A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
ΔU =
-Wint
Consdier the work of of
interaction is W =m*g*h - equation -1
and the Potential energy U.
Final Potential energy Uf =0
, And the Initial Potential Energy Ui =m*g*h
<span>Now we will write the
equation for a Change in Potential energy ΔU,</span>
ΔU = Uf
- Ui
= 0-m*g*h
<span> ΔU = -m*g*h --Equation 2</span>
Now compare the both equation
<span>Wint = -ΔU</span>
we can rewrite the above
equation
ΔU =
-W.
<span>So our Answer is ΔU = -W. .</span>
<span> </span>
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:
Shawn's speed relative to Susan's speed = 10 mph
Resultant velocity = 82.32 mph
Explanation:
The given data :-
i) Susan driving in north and speed of Susan is ( v₁ ) = 53 mph.
ii) Shawn driving in east and speed of Shawn is ( v₂ ) = 63 mph.
iii) The speed of both Susan and Shawn is relative to earth.
iv) The angle between Susan in north and Shawn in east is 90°.
We have to find Shawn's speed relative to Susan's speed.
v₂₁ = v₂ - v₁ = 63 - 53 = 10 mph
Resultant velocity,
v = 82.32 mph
