Answer:
The effect of lowering the condenser pressure on different parameters is explained below.
Explanation:
The simple ideal Rankine cycle is shown in figure.
Effect of lowering the condenser pressure on
(a). Pump work input :- By lowering the condenser pressure the pump work increased.
(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.
(c). Heat supplied :- Heat supplied increases.
(d). Heat rejected :- The heat rejected may increased or decreased.
(e). Efficiency :- Cycle efficiency is increased.
(f). Moisture content at turbine exit :- Moisture content increases.
The work done on the wagon is 3549 J
Explanation:
The work done by a force when moving an object is given by
where
:
F is the magnitude of the force
d is the displacement
is the angle between the direction of the force and of the displacement
In this problem we have the following data:
F = 87 N is the magnitude of the force
d = 44 m is the displacement of the wagon
is the angle between the direction of the force and the displacement
Substituting, we find the work done

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Answer:
If they are metallic spheres they are connected to earth and a charged body approaches
non- metallic (insulating) spheres in this case are charged by rubbing
Explanation:
For fillers, there are two fundamental methods, depending on the type of material.
If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.
If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.
We are given: Final velocity (
)=20 m/s .
Time t= 2.51 s and
distance s = 82.9 m.
We know, equation of motion

Let us plug values of final velocity, and time in above equation.


Subtracting 2.51a from both sides, we get
-----------equation(1)
Using another equation of motion

Plugging values of vi =20-2.51a, t=2.51 and distnace s=82.9 in this equation.
We get,

Now, we need to solve it for a.
20-20+2.51a=165.8a.
-163.29a=0
a=0.
So, the acceleration would be 0 m/s^2.
The key projectile motion is that gravity allows downward only