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poizon [28]
2 years ago
11

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
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