Answer:
The kinetic energy of the weight is 344.5 J
Explanation:
Given that:
Force = F = 65 newton
distance = d = 5.3 meters
We have to find change in kinetic energy ΔK.E
Now we know that, initially kinetic energy was 0 So the formula we use will be:
Work done = Change in kinetic energy
Mathematically,
W = ΔK.E
As we know W = F . d and ΔK.E = K.E(final) - K.E(initial)
So by putting values:
F . d = K.E(final) - K.E(initial)
F . d = K.E(final)
As K.E(initial) is 0 so by putting values of F and d
(65)* (5.3) = K.E(final)
344.5 J = K.E(final)
So the change in K.E will also be 344.5 J
i hope it will help you!
Answer:
The required wall thickness is
m
Explanation:
Given:
Fluid density

Diameter of tank
m
Length of tank
m
F.S = 4
For A-36 steel yield stress
MPa,
Allowable stress 
MPa
Pressure force is given by,


Pa
Now for a vertical pipe,

Where
required thickness


m
Therefore, the required wall thickness is
m
Answer:
The required pumping head is 1344.55 m and the pumping power is 236.96 kW
Explanation:
The energy equation is equal to:

For the pipe 1, the flow velocity is:

Q = 18 L/s = 0.018 m³/s
D = 6 cm = 0.06 m

The Reynold´s number is:


Using the graph of Moody, I will select the f value at 0.0043 and 335339.4, as 0.02941
The head of pipe 1 is:

For the pipe 2, the flow velocity is:

The Reynold´s number is:


The head of pipe 1 is:

The total head is:
hi = 1326.18 + 21.3 = 1347.48 m
The required pump head is:

The required pumping power is:

Answer:
Days: 6.9444 days
Production rate: 547.2035 ft²/s
Explanation:
the solution is attached in the Word file
Answer:
Mechanical Engineering
Chemical Engineering
Civil Engineering
Explanation:
I got it from my old homework And I learn those at school ( Thank You For The Points)