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Doss [256]
3 years ago
8

What is the potential consequence of online impulsive buying?

Engineering
2 answers:
cricket20 [7]3 years ago
7 0
Probably B because I impulsively shop online and b is true for me. I’m not 100% sure though.
Nastasia [14]3 years ago
3 0
I’m pretty sure it’s B
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Question 2
ValentinkaMS [17]

Answer:

r h ch y TC GG ch hh h h

No se

Explanation:

No seibibk h j h uh t t t

3 0
2 years ago
Someone claims that in fully developed turbulent flow in a tube, the shear stress is a maximum at the tube surface. Is this clai
Alika [10]

Answer:

Yes this claim is correct.

Explanation:

The shear stress at any point is proportional to the velocity gradient at any that point. Since the fluid that is in contact with the pipe wall shall have zero velocity due to no flow boundary condition and if we move small distance away from the wall the velocity will have a non zero value thus a maximum gradient will exist at the surface of the pipe hence correspondingly the shear stresses will also be maximum.

5 0
4 years ago
Engineer Smith, who is licensed in several jurisdictions, recently had his license acted upon by a licensing authority in one of
Nimfa-mama [501]

Answer:

Assuming the infraction in the other jurisdiction is an infraction in Florida, having approximately the same penalty imposed by the FBPE as was imposed in the other jurisdiction

Explanation:

The idea of 8 hours continuation in the options is irrelevant to this case. Rather, Engineer Smith should expect about the same penalty imposition.

3 0
3 years ago
A venturi meter is to be installed in a 63 mm bore section of a piping system to measure the flow rate of water in it. From spac
Alex Ar [27]

Answer:

Throat diameter d_2=28.60 mm

Explanation:

 Bore diameter d_1=63mm  ⇒A_1=3.09\times 10^{-3} m^2

Manometric deflection x=235 mm

Flow rate Q=240 Lt/min⇒ Q=.004\frac{m^3}{s}

Coefficient of discharge C_d=0.8

We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

A_2=6.42\times 10^{-4} m^2

 ⇒d_2=28.60 mm

So throat diameter d_2=28.60 mm

     

4 0
3 years ago
Pls help me thank you ​
sergey [27]

Answer:   1.  Hand Plane

                 2. Rift Sawn

Explanation:  I'm not Sure tho hope it helps

8 0
2 years ago
Read 2 more answers
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