1. You are driving to the grocery store at 20 m/s. You are 110 m from an intersection when the traffic light turns red. Assume t
1 answer:
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Answer:
height of the water rise in tank is 10ft
Explanation:
Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)
-------(1)
substitute
0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)
Applying bernoulli's equation between tank surface (3) and orifice exit (4)
substitute
0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g
At equillibrium Fow rate at point 2 is equal to flow rate at point 4
Q₂ = Q₄
A₂V₂ = A₃V₃
The diameter of the orifice and the siphon are equal , hence there area should be the same
substitute A₂ for A₃
for V₂
for V₄
A₂V₂ = A₃V₃
Therefore ,height of the water rise in tank is 10ft
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.