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wariber [46]
3 years ago
9

Which statement about tensile stress is true? A. Forces that act perpendicular to the surface and pull an object apart exert a t

ensile stress on the object. B. Forces that act perpendicular to the surface and squeeze an object exert a tensile stress on the object. C. Forces that act parallel to the surface exert a tensile stress on the object. D. Forces that decrease the length of the material exert a tensile stress on the object.
Engineering
1 answer:
svp [43]3 years ago
3 0

Answer:

A. Forces that act perpendicular to the surface and pull an object apart exert a tensile stress on the object.

Explanation:

Tensile stress is referred as a deforming force, in which force acts perpendicular to the surface and pull an object apart, attempting to elongate it.

The tensile stress is a type of normal stress, in which a perpendicular force creates the stress to an object’s surface.

Hence, the correct option is "A."

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Answer:

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6 0
3 years ago
Out
olchik [2.2K]

ave you ever seen a Rube Goldberg machine in action? You probably have, even if you didn’t know what it was. A Rube Goldberg machine is a contraption that uses a chain reaction to carry out a simple task. It performs a very basic job in a complicated way.

7 0
3 years ago
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
Determine the work done by an engine shaft rotating at 2500 rpm delivering an output torque of 4.5 N.m over a period of 30 secon
balu736 [363]

Answer:

work done= 2.12 kJ

Explanation:

Given

N=2500 rpm

T=4.5 N.m

Period ,t= 30 s

torque =\frac{power}{2\pi N}

power=2\pi N\times T

P=2\times \pi \times2500 \times 4.5

P=70,685W

P=70.685 KW

power=\frac{work done}{time}

work done = power * time

                  = 70.685*30=2120.55J

                  = 2.12 kJ

7 0
3 years ago
Refrigerant 134a is the working fluid in a vaporcompression heat pump that provides 35 kW to heat a dwelling on a day when the o
Dennis_Churaev [7]

Answer:

Hello, dear.

For the answer please see the explanation below.

Explanation:

The compressor power is:

2.212kW

(b) The refrigeration capacity is:

3.62 tons

(c) The coefficient of performance is:

5.75

5 0
3 years ago
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