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Umnica [9.8K]
3 years ago
11

What are the three main areas of bioengineering?

Engineering
1 answer:
V125BC [204]3 years ago
8 0

Answer:Surgical Instruments and Medical Devices.

Tissue Engineering.

Biomaterials

Explanation:

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We can process oil into a lot of useful fuels to run our cars, trucks, and even airplanes. Oil is used for making lots of other
Ostrovityanka [42]

Answer:

Explanation:

Products of oil in our everyday life:

(1) Petro-Chemical Feedstock: These are by product of Refining of Oil which it is used extensively to make PET bottles, Paints, Polyester Shirts, Pocket combs e.t.c

(2) Asphalt : Used extensively to make Motor Road, highways

(3) Plastics : we use plastics in our everyday life, this is also a product of Refining of crude oil e.g PVC, Telephone casing, Tapes e.t.c

(4) Lubricating Oil/Grease : This is another product from crude oil Fractional Distillation.

(5) Propane/ Cooking Gas: This is also a product from oil which is used in our everyday life for cooking, grilling etc.

4 0
3 years ago
If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab
12345 [234]

Answer:

Explanation:

load = 4500lb                   lift height= 30 ft

time =15 s

velocity=\frac{30}{15} ft/s

velocity=2 ft/s

power = force\times velocity

power={4500}\times2

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= \frac{9000}{550} =16.36 hp

5 0
3 years ago
Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload
Vinil7 [7]

Answer:

1) V_o = 10 liters

2) V_o = 12.26 liters

Explanation:

For isothermal process n =1

V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}

V_o = 10 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.03

actual \ volume = c1\times 10 = 10.3 liters

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}

V_o  = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}

V_o = 12.26 liters

calculate pressure ratio to determine correction factor

\frac{p_2}{p_1} =\frac{180}{80} = 2.25

correction factor for calculate dpressure ration  for isothermal process is

c1 = 1.15

actual \volume = c1\times 10 = 11.5 liters

8 0
3 years ago
In response to the market revolution:the legal system worked with local governments to find better ways to regulate entrepreneur
Nataliya [291]

Answer:

Local judges protected businessmen from paying property damages associated with factory construction and from workers seeking to unionize.

Explanation:

The Market Revolution is the name given to change in the economy that occurred in the 19th century. This drastic change led to various important changes in the United States and across the world. During this period, capitalism became more entrenched and society became, for the first time, predominantly capitalist. This gave businessmen great power, as they played an increasingly important role when it came to economic growth. The power that they had influenced society deeply, including legislation. Judges often protected businessmen from paying property damages that were associated with their business enterprises. Moreover, workers had few rights and protections, and judges prevented them from unionizing.

5 0
3 years ago
A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu
Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,  

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

  • Negligible heat loss through the insulation
  • Steady state system
  • One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}

During steady state

\frac{dT}{dt} = 0 which gives k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0

From which we have \frac{d^{2}T }{dx^{2} }  = -\frac{q'_{G}}{k}

Considering the boundary condition at x =0 where there is no heat loss

 \frac{dT}{dt} = 0 also at the other end of the plane wall we have

-k\frac{dT }{dx } = hc (T - T∞) at point x = L

Integrating the equation we have

\frac{dT }{dx }  = \frac{q'_{G}}{k} x+ C_{1} from which C₁ is evaluated from the first boundary condition thus

0 = \frac{q'_{G}}{k} (0)+ C_{1}  from which C₁ = 0

From the second integration we have

T  = -\frac{q'_{G}}{2k} x^{2} + C_{2}

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k}  + C_{2}-T∞) → C₂ = q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞

T(x) = \frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞ and T(x) = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )

∴ Tmax → when x = 0 = T∞ + \frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))

Substituting the values we get

T = 167 ° C

4 0
3 years ago
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