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Aleksandr-060686 [28]
3 years ago
6

I NEED HELP ASAP!!!! IM GIVING POINTS!! Which option identifies what the engineer will most likely do in the following scenario?

An engineer is summoned to review a fluid power system where there is a significant amount of energy being lost to waste. In the system, the engineer notices that the tubes through which the fluid travels are smooth. A) The engineer will create a regenerative braking system. B)The engineer will contact the CCEFP for assistance in solving this issue. C)The engineer will add microstructures to decrease friction in the fluid system. D) The engineer will add an additional pump to increase the power in the system.
Engineering
2 answers:
monitta3 years ago
8 0

Answer:

D

Explanation:

Dimas [21]3 years ago
3 0

Answer:

d,d,d,d,d,d,d,d,d,d,d

Explanation:

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Answer: C) a&c

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5 0
1 year ago
An inflatable structure has the shape of a half-circular cylinder with hemispherical ends. The structure has a radius of 40 ft w
NARA [144]

Find the given attachment

4 0
3 years ago
A driver complains that his front tires are wearing
Margarita [4]

Answer:

The correct option is;

Neither A nor B

Explanation:

The location of the where the thread wears in tire that has too high inflation is at the thread pattern center due to the reduced size of the contact patch with the load of the car resting on the central portion of the tire's contact surface

When the wear occurs at the outer edges of the tire, the load of the car rests on the outer edges as the contact patch increases due to the tire being under-inflated

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8 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 78 MPa (70.98 ksi). If the plate is
Reika [66]

Answer:

minimum length of a surface crack is 15.043 mm

Explanation:

given data

strain fracture toughness K = 78 MPa

tensile stress = 345 MPa

Y = 1.04

to find out

minimum length of a surface crack

solution

we find here length of critical interior flaw from formula that is

α  =  \frac{1}{\pi} (\frac{K}{\sigma Y})^2     ....................1

put here value we get

α  =  \frac{1}{\pi} (\frac{78*\sqrt{10^3} }{345*1.04})^2

α  = 15.043 mm

so minimum length of a surface crack is 15.043 mm

7 0
3 years ago
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