A hydrauliic jack is rated at 5000 pound capacity. The area of the large piston on the jack is 4.45 Square inches. Calculate the
1 answer:
Answer:
1123.6 pounds/ square inch.
Explanation:
Fluid pressure is the ratio of force or weight applied by the fluid per unit area.
i.e Fluid pressure =
The maximum load of the jack is obtained at its maximum capacity = 5000 pounds
Area of the large piston on the jack = 4.45 square inches
Thus,
Fluid pressure =
= 1123.5955
Fluid pressure = 1123.6 pounds/ square inch
Thu, the fluid pressure in the jack at maximum load is 1123.6 pounds/ square inch.
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Answer:
Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:
For the number of sinosoidial signals;
Power can be expressed as:
As such,
For x(t), Power
For the number of sinosoidial signals;
For x(t), the RMS value is as follows:
Answer:
Not subject to detection
Explanation:
Assuming the value of strain fracture toughness is 77 Mpa
The design stress is half hence
Critical flaw size,
Where Y is dimensionless parameter, is applied stress,
is plane strain fracture toughness,
is critical length of surface crack
The critical length of surface crack is therefore 3.85 mm, which is less than detection apparatus size given as 4 mm
Since the critical flaw size is less than the resolution limit of flaw detection apparatus, the critical flaw for this plate is not subjected to detection.