Answer:
<em>The temperature will be greater than 25°C</em>
Explanation:
In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.
mathematically
Change in the internal energy of a system ΔU = ΔQ + ΔW
in an adiabatic process, ΔQ = 0
therefore
ΔU = ΔW
where ΔQ is the change in heat into the system
ΔW is the work done by or done on the system
when work is done on the system, it is conventionally negative, and vice versa.
also W = pΔv
where p is the pressure, and
Δv = change in volume of the system.
In this case,<em> work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C </em>
Answer:
Driving test
Explanation:
Usually according to laws in countries worldwide, to be licenced to drive, one is required to go through a driving school to learn the ethics and rules of driving.
Answer:
a. 9.58kgs b. 340.32KJ
Explanation:
Volume of tank= 2m³
Initial Pressure Pi= 100KPa
Initial Temperature Ti= 22 C= 295K
Line Pressure P₁= 600 KPa
Line Temperature T= 22 C= 295K
Final Pressure P2= 600 KPa
Final Temperature T2= 77 C= 350K
Use Ideal Gas Equation
PV= mRT
P₁V₁= m₁RT₁
m₁= (100 x 2)/(0.287 x 295) = 2.3622kg
P₂V₂= m₂RT₂
m₂= (600 x 2)/(0.287 x 350) = 11.946 kg
Since valve is closed and no mass leave
m₁ + mi = m₂ + me
as per above condition me= 0
mi= m₂ - m₁ = 11.946 - 2.3622 = 9.5838kg
Applying energy equation
m₁u₁ + mihi + Q = m₂u₂ + mehe + W
me and W=0
m₁u₁ + mihi + Q = m₂u₂
m₁CvT₁ + miCpTi + Q = m₂CvT₂
Q = m₂CvT₂- m₁CvT₁ - miCpTi
Q = (11.946 x 0.717 350) - (2.3622 x 0.717 x 295) - (9.5868 x 1.004 x 295)
Q = -340.321 KJ (Negative sign doesn't matter as energy is not a vector quantity)
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
