Answer:
Explanation:
The speed of the water in the large section of the pipe is not stated
so i will assume 36m/s
(if its not the said speed, input the figure of your speed and you get it right)
Continuity equation is applicable for ideal, incompressible liquids
Q the flux of water that is Av with A the cross section area and v the velocity,
so,
the diameter decreases 86% so
Thus, speed in smaller section is 48.6 m/s
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?
0.57 m
0.64 m
0.80 m
1.25 m
Data:
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
Answer:
The displacement of the spring = 0.8 m (or 0.80 m)
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I Hope this helps, greetings ... Dexteright02! =)