I'm not accurately sure if you're asking for why the bulb of a thermometer is in a cylindrical shape. So let me continue. The shape of the which is thin and cylindrical in the shape is the increase of the effect of mercury in the tube to rise and fall depending on the contact temperature.
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 law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.
If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is
W = k[{x₂-x₁)² - (x₁-xn)²],
where
xn is the natural length
x₁ is the stretched length
x₂ is also the stretched length when stretched even further than x₁
In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.
Then, substituting the values,
2 J = k (0.11² -0²)
k = 165.29 N/m
Finally, we use the value of k to the Hooke's Law to determine the Force.
F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂
Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)
We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa
Therefore your answer is: 120kPa
<span>The elevator must be accelerating
</span>
This is the case because the tension is not equivalent to the load's weight, which means there must another force acting on the system.