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Snowcat [4.5K]
3 years ago
14

I need help with this problem!

Mathematics
2 answers:
finlep [7]3 years ago
7 0

Answer:

104

Step-by-step explanation:

20 + (20*4) + (20*0.2)

= 20 + 80 + 4

= 104

Paha777 [63]3 years ago
5 0

Answer:

104

Step-by-step explanation:

100%= 1.00

so

420%= 4.20

Multiply 4.20×20

84

now add the original 20

104. She has 104 boxes in total.

Hope this helps! If you have any questions on how I got my answer feel free to ask. Stay safe!

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Ammonia at 70 F with a quality of 50% and a total mass of 4.5 lbm is in a rigid tank with an outlet valve at the bottom. How muc
Inessa05 [86]

Answer:

0.10865 killograms

Step-by-step explanation:

calculating the liquid mass of ammonia removed through the bottom value from a rigid tank at constant temperature.

Given:

temperature: T=70 F

quality : 50% = 0.5

initial mass: m1= 4.5 lbm

to find the removed liquid mass first we have to find total volume from which we can find remaining mass. as the tang is rigid the temperature and volume remains constant.

by taking the difference of mass we can determine the mass of liquid removed.

 we have two phases at temperature T= 70 F with specific volume for liquid  vf=0.02631 ft^3/lbm  and specific volume for vapor is  vg=2.3098 ft^3/lbm .

The Volume in the initial state is given by, (Using definition of specific volume)

                                          V= m1v1

 using v1=x(vf+vg)

                                    V= m1x(vf+vg)

substituting m1= 4.5 lbm\\ , vf= 0.02631 ft^3/lbm , vg=2.3098 ft^3/lbm

we get

        V= (4.5 lbm)(0.5)(0.02631 ft^3/lbm +2.3098 ft^3/lbm)  

finally          V=5.2625 ft^3  

we know the formula to find liquid mass is

mass =density *volume

density of ammonia is  0.73 kg/m^3

inserting the values into the formula we get the value for liquid mass removed through the valve.

m = (0.73 kg/m^3)*(5.25625 ft^3)

the final answer is

                           m= 0.10865 kg

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