Answer:
f = 628.32 lb
t = 2513.28 lb-inc
Explanation:
given data:
θ = 45°
outside radius = 6 inch
inside radius = 4 inch
coefficient of friction = 0.4
max pressure = 100 psi
a) determine force required for applying one pad
f =
f =
f = 628.32 lb
b) torque capacity (t)
t =
t = 0.4 *628.32*5
torque = 1256.64 lb-inc
for both pad = 2 * 1256.64 =2513.28 lb-inc
Answer:
The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Explanation:
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
assume from specifications that k = 5v/5cm
= 1v/cm
u^2 = (0.0025*2)^(2) + (0.005*10*2)^2 + (0.0025*2)^2
= 0.01225v
v = 2v * 0.001
= 0.002v
uncertainty in a nominal displacement
= (u^2 + v^2)^(1/2)
= ((0.01225)^2 + (0.002)^2)^(1/2)
= 0.0124 cm
Therefore, The Estimated uncertainty in a nominal displacement of 2 cm at the design stage is plus or minus 0.0124cm
Answer:
“How can I help?”
Though a second could be “Don’t tell me; show me.”, but that is very authoritative , while conventional personalities types don't typically like to be in an authoritative position.
Explanation:
Chain of Command and Unity of Command
Information and Intelligence Management
Answer:
the final temperature would have been 2.81 °C if the pressure drop was 2 psi
Explanation:
if we assume that is no change in volume, there are not leaks present and also that the gas inside the football behaves as an ideal gas, we have:
initial state) P1 V = n R T1
final state) P2 V = n R T2
where P = absolute pressure , V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant T= absolute temperature
therefore since V= constant (constant volume) and n= constant ( no leaks), if we divide both equations
P2/P1 = T2/T1
therefore
T2 = T1 *(P2/P1)
since P1 absolute = P1 relative + P atmospheric (14.7 psi) = 12.5 psi + 14.7 psi = 27.2 psia
also P2 = P1 - 2 psi = 25.2 psia
T1 = 24.7°C + 273 °C = 297.7 K
therefore
T2 = T1 * (P2/P1) = 297.7 K ( 25.2 psia/27.2 psia) = 275.81 K
thus T2 = 275.81 K = 2.81 °C