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Lady_Fox [76]
3 years ago
13

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

pm and transmits 40 hp through a gear. The hub upon which the gear is seated is 1.5 in wide. You have in stock 1137 OQT 1300 carbon steel that you are considering using as a key to seat the gear hub. What dimensions will the key need to be? Is this material appropriate for the application? If not, select a more appropriate material.

Engineering
2 answers:
MrMuchimi3 years ago
8 0

Answer:

The key dimensions; (4.54mm× 6.75mm×4.5mm)

Explanation

Check attachment

velikii [3]3 years ago
4 0
Answer:










Explanation:



4140-40 I’d pick wood




I hope this helps! :)
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Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti
12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

4 0
2 years ago
Establishes general guidelines concerning licensing and vehicle
gavmur [86]

Answer:

A. National Highway Safety Act

Explanation:

The National Highway Safety Act establishes general guidelines concerning licensing, vehicle registration and inspection, and traffic laws for state regulations. The act was made in 1966 to reduce the amount of death on the highway as a result of increase in deaths by 30% between 1960 and 1965

National Traffic and Motor Vehicle Safety Act regulates vehicle manufacturers  by ensuring national safety standards and issuance recalls for defective vehicles

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5 0
3 years ago
Read 2 more answers
Why does the auto industry prefer uniform (national) standards for automobile emissions as opposed to regionally varying standar
WINSTONCH [101]

Answer:

Explanation:

For automobile emission, a uniform standard is preferred, because no unnecessary advantage is given by it to any company that is located in particular states where the regional standards are less severe.

Since pollution has its impact across the states and in the whole of the USA, then there should be uniform standards across all the states. It will also invalidate the impact of regional standards as a factor in the selection of plant locations for the automobile company. It means that a state offering less valid emission standards, will attract more companies to herself and it will be against the other states who care more about the natural environment. It can make more states to opt for the permissive emission standards, that will be more harmful to the USA as a country, than the good. So, a uniform standard is preferred to eliminate it as a factor in plant location decisions.

Yes, uniform standards are beneficial to everyone, because it will bring effective control upon the pollution level because there will be no state where the culprit firm can hide. Besides, it is more effective as efforts done towards environment conservation.

3 0
3 years ago
What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
3 years ago
How will the delay and active power per device change as you increase the doping density of both the N- and the P-MOSFET?
Murljashka [212]

Answer:

hello your question is incomplete attached below is the missing part of the  question

Consider an inverter operating a power supply voltage VDD. Assume that matched condition for this inverter. Make the necessary assumptions to get to an answer for the following questions.

answer : Nd ∝ rt

Explanation:

Determine how the delay and active power per device will change as the doping density of N- and P-MOSFET increases

Pactive ( active power ) = Efs * F

Pactive = \frac{q^2Nd^2*Xn^2}{6Eo} * f

also note that ; Pactive ∝ Nd2 (

tD = K . \frac{Vdd}{(Vdd - Vt )^2}  since K = constant

Hence : Nd ∝ rt

5 0
3 years ago
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