To reduce the internal energy U of an ideal gas you need the gas to lose heat and/or do work.
Answer:
a)Work done by fireman= 2.15 Btu
b) Time t= 0.86 sec
Explanation:
Given that
Weight = 280 lbf
We know that 1 lbf = 4.44 N
so 280 lbf = 1245.5 N
Weight =1245.5 N
Height h = 60 ft
We know that
1 ft = 0.3048 m
So 60 ft = 18.28 m
h =18.28 m
Power = 3.5 hp
We know that
1 hp =0.74 KW
So 3.5 hp = 2.61 KW
Power = 2.61 KJ/s
So the work done by fireman = Weight x h
Now by putting the values
Work done by fireman= 1245.5 x 18.28 J
Work done by fireman= 2267.74 J
Work done by fireman= 2.26774 KJ
We know that 1 Btu= 1.05 KJ
So 2.266 KJ = 2.15 Btu
Work done by fireman= 2.15 Btu
We know that ,rate of work is called power.
Power x time = work
2.61 x t = 2.26
So t= 0.86 sec
Answer:
Explanation:
From the position coordinates given , it appears that the ball moves simultaneously along x and y direction.
Displacement along x direction in one second = 4.4 - 1.8 = 2.6 m
So velocity along x direction V_x = 
Similarly velocity along y direction V_y(1) = 
In the next phase velocity changes both in x and y direction.
velocity in x - direction V_x(2) = [tex]\frac{2}{s}[/tex
Velocity in Y- direction V_y(2) = [tex]\frac{3.1}{s}[/tex
Acceleration in x direction = change of velocity in x direction
= ( 2 - 2.6 ) = -.6 m s⁻²
Acceleration in y direction = ( 3.1 - 2.6 ) = 0.5 m s⁻²
Total acceleration =\sqrt{( -.6 )² + ( .5 )²}
= .78 ms⁻²
Answer:

Explanation:
The capacitor of a parallel-plate capacitor is given by:

where
A is the area of each plate
d is the separation between the plates
is the vacuum permittivity
The energy stored in a capacitor instead is given by

where
Q is the charge stored in each plate
Substituting the expression we found for C inside the last formula,

And re-arranging it

Now if we substitute

We find the charge stored on the capacitor:

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