Answer:
Explanation:
a )
one kg of coal gives energy of 27 x 10⁶ J
75 kg of coal gives energy of 27 x 10⁶ x 75 J
So rate which energy is coming out of coal per second
= 27 x 10⁶ x 75 J
= 2025 x 10⁶ J /s
2025 million watts .
b ) energy output = 800 million watts
efficiency = (800 / 2025) x 100
= 39.5 % .
Answer:
24.3KW
Explanation:
A)The kinetic energy is changing, the potential energy is changing and the chemical energy in form of fuel powering the engine also is changing
The kinetic energy is increasing as the body gain speed, the potential energy also increases as the body gain height against gravity and the chemical energy in form of fuel decreases as the body burn the fuel to create a lifting force
B) The workdone by the lifting force = the change in kinetic energy + the change in potential energy
C)The time taken in seconds to do the work is the variable needed
D) average power generated by the lifting force = (change in kinetic energy + change in potential energy) / time taken in seconds
Average power = 1/2 * m(mass) (Vf-Vi)^2 + mg(hf-hi) /t where vf is final speed and vi is initial speed at rest = 0, similarly, hf = final height and hi = initial height.
Average power = 1/2*810*7^2 + 810*9.81*8.2/3.5s
Average power = (19845+65158.02)/3.5 = 24286.577 approx 24.3kW
Answer:
0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli
1.1m to 0.04125 m
Explanation:
T= 1/f=
if f= 300Hz then T = 1/300 =0.003333 s
if f= 8000 then T= 1/8000 = 0.000125s
now v=f×wave length
or wavelength = speed/ frequency
when f = 300 Hz
wavelength = 330/300=1.1 m
wavelength = 330/8000 = 0.04125m
note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s
Answer:
432 units
Explanation:
Let the charges be q and Q separated by a distance r. The electrostatic force , F = kqQ/r² = 72 units. If q = 2q and Q = 3Q, then the new electrostatic force is
F = k × 2q × 3Q/r² = 6kqQ/r² = 6 × 72 = 432 units
Using
KE = ½mv² = ½×1500×19×19 = 270750 joules
