How can philosophy help you become a productive citizen
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Consider the diagram below.
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
m₁ = 8.5 x 10⁶ kg, the starship's mass
m₂ = 10⁴ kg, the shuttlecraft's mass
a₁ = acceleration of the starship
a₂ = the acceleration of the shuttle
F = 4 x 10⁴ N, the force exerted
Let y represent the distance covered by the starship
Let x denote the distance covered by the shuttlecraft
If t indicates the travel time, then
y = 0.5a₁t² (1)
x = 0.5a₂t² (2)
F = m₁a₁ = m₂a₂ (3)
Additionally,
x + y = 14000 m (4)
From (2), we derive
a₁ = (4 x 10⁴ N)/(8.5 x 10⁶ kg) = 4.706 x 10⁻³ m/s²
a₂ = (4 x 10⁴ N)/(10⁴ kg) = 4 m/s²
From (1), (2) and (4), we find
0.5*(t s)²*(4 + 4.706 x 10⁻³ m/s²) = 14000 m
2.002353t² = 14000
t² = 6991.774 s²
t = 83.617 s
Thus
x = 0.5*4*6991.774 = 13984 m = 13.984 km
y = 0.5*4.706 x 10⁻³*6991.774 = 16.452 m
The starship moves roughly 16.5 meters while towing the shuttlecraft by 13.98 kilometers.
Result: The starship shifts by 16.5 m (to the nearest tenth)
Answer:
a) W =400 kJ
b) W = 0 kJ
c) W =-160.944 KJ
Explanation:
Given
Process 1 ---> 2
Pressure at point (1) P1 = 10 bar = P2
Volume at point (1) V1 = 1 m^3
Volume at point (2) V2 =4 m^3
For Process 2 ---> 3, where V = constant
Pressure at point (3) P3 = 10 bar
Volume at point (3) V3 = 4 m^3
Process 3 ---> 1 defined as PV = constant.
Required
Sketch the processes on the PV coordinates To calculate the work in kJ
Work is calculated by W=
a=V2
b=V1
x=Pdx=dV
For Process 1 ---> 2 where P3 = P4 = 5 barW=
a=V3
b=V2x=4dx=dV
substituting the values here into the integral givesW=400 kJFor Process 2 ---> 3
As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ
For Process 3 ---> 1 By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1
W=a=V1
x=1V^-1dx=dVsubstituting values into this integral results in
W=| ln V | limits a and b= -160.944 KJ