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It will not, it would be 2 m from the top of the tank. If I use archimede's theory, if i put the cube into the water, the water level will rise by the volume of the cube which is 8m cubed. Since the base area of the tank is 16 m squared, the height would only increase .5m, making it 2 m from the top of the tank.
Grade 7 Once again, I will just do the math. ABCD = 560 cm^2 7 Rectangles I will call one side length x and the other y. Here is some useful information that can be identified through the setup: 5x = 1y = AB/DC 2x + 1y = 7x = AD/BC I can now just use ratios in a way to find the answer. 5x x 7x = 560cm^2 For this to work, x should be a small value. 5 x 7 = 35 10 x 14 = 140 15 x 21 = 315 20 x 28 = 560 x = 4 and the large rectangle has dimensions of 20 and 28. That means one side length of the small rectangle is 4 and the other is 4x5 or 20. The dimensions of the small rectangle is 4 cm and 20 cm.
Grade 8 I will once again go straight to the math. The volume of the tank is 5 x 4 x 4 = 80 m^3. The volume of the water filled is 2.5 x 4 x 4 = 40 m^3. The volume of the cube is 2 x 2 x 2 = 8 m^3. The volume would then increase to 8 + 40 = 48 m^3. The volume of the water filled is now 48/(4 x 4) = 48/16 = 3 m. It's 3 m high so the water would reach 5 - 3 = 2 m below the top of the tank.
Grade 7 POTW: Info: - Rectangle made up of 7 smaller rectangle - Total Area is 560cm^2. - Find demensions of smaller rectangles.
Since there are 7 rectangles to made a rectangle that has an area of 560cm^2, you can divide 560 by 7 to find the area of one rectangle. 560/7=80. The area of one smaller rectangle is 80. Since we do not know even one side of the smaller rectangle, we can make the side lengths any factor of 80. The possible side lengths are: 1,2,4,5,8,10,16,20,40,80. The side lengths can be any of those numbers and its factor pair. :D (Did I do this wrong? Seems kind of easy for a POTW........)
Grade 8 POTW (because why not): Info: - Tank: 4x4x5 - Water is filled up to 2.5 m - Cube (2x2x2) is thrown in. - Find if the water will reach the top of the tank - If not, how far below is the water from the top.
I don't know if this is the right way, but the way to do this is to find the volume of the tank that has water, find the volume of the cube, "form" it to find on top of the tank's water level. Make sense? Anyways, the volume of the cube is 8m^3 and the volume of the tank's water is 40 m^3. Then, if the cube is thrown in, the water level will rise the same volume as the cube. So, I need to change some of the cube's dimensions so that the area is the same, but it is 4 x 4 instead. This may sound "impossible" but the height can be a decimal. To make this new "cube", the height can be 0.5 So, to find the height of the new "water level", I add 2.5 and 0.5 (3). The height of the new "water level"is 3. The water does not reach the top of the tank and it is 2 m below the top of the tank. :DDD
The water doesn't reach the top of the tank, and the water's height is 3m, which means that the water height is 2m below the top of the tank. I did my work on paper.
We can see that there multiple components to the question: - A tank with the dimensions of 4*4*5 - Water with the capacity of 4*4*2.5 - A cube with the dimensions of 2*2*2
From here, we can find the volume or capacity of each of the following items. Cube: 2*2*2 = 8 m^3 Tank: 4*4*5 = 80 m^3 Water: 4*4*2.5 = 40 m^3 By dropping the cube inside the water filled tank, it will displace a certain amount of the water. To the find the remaining height, we can use archimedes' principle (grade 8 science). Since the base area is 16 m^2, we can divide that by the total volume of the water plus the volume of the cube. 48/16 = 3 That would be the new height. This means that the water does not reach the top of the tank, making it 2 m off from the reaching it.
28 x 20 = 560 560/7 = 80 Divide by 7 because there are 7 rectangles for area. 4*20 works as dimensions so one side would be 20 cm. The other side would then be 20 + 4 + 4 = 28. 28*20 = 560.
Sorry I forgot to do this one! To save time from typing it, I did it on paper and used a calculator. The water is 2m below from the top part of the tank.
For this question, we need to use our knowledge of Archimedes Principle. But first, let's get some general measurements down..
Volume of the cube: 8 m ^3 Volume of the tank: 80 m ^3 Volume of the water inside: 40 m^3
So, now that we know those measurements, we can move on to the actual question. Since the level of the water is going to rise once the cube is placed inside (science knowledge), we need to add the volume of the cube to the volume of the water inside, which gives us 48 m ^3. Finally, all we have to do now is divide that by the area of the base of the tank which is 16 m ^2. So, the level of he water would be 3 m once the cube is placed in.
However, since the tank is 5 m high, there would still be 2 m of space left between the top of the tank and the top of the water. That is my answer =)
Volume of tank: V = L*W*H V = 4*4*5 V = 80m^3 The volume of the tank is 80m^3
Volume of water: V = 4*4*2.5 V = 40 The volume of the water inside the tank is 40m^3
Volume of cube: V = 2*2*2 V = 8 The volume of the cube is 8m^3.
When you place the cube in the tank, the water level rises due to displacement. If we add the volume of the cube to the volume of the water in the tank, we will get the level of the water with the cube inside.
40 + 8 = 48 Next, I need to find the dimensions. I know that 2 of the sides are 4m and 4m. 48/4/4 = 3 The water level would be 3 m high.
The water would not reach the top of the tank and there is 2m left until it reaches the top.
It will not, it would be 2 m from the top of the tank. If I use archimede's theory, if i put the cube into the water, the water level will rise by the volume of the cube which is 8m cubed. Since the base area of the tank is 16 m squared, the height would only increase .5m, making it 2 m from the top of the tank.
ReplyDeleteGrade 7
ReplyDeleteOnce again, I will just do the math.
ABCD = 560 cm^2
7 Rectangles
I will call one side length x and the other y. Here is some useful information that can be identified through the setup:
5x = 1y = AB/DC
2x + 1y = 7x = AD/BC
I can now just use ratios in a way to find the answer.
5x x 7x = 560cm^2
For this to work, x should be a small value.
5 x 7 = 35
10 x 14 = 140
15 x 21 = 315
20 x 28 = 560
x = 4 and the large rectangle has dimensions of 20 and 28. That means one side length of the small rectangle is 4 and the other is 4x5 or 20.
The dimensions of the small rectangle is 4 cm and 20 cm.
Grade 8
ReplyDeleteI will once again go straight to the math.
The volume of the tank is 5 x 4 x 4 = 80 m^3.
The volume of the water filled is 2.5 x 4 x 4 = 40 m^3.
The volume of the cube is 2 x 2 x 2 = 8 m^3.
The volume would then increase to 8 + 40 = 48 m^3.
The volume of the water filled is now 48/(4 x 4) = 48/16 = 3 m.
It's 3 m high so the water would reach 5 - 3 = 2 m below the top of the tank.
560 / 7 = 80
ReplyDelete/7 because there are 7 rectangles just like the question says
each rectangle = 80 cm squared
the dimensions that looks to be right would be 4 x 20.
I know no math problems are to scale but still......
now lets test it:
one side would be 20cm
the other would be 20 + 4 + 4 = 28
28 x 20 = 560
so hurray...
therefore the dimension of the inner rectangles are 4 cm by 20 cm
Grade 7 POTW:
ReplyDeleteInfo:
- Rectangle made up of 7 smaller rectangle
- Total Area is 560cm^2.
- Find demensions of smaller rectangles.
Since there are 7 rectangles to made a rectangle that has an area of 560cm^2, you can divide 560 by 7 to find the area of one rectangle. 560/7=80. The area of one smaller rectangle is 80. Since we do not know even one side of the smaller rectangle, we can make the side lengths any factor of 80. The possible side lengths are:
1,2,4,5,8,10,16,20,40,80. The side lengths can be any of those numbers and its factor pair.
:D
(Did I do this wrong? Seems kind of easy for a POTW........)
Everything was right so far except for the parts that we don't know the side lengths. We have to find it out. Not everything will be given like that.
DeleteLook at the solution provided in the POTW #18 posting Fiona!
DeleteGrade 8 POTW (because why not):
ReplyDeleteInfo:
- Tank: 4x4x5
- Water is filled up to 2.5 m
- Cube (2x2x2) is thrown in.
- Find if the water will reach the top of the tank
- If not, how far below is the water from the top.
I don't know if this is the right way, but the way to do this is to find the volume of the tank that has water, find the volume of the cube, "form" it to find on top of the tank's water level. Make sense? Anyways, the volume of the cube is 8m^3 and the volume of the tank's water is 40 m^3. Then, if the cube is thrown in, the water level will rise the same volume as the cube. So, I need to change some of the cube's dimensions so that the area is the same, but it is 4 x 4 instead. This may sound "impossible" but the height can be a decimal. To make this new "cube", the height can be 0.5
So, to find the height of the new "water level", I add 2.5 and 0.5 (3). The height of the new "water level"is 3.
The water does not reach the top of the tank and it is 2 m below the top of the tank.
:DDD
Grade 8 POTW
ReplyDeleteThe water doesn't reach the top of the tank, and the water's height is 3m, which means that the water height is 2m below the top of the tank. I did my work on paper.
Grade 8 POTW:
ReplyDeleteWe can see that there multiple components to the question:
- A tank with the dimensions of 4*4*5
- Water with the capacity of 4*4*2.5
- A cube with the dimensions of 2*2*2
From here, we can find the volume or capacity of each of the following items.
Cube: 2*2*2 = 8 m^3
Tank: 4*4*5 = 80 m^3
Water: 4*4*2.5 = 40 m^3
By dropping the cube inside the water filled tank, it will displace a certain amount of the water. To the find the remaining height, we can use archimedes' principle (grade 8 science).
Since the base area is 16 m^2, we can divide that by the total volume of the water plus the volume of the cube.
48/16 = 3
That would be the new height. This means that the water does not reach the top of the tank, making it 2 m off from the reaching it.
28 x 20 = 560
ReplyDelete560/7 = 80
Divide by 7 because there are 7 rectangles for area.
4*20 works as dimensions so one side would be 20 cm. The other side would then be 20 + 4 + 4 = 28. 28*20 = 560.
What are the two sides?
DeleteBased on Archimedes' principal and displacement, the water should be 2 metres from the top of the tank
ReplyDeleteSorry I forgot to do this one! To save time from typing it, I did it on paper and used a calculator. The water is 2m below from the top part of the tank.
ReplyDeletePOTW: (Sorry it's late!)
ReplyDeleteFor this question, we need to use our knowledge of Archimedes Principle. But first, let's get some general measurements down..
Volume of the cube: 8 m ^3
Volume of the tank: 80 m ^3
Volume of the water inside: 40 m^3
So, now that we know those measurements, we can move on to the actual question. Since the level of the water is going to rise once the cube is placed inside (science knowledge), we need to add the volume of the cube to the volume of the water inside, which gives us 48 m ^3. Finally, all we have to do now is divide that by the area of the base of the tank which is 16 m ^2. So, the level of he water would be 3 m once the cube is placed in.
However, since the tank is 5 m high, there would still be 2 m of space left between the top of the tank and the top of the water. That is my answer =)
Grade 8 POTW:
ReplyDeleteVolume of tank:
V = L*W*H
V = 4*4*5
V = 80m^3
The volume of the tank is 80m^3
Volume of water:
V = 4*4*2.5
V = 40
The volume of the water inside the tank is 40m^3
Volume of cube:
V = 2*2*2
V = 8
The volume of the cube is 8m^3.
When you place the cube in the tank, the water level rises due to displacement.
If we add the volume of the cube to the volume of the water in the tank, we will get the level of the water with the cube inside.
40 + 8 = 48
Next, I need to find the dimensions.
I know that 2 of the sides are 4m and 4m.
48/4/4 = 3
The water level would be 3 m high.
The water would not reach the top of the tank and there is 2m left until it reaches the top.