Suppose that the distance an aircraft travels along a runway before takeoff is given by Upper D equals (5 divided by 3 )t square
1 answer:
Answer:
a. Time=25seconds
b.distance=1041.67m
Explanation:
a.The equation for in terms of m/s is
after conversion.
To find when speed reaches 300km/hr=83.33m/s, we find and solve for
b. From a, above we already have our t=25seconds as the time it takes before the plane is airborne.
#To find distance travelled in that time , we substitute for in our distance equation:
Hence the distance of the plane before it gets airborne is 1041.67m
You might be interested in
Answer:
a) ΔEC=-23.4kW
b)W=12106.2kW
c)
Explanation:
A)
The kinetic energy is defined as:
(vel is the velocity, to differentiate with v, specific volume).
The kinetic energy change will be: Δ ()=
Δ ()=
Where 1 and 2 subscripts mean initial and final state respectively.
Δ()=
This amount is negative because the steam is losing that energy.
B)
Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).
We already know the last quantity: =
Δ (
)=23400W
For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that,
The exit state is a liquid-vapor mixture, so its enthalpy is:
Finally, the work can be obtained:
C) For the area, consider the equation of mass flow:
where
is the density, and A the area. The density is the inverse of the specific volume, so
The specific volume of the inlet steam can be read also from the steam tables, and its value is: , so: