Answer:
T₂=659.25 K
Explanation:
Given that
W= 25 J
Qr = 20 J
T₁ = 20⁰ = 20 +273 = 293 K
The minimum temperature of the hot reservoir = T₂
If the engine is Carnot engine then
Qa= W+ Qr
Qa=25 + 20 J
Qa= 45 J
T₂=659.25 K
Therefore the temperature of hot reservoir will be 659.25 K
<h3><u>L</u><u>A</u><u>P</u><u>L</u><u>A</u><u>C</u><u>E</u><u> </u><u>L</u><u>A</u><u>W</u><u> </u><u>:</u><u>-</u></h3>
Laplace's law for the gauge pressure inside a cylindrical membrane is given by
<h3>ΔP = γ/r</h3>
<u>W</u><u>h</u><u>e</u><u>r</u><u>e</u>
γ => surface tension
r => the radius of the cylinder
Note : there is an inverse relation between the pressure and the radius.
Answer:
V = 4.826m/s, 716N
Explanation:
At the lowest swinging point, the net force acting on the child is equal to the centripetal force and it could be represented as
F = mv^2/r
2T-mg =mv^2/r
r(2T-mg) = mv^2
Divide both sides by m
r(2T-mg)/m = mv^2/m
r(2T/m-g) = v^2
V= √ r(2T/m-g)
Where v is the velocity
r is the length of the chain
m is the mass of the child in kg
T is the tension in Newton
g is the acceleration due to gravity
Given that g = 9.8m/s^2
T = 358N
m = 41.0kg
r = 3.04m
Substituting the values into the formula
V = √ 3.04( 2*358/41 -9.89
V = √ 3.04 ( 716/41 - 9.8 )
V = √3.04 ( 17.463-9.8 )
V = √3.04( 7.6634)
V = √23.2967
V = 4.826m/s
For the second part which is the tension in the two chains
N - m*g = m*(v^2 / r)
N - (41)*(9.81) = (41)*(4.826^2 / 3.04)
N - 402.21 = 41×7.66
N - 402.21 = 314.112
N = 402.21 + 314.112
N = 716.332 newton
Approximately = 716N
Or alternatively, since there are two chains holding the swing, of which each chain is acted upon by a 358N tension. Hence = 2T
2*358 = 716N