Answer:
μ = 0.136
Explanation:
given,
velocity of the car = 20 m/s
radius of the track = 300 m
mass of the car = 2000 kg
centrifugal force
F c = 2666. 67 N
F f= μ N
F f = μ m g
2666.67 = μ × 2000 × 9.8
μ = 0.136
so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution
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90 inches
Explanation:
According to OSHA requirement, the distance around safety showers and eyewash should be between 82-96 inches off the flow. This will allow for maximum diameter of spray.
Learn More
Safety distance around safety showers:brainly.com/question/11123362
Keywords: distance, fire sprinkler head, safety showers, eyewash units,heating and cooling units
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Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N
