Answer:
the restoring force is = 3/4NKT
Explanation:
check the attached files for answer.
Answer:
Macronutrients are simply nutrients the body needs in a very high amount e.g Carbohydrate.
MicroNutrients are simply nutrients the body needs but in little amount e.g Minerals.
Explanation:
So for further breakdown:
What are nutrients? Nutrients are essential elements that nourish the body in different capacities. We as humans get most of out nutrients from the food and water we ingest.
Now about Macro Nutrients: From the prefix "Macro" which means large, we can infer that macro nutrients are elements need by the body for the fundamental processes of the body, deficiency in this nutrients are very easy to spot. Examples are: Carbohydrates, Protein, Fats amd Water.
Micro Nutrients: In relation to macro nutrients this are elements that the body needs but are not needed in Large quantities. They mostly work like supporting nutrients. Most chemical activities like reaction that occur in the body are a function of micro nutrients. Defiencies in micrp nutrients may take some time to spot e.g Minerals and Vitamins
In regards to exercise: Macro nutrients are the essential ones here since they are the ones that generate energy. PS: micro nutrients dont generate energy.
In regards to rest: Both the Macro and Micro Nutrients are essentail for the overall well being of the body.
Answer:
Most hydraulic systems develops pressure surges that may surpass settings valve. by exposing the hose surge to pressure above the maximum operating pressure will shorten the hose life.
Explanation:
Solution
Almost all hydraulic systems creates pressure surges that may exceed relief valve settings. exposing the hose surge to pressure above the maximum operating pressure shortens the hose life.
In systems where pressure peaks are severe, select or pick a hose with higher maximum operating pressure or choose a spiral reinforced hose specifically designed for severe pulsing applications.
Generally, hoses are designed or created to accommodate pressure surges and have operating pressures that is equal to 25% of the hose minimum pressure burst.
Answer:
a) 5.2 kPa
b) 49.3%
Explanation:
Given data:
Thermal efficiency ( л ) = 56.9% = 0.569
minimum pressure ( P1 ) = 100 kpa
<u>a) Determine the pressure at inlet to expansion process</u>
P2 = ?
r = 1.4
efficiency = 1 - [ 1 / (rp)
]
0.569 = 1 - [ 1 / (rp)^0.4/1.4
1 - 0.569 = 1 / (rp)^0.285
∴ (rp)^0.285 = 0.431
rp = 0.0522
note : rp = P2 / P1
therefore P2 = rp * P1 = 0.0522 * 100 kpa
= 5.2 kPa
b) Thermal efficiency
Л = 1 - [ 1 / ( 10.9 )^0.285 ]
= 0.493 = 49.3%
Answer:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D)
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness,
is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
![Y=\frac {K}{\sigma_c \sqrt {a\pi}}= \frac {40}{300\sqrt {(0.002*\pi)}}=1.682](https://tex.z-dn.net/?f=Y%3D%5Cfrac%20%7BK%7D%7B%5Csigma_c%20%5Csqrt%20%7Ba%5Cpi%7D%7D%3D%20%5Cfrac%20%7B40%7D%7B300%5Csqrt%20%7B%280.002%2A%5Cpi%29%7D%7D%3D1.682)
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for
while a is taken as 0.003m and Y is already known
![K=260*1.682*\sqrt {0.003*\pi}=42.455 Mpa](https://tex.z-dn.net/?f=K%3D260%2A1.682%2A%5Csqrt%20%7B0.003%2A%5Cpi%7D%3D42.455%20Mpa)
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material