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Brilliant_brown [7]
3 years ago
8

Can someone pls give me the answer to this?

Engineering
1 answer:
olganol [36]3 years ago
7 0

I think option c 12 is currect

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.Name the control group ___________________________ 2.Name the experimental group ____________________ 3.Name the independent va
PIT_PIT [208]

Answer:

please

Explanation:

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3 years ago
The velocity distribution for laminar flow between parallel plates is given by u umax = 1 − ( 2y h ) 2 Where h is the distance s
Lynna [10]

Answer:

Explanation:

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5 0
4 years ago
Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.
marin [14]

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

8 0
3 years ago
A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.
Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature T_1=8^{\circ}C

T_2=28^{\circ}C

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ

We know that COP=\frac{Q_1}{W}

1.5=\frac{Q_1}{450}

Q_1=675 W=0.675 KW

so time required to cool the watermelon is

t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec  

4 0
3 years ago
How can I start a new project on solar panels building?​
Alborosie

Answer:so you can start by thinking about it and creating ideas that work well with it

Explanation:

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3 0
3 years ago
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