Answer:
Explanation:
Since the roundabout is rotating with uniform velocity ,
input power = frictional power
frictional power = 2.5 kW
frictional torque x angular velocity = 2.5 kW
frictional torque x .47 = 2.5 kW
frictional torque = 2.5 / .47 kN .m
= 5.32 kN . m
= 5 kN.m
b )
When power is switched off , it will decelerate because of frictional torque .
<span>3933 watts
At 100 C (boiling point of water), it's density is 0.9584 g/cm^3. The volume of water lost is pi * 12.5^2 * 10 = 4908.738521 cm^3
The mass of water boiled off is 4908.738521 * 0.9584 = 4704.534999 grams.
Rounding to 4 significant figures gives me 4705 grams of water.
The heat of vaporization for water is 2257 J/g. So the total energy applied is
2257 J/g * 4705 g = 10619185 J
Now we need to divide that by how many seconds we've spent boiling water. That would be 45 * 60 = 2700 seconds.
Finally, the rate of heat transfer in Joules per second will be the total number of joules divided by the total number of seconds. So
10619185 J / 2700 s = 3933 J/s = 3933 (kg m^2/s^2)/s = 3933 (kg m^2/s^3)
= 3933 watts</span>
Answer:
The Most Famous Astronomers of All Time. Karl Tate, SPACE.com. ...
Claudius Ptolemy. Bartolomeu Velho, Public Domain. ...
Nicolaus Copernicus. Public Domain. ...
Johannes Kepler. NASA Goddard Space Flight Center Sun-Earth Day. ...
Galileo Galilei. NASA
