Answer:
the energy required for the extension is 12.25 J
Explanation:
Given;
force constant of trampoline spring, k = 800 N/m
extension of trampoline spring, x = 17.5 cm = 0.175 m
The energy required for the extension is calculated as;
E = ¹/₂kx²
E = 0.5 x 800 x 0.175²
E = 12.25 J
Therefore, the energy required for the extension is 12.25 J
Answer:
x2 = 0.99
Explanation:
from superheated water table
at pressure p1 = 0.6MPa and temperature 200 degree celcius
h1 = 2850.6 kJ/kg
From energy equation we have following relation
![2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]](https://tex.z-dn.net/?f=2850.6%20%2B%20%5B%5Cfrac%7B50%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D%20%3D%20h2%20%2B%5B%20%5Cfrac%7B600%5E2%7D%7B2%7D%20%2A%20%5Cfrac%7B1%20kJ%2Fkg%7D%7B1000%20m%5E2%2FS%5E2%7D%5D)
h2 = 2671.85 kJ/kg
from superheated water table
at pressure p2 = 0.15MPa
specific enthalpy of fluid hf = 467.13 kJ/kg
enthalpy change hfg = 2226.0 kJ/kg
specific enthalpy of the saturated gas hg = 2693.1 kJ/kg
as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have
quality of steam x2
h2 = hf + x2(hfg)
2671.85 = 467.13 +x2*2226.0
x2 = 0.99
Stress = force / area
= 4.8 x 10³ / 1.2 x 10^-4
= 4 x 10⁷ N /m²
YOUNGS MODULAS = stress / strain
= 4 x 10⁷ / 1.2 x 10^11
= 3.3 x 10^-4
INCREMENT OF LENGTH = longitudinal length x intitial length
= ( 3.3 x 10^-4 ) x 4
= 13.2 x 10^-4 m
= 13.2 mm
mark ‼️ it brainliest if it helps you ❤️
Answer:
a) Fi = 85.76 N
b) Fi = 87.8 N
Explanation:
Given:-
- Density of hydraulic oil, ρ = 804 kg/m^3
- The radius of input piston, ri = 0.00861 m
- The radius of output piston, ro = 0.141 m
Find:-
What input force F is needed to support the 23000-N combined weight of a car and the output plunger, when
(a) the bottom surfaces of the piston and plunger are at the same level
(b) the bottom surface of the output plunger is 1.10 m above that of the input plunger?
Solution:-
For part a.
- We see that both plungers are equal levels or there is no pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers to be equal. This part is an application of Pascal's Law:
Pi = Po
Fi / Ai = Fo / Ao
Fi = Ai / Ao * Fo
Fi = (ri/ro)^2 * Fo
Fi = ( 0.00861 / 0.141 )^2 * 23000
Fi = 85.76 N
For part b.
- We see that both plungers are at different levels so there is pressure due to elevation head. So we are only dealing with static pressure exerted by the hydraulic oil on both plungers plus the differential in heads. This part is an application of Bernoulli's Equation:
Pi = Po + ρ*g*h
Where, h = Elevation head = 1.10m
Fi = (ri/ro)^2 * Fo + ρ*g*h*π*ri^2
Fi = 85.76 + (804*9.81*1.1*3.142*0.00861^2)
Fi = 87.8 N
