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2 years ago

Which items are NOT found on a

Engineering

1 answer:

Alja [10]2 years ago

3 0

Answer:

None of the above cause thats what i put

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A 1.00 liter solution contains 0.46 M hydrocyanic acid and 0.35 M potassium cyanide If 25.0 mL of water are added to this system

victus00 [196]

I won leader solution contain 0.46 mL of hydronic I said of 0.3 potassium

4 0

2 years ago

Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t

Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

$h=\frac{2\sigma cos(\alpha )}{\rho gd}$

where

$\sigma$ is the surface tension of the liquid

$\alpha$ is the contact angle of the liquid

$\rho$ is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

$\sigma =7.28\times 10^{-2}N/m$

$\alpha =0$

$\rho _{w}=1000kg/m^3$

Applying the values we get

$h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm$

Part b) When the liquid is mercury:

For mercury and glass we have

$\sigma =485.5\times 10^{-3}N/m$

$\alpha =138^o$

$\rho _{w}=13.6\times 10^{3}kg/m^3$

Applying the values we get

$h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm$

The negative sign indicates that there is depression in mercury in the tube.

4 0

3 years ago

Kam

jolli1 [7]

I think D i’m not sure

7 0

3 years ago

Water flows with an average speed of 6.5 ft/s in a rectangular channel having a width of 5 ft The depth of the water is 2 ft.

lisov135 [29]

Answer:

specific energy = 2.65 ft

y2 = 1.48 ft

Explanation:

given data

average speed v = 6.5 ft/s

width = 5 ft

depth of the water y = 2 ft

solution

we get here specific energy that is express as

specific energy = y + $\frac{v^2}{2g}$ ...............1

put here value and we get

specific energy = $2 + \frac{6.5^2}{2\times 9.8\times 3.281}$

specific energy = 2.65 ft

and

alternate depth is

y2 = $\frac{y1}{2} \times (-1+\sqrt{1+8Fr^2})$

and

here Fr² = $\frac{v1}{\sqrt{gy}} = \frac{6.5}{\sqrt{32.8\times 2}}$

Fr² = 0.8025

put here value and we get

y2 = $\frac{2}{2} \times (-1+\sqrt{1+8\times 0.8025^2})$

y2 = 1.48 ft

7 0

3 years ago

Which items are NOT found on a

Alja [10]

Answer:

None of the above cause thats what i put

3 0

2 years ago