The job that the fan is designed and built to do is to convert the electrical energy it uses into the kinetic (motion) energy of moving air.
I can't really guarantee that it accomplishes that with MOST of the electrical energy it uses, because I don't know how efficient your fan is. For example, if it's a really old fan, and one blade has the end broken off, and a lot of dust and mosquitoes have gotten into the motor, and it shakes and vibrates and makes a lot of noise when it's running, then it's converting a lot of the electrical energy into thermal energy (it gets hot when it runs) and some into sound energy too.
If you can live without the word "most" in the question, then we can assume that the fan is well designed and running like a top, and the answer is definitely choice-B .
Answer:
Petroleum
Explanation:
Water has a boiling point of 100°C meaning it will start to boil once the temperature reach 100° Celsius meanwhile
Petroleum's boiling point will change based on which petroleum product it is.
For example
LPG (which is a petroleum product)will boil at 40°C.
Answer : The rate of heat transfer to the water is, 37.92 kJ/min
Explanation : Given,
Time = 10 min
Mass of water = 200 g
Latent heat of fusion of water = 334 J/g
Latent heat of vaporization of water = 2230 J/g
Now we have to calculate the rate of heat transfer to the water.

Now put all the given values in the above formula, we get:


Thus, the rate of heat transfer to the water is, 37.92 kJ/min
Answer:
The expression for the initial speed of the fired projectile is:
![\displaystyle v_0=\frac{M+m}{m}(2gL[1-cos(\theta)]^{\frac{1}{2}})](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v_0%3D%5Cfrac%7BM%2Bm%7D%7Bm%7D%282gL%5B1-cos%28%5Ctheta%29%5D%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%29)
And the initial speed ratio for the 9.0mm/44-caliber bullet is 1.773.
Explanation:
For the expression for the initial speed of the projectile, we can separate the problem in two phases. The first one is the moment before and after the impact. The second phase is the rising of the ballistic pendulum.
First Phase: Impact
In the process of the impact, the net external forces acting in the system bullet-pendulum are null. Therefore the linear momentum remains even (Conservation of linear momentum). This means:
(1)
Second Phase: pendular movement
After the impact, there isn't any non-conservative force doing work in al the process. Therefore the mechanical energy remains constant (Conservation Of Mechanical Energy). Therefore:
(2)
The height of the pendulum respect L and θ is:
(3)
Using equations (1),(2) and (3):
(4)
The initial speed ratio for the 9.0mm/44-caliber bullet is obtained using equation (4):

Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.