Can you guys please help me out and it would me great if you could explain it too!!
2 answers:
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Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
Answer:
1) The mass of the student is approximately 73.39 kg
2) The net force on the student is approximately 947.523 N
3) The value the scale will read is approximately 96.59 kg
Explanation:
The given parameters are;
The weight of the student = 720 N
The speed at which the elevator is decreasing = 3.1 m/s²
1) The weight of the student = The mass of the student × The acceleration due to gravity
The acceleration due to gravity is a constant = 9.81 m/s²
Substituting the known values gives;
720 N = The mass of the student × 9.81 m/s²
∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg
2) The forces acting on the student are;
i) The force of gravity which is the weight of the student acting downwards
ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student
The net force, = The weight of the student + The inertia force of the slowing elevator
∴ The net force, = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N
3) The scale will read the mass of the student as follows;
Mass reading of student on the scale = Force on scale/9.81
∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg
The value the scale will read = 96.59 kg.