80000 Joule is the change in the internal energy of the gas.
<h3>In Thermodynamics, work done by the gas during expansion at constant pressure:</h3>
ΔW = -pdV
ΔW = -pd (V₂ -V₁)
ΔW = - 1.65×10⁵ pa (0.320m³ - 0.110m³)
= - 0.35×10⁵ pa.m³
= - 35000 (N/m³)(m³)
= -35000 Nm
ΔW = -35000 Joule
Therefore, work done by the system = -35000 Joule
<h3>Change in the internal energy of the gas,</h3>
ΔV = ΔQ + ΔW
Given:
ΔQ = 1.15×10⁵ Joule
ΔW = -35000 Joule
ΔU = 1.15×10⁵ Joule - 35000 Joule
= 80000 Joule.
Therefore, the change in the internal energy of the gas= 80000 Joule.
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I think it’s output because output work is work done by a machine
Answer:
P₁- P₂ = 91.1 10³ Pa
Explanation:
For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)
In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference
For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m
P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)
P₁-P₂ = 22.5 10³ + 68.6 10³
P₁- P₂ = 91.1 10³ Pa
Answer:
The work done on the sled by friction, W = - 4593.75 J
Explanation:
Given data,
The combined mass of sled and the boy, m = 75 kg
The displacement of the boy, S = 25 m
The coefficient of the friction, u = 0.25
The frictional force acting on the boy,
<em>F = u η</em>
Where,
η - is the normal force acting on the boy (mg)
Substituting the values,
F = 0.25 x 75 x 9.8
= 183.75 N
Since the direction of the frictional force is against the direction of motion
F = - 183.75 N
The work done on the sled by friction,
W = F x S
= - 183.75 x 25
= - 4593.75 J
Hence, the work done on the sled by friction, W = - 4593.75 J
