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kogti [31]
3 years ago
6

Someone help I'm stuck on question 8 and 12

Physics
1 answer:
Kaylis [27]3 years ago
4 0
Sippen lein an hr later is theanswer to both 

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Elements from opposite sides of the periodic table tend to form ________.
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Ionic compounds is your answer. What happens is one atom donates electron(s) to the other atom, making one positive and the other negative. The opposite atoms attract, forming an ionic bond. 

Hope this helps! :)
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HELPPPPPPPP PLSSSSSSSS A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2
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Explanation:

6.2 miles

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The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete
oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\

make shear stress (τ) the subject of the formula

\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of  0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2

Part (b) shear stress at a distance of  0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

\tau = \frac{\delta P *r}{2L}  = \frac{86.4 *0}{2*12} =0

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3 years ago
Solving a physics problem is . . .
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C. A step-by-step process that takes time, and is essential for learning physics concepts.
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Examples of uniform speed​
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A body is said to be moving with uniform speed, if it covers equal distances in equal intervals of time. ...

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