        
        
        
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)
...(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:
net force on the skier = mg sin 39 - μ mg cos39 
mg ( sin39 - μ cos39 ) 
= 73 x 9.8 ( .629 - .116)
= 367 N 
impulse = net force x time = change in momentum .
= 367 x 5 = 1835 kg m /s 
velocity of the skier after 5 s = 1835 / 73 
= 25.13 m /s 
b ) 
net force becomes zero 
mg ( sin39 - μ cos39 ) = 0 
μ = tan39 
= .81 
c )
net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s
so he will have speed of 25.13 m /s after 5 s . 
 
        
             
        
        
        
Calculated weight (by experimentally) of Earth is 5.972 × 10²⁴ N
Hope this helps!