Response:
1) g=31.87ft/s^2
2)m=120
W=119.64lbf
Clarification:
initial segment
the weight of an object with a specified mass is determined using the equation
W=mg
we need to convert 120lb into slug
m=120lbx1slug/32.147lb=3.733slug
calculating for g
g=W/m
g=119/3.733=31.87ft/s^2
subsequent segment
the mass remains unchanged
m=120lb=3.733slug
Weight
W=3.733slug*32.05ft/s^2=119.64lbf
Answer:
Ps=19.62N
Explanation:
A thorough explanation of the answer can be found in the attached files.
Response:
The solution to this question is 1273885.3 ∅
Clarification:
The first step is to ascertain the required hydraulic flow rate liquid based on the working pressure if a cylinder with a piston diameter of 100 mm is utilized.
Given that,
The distance = 50mm
The time t =10 seconds
The force F = 10kN
The piston diameter = 100mm
The pressure = F/A
10 * 10^3/Δ/Δ
P = 1273885.3503 pa
Subsequently
Power = work/time = Force * distance /time
= 10 * 1000 * 0.050/10
which amounts to =50 watt
Power =∅ΔP
50 = 1273885.3 ∅
Answer:
Explanation:
In a steady-state condition, the total volume exiting matches the total volume entering, and
the total amount of salt entering equals the total amount of salt exiting.
The volume entering each minute is = 2000 + 2 x 10³ x 60
= 122000 L.
The total salt entering per minute = 1200 x 2000 + 20 x 2 x 10³ x 60
= 2400000 + 2400000 mg
= 4800000 mg.
The volume of water exiting each minute = 122000 L.
The total salt exiting each minute = 4800000 mg.
The concentration of salt in the exiting water = 4800000 / 122000
= 39.344 mg / L.
