The Main Show Tank Calculations In Physics
Calculations-
1. Main Show Tank Calculation:
The main show tanks radius is 70 feet. Based on that the tank is a sphere-shaped and we have to find the volume, I used the formula for the sphere volume which is: V= 4/3(3.14 x r^3)
V= 4/3(3.14 x 70^3)
V= 4/3(1,077,020)
V= 1,436,026.66666...
V= 1,436,027
But I have to find the volume of the quarter sphere-shaped, so:
V= 1/4 x 1,436,027
V= 359,007
The volume of the quarter-shaped tank is
359,007 cubic feet
Holding Tank Calculations-
2. I used the formula of volume of cylinders. The cylinders are cut in two congruent halves, with a height of 120 and a radius of 15.
V= 3.14 x r^2 x h
V= 3.14 x 15^2 x 120
V= 3.14 x 225 x 120
V= 84,780. (both tanks)
Both of the tanks volume is 84,780 cubic …show more content…
Reflection
5. If each dimension is made 6 times smaller, the new dimensions would be: radius = 70/6 r = 11.67 Because is the quarter of the volume:
V= 4/3 (3.14 x r^3) V= 6,653.97175176/ 4
V= 4/3 (3.14 x 11.67^3) V= 1,663.49293794
V= 6,653.97175176 V= 1,663 cubic feet.
The actual volume of the tank is 359,007 and the volume of the mock-up is 1,663 cubic feet, so
359,007/ 1,663 = 215.879134095009 rounded to 216
The volume of the mock up is 216 times smaller than the actual volume
6. The volume of the actual tank is 1.9 % of the mock-up of the tank.
Volume of the actual tank is 359,007
Volume of the mock up is 6,654.
P/T = %/100
6,654/359,007 = % / 100
6,654 x 100 / 359,007 = %
1.853445754539605 = %
1.9 = %
7. Based on the holding tanks beings cylinders and the base of them being curved, if we do a cross section parallel to the base the shape would be a sphere.
8. The chart shows that the grams of salt per liter of seawater is decreasing and the density is decreasing too. To solve the problem I would put 35 g of salt and keep the density in 3.5 % in each tank, which is normal condition to maintain aquatic
1. Main Show Tank Calculation:
The main show tanks radius is 70 feet. Based on that the tank is a sphere-shaped and we have to find the volume, I used the formula for the sphere volume which is: V= 4/3(3.14 x r^3)
V= 4/3(3.14 x 70^3)
V= 4/3(1,077,020)
V= 1,436,026.66666...
V= 1,436,027
But I have to find the volume of the quarter sphere-shaped, so:
V= 1/4 x 1,436,027
V= 359,007
The volume of the quarter-shaped tank is
359,007 cubic feet
Holding Tank Calculations-
2. I used the formula of volume of cylinders. The cylinders are cut in two congruent halves, with a height of 120 and a radius of 15.
V= 3.14 x r^2 x h
V= 3.14 x 15^2 x 120
V= 3.14 x 225 x 120
V= 84,780. (both tanks)
Both of the tanks volume is 84,780 cubic …show more content…
Reflection
5. If each dimension is made 6 times smaller, the new dimensions would be: radius = 70/6 r = 11.67 Because is the quarter of the volume:
V= 4/3 (3.14 x r^3) V= 6,653.97175176/ 4
V= 4/3 (3.14 x 11.67^3) V= 1,663.49293794
V= 6,653.97175176 V= 1,663 cubic feet.
The actual volume of the tank is 359,007 and the volume of the mock-up is 1,663 cubic feet, so
359,007/ 1,663 = 215.879134095009 rounded to 216
The volume of the mock up is 216 times smaller than the actual volume
6. The volume of the actual tank is 1.9 % of the mock-up of the tank.
Volume of the actual tank is 359,007
Volume of the mock up is 6,654.
P/T = %/100
6,654/359,007 = % / 100
6,654 x 100 / 359,007 = %
1.853445754539605 = %
1.9 = %
7. Based on the holding tanks beings cylinders and the base of them being curved, if we do a cross section parallel to the base the shape would be a sphere.
8. The chart shows that the grams of salt per liter of seawater is decreasing and the density is decreasing too. To solve the problem I would put 35 g of salt and keep the density in 3.5 % in each tank, which is normal condition to maintain aquatic