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3 years ago

Automotive service P2 Wastewater Management and Handling Spins

Engineering

1 answer:

zvonat [6]3 years ago

6 0

True you should never discharge wast water

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Side milling cutter is an example of ______ milling cutter.

dusya [7]

Answer:

special type

Explanation:

As per the classification of milling cutters. This cutter can handle deep and long open slots in a more comfortable manner, which increase the productivity.

6 0

3 years ago

Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

mixas84 [53]

Answer:

work done = 48.88 × $10^{9}$ J

Explanation:

given data

mass = 100 kN

velocity = 310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J

work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J

work done = 48.88 × $10^{9}$ J

6 0

3 years ago

According to the video, what are examples of systems that Stationary Engineers oversee? Check all that apply. electrical systems

garik1379 [7]

Answer:

electrial systems

fire systems

heating systems

air systems

Explanation:

3 0

3 years ago

Read 2 more answers

Imagine you are making a pizza, you start of with the dough for the crust in a large ball. You then begin to roll out the pizza

DochEvi [55]

Answer: The force exerted on the dough.

Explanation:

The force is responsible for stimulating the stress.

Recall that:

stress= Tensile force/area.

5 0

3 years ago

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the

Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$ ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

$\sigma a$ = $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$ ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$ ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$

solve it we get

FOS = 1.5432

6 0

3 years ago