It would be not be able to move yet it would be in the air
Answer:
229,098.96 J
Explanation:
mass of water (m) = 456 g = 0.456 kg
initial temperature (T) = 25 degrees
final temperature (t) = - 10 degrees
specific heat of ice = 2090 J/kg
latent heat of fusion =33.5 x 10^(4) J/kg
specific heat of water = 4186 J/kg
for the water to be converted to ice it must undergo three stages:
- the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp
Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J
- the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp
Q = 0.456 x 33.5 x 10^(4) = 152760 J
- the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp
Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J
The quantity of heat removed from all three stages would be added to get the total heat removed.
Q total = 66,808.56 + 152,760 + 9,530.4 = 229,098.96 J
Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90
to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.