#$&*
course Mth 173
10/7 16:00
Question Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
• If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
• If there are 1.323 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?
Answer
For a sand pile we know that the volume of the sand pile is directly proportional to the cube of the height of the cone, thus we have V = a( h^3), where a is a proportionality constant.
Also we know density = mass / volume, and since density remains the same and thus mass is directly proportional to the volume. Thus mass will be directly proportional to the cube of the height.
Thus m = k(h^3) where k is the proportionality constant , m is the mass and h is the height.
We know that for h = 2.1 meters we have mass of 5000.939 kg, thus
5000.939 = k ( 2.1 ^ 3 ) = k * 9.261
Thus k = 539.99989 kg / cubic meter which is approximately k = 540 kg / cubic meter.
Thus in order to calculate mass for a sand pile of 6.1 meters high
We know k and h = 6.1 meters
Thus m = 540 kg / cubic meter * ( 6.1 ^ 3 cubic meter ) = 540 kg/ cubic meter * 226.981 cubic meter = 122569.74 kg is the mass of the sand pile 6.1 meters high.
Grains of sand exposed on the surface of first sand pile = 1.323 billion
We saw that the volume was proportional to the cube of height, the surface area of first sand pile will be proportional to the square of the height.
The surface area will also be directly proportional to the number of grains exposed thus the number will be directly proportional to the square of height.
Let number of grains exposed be N
Thus we have N = k1 * ( h^2 ) , where k1 is the number of grains exposed function proportionality.
Thus for h = 2.1 meters we have N = 1.323 billion.
Thus 1.323 billion = k1 * ( 2.1 * 2.1 square meter )
k1 = 0.3 billion / square meter
next for height h = 6.1 meters to calculate grains of sand exposed we use the value of k1
thus we have N(6.1) = 0.3 billion / square meter * ( 6.1 * 6.1 square meter )
thus N(6.1) = 11.163 billion
thus we would expect 11.163 billion sand grains exposed on the surface of the second sand pile.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#$&*
course Mth 173
10/7 16:00
Question Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
• If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
• If there are 1.323 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?
Answer
For a sand pile we know that the volume of the sand pile is directly proportional to the cube of the height of the cone, thus we have V = a( h^3), where a is a proportionality constant.
Also we know density = mass / volume, and since density remains the same and thus mass is directly proportional to the volume. Thus mass will be directly proportional to the cube of the height.
Thus m = k(h^3) where k is the proportionality constant , m is the mass and h is the height.
We know that for h = 2.1 meters we have mass of 5000.939 kg, thus
5000.939 = k ( 2.1 ^ 3 ) = k * 9.261
Thus k = 539.99989 kg / cubic meter which is approximately k = 540 kg / cubic meter.
Thus in order to calculate mass for a sand pile of 6.1 meters high
We know k and h = 6.1 meters
Thus m = 540 kg / cubic meter * ( 6.1 ^ 3 cubic meter ) = 540 kg/ cubic meter * 226.981 cubic meter = 122569.74 kg is the mass of the sand pile 6.1 meters high.
Grains of sand exposed on the surface of first sand pile = 1.323 billion
We saw that the volume was proportional to the cube of height, the surface area of first sand pile will be proportional to the square of the height.
The surface area will also be directly proportional to the number of grains exposed thus the number will be directly proportional to the square of height.
Let number of grains exposed be N
Thus we have N = k1 * ( h^2 ) , where k1 is the number of grains exposed function proportionality.
Thus for h = 2.1 meters we have N = 1.323 billion.
Thus 1.323 billion = k1 * ( 2.1 * 2.1 square meter )
k1 = 0.3 billion / square meter
next for height h = 6.1 meters to calculate grains of sand exposed we use the value of k1
thus we have N(6.1) = 0.3 billion / square meter * ( 6.1 * 6.1 square meter )
thus N(6.1) = 11.163 billion
thus we would expect 11.163 billion sand grains exposed on the surface of the second sand pile.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!
#$&*
course Mth 173
10/7 16:00
Question Solve using proportionalities by stating the appropriate proportionality law and finding the proportionality constant:
• If a sand pile 2.1 meters high has a mass of 5000.939 kg, then what would we expect to be the mass of a geometrically similar sand pile 6.1 meters high?
• If there are 1.323 billion grains of sand exposed on the surface of the first sand pile, how many grains of sand we expect to be exposed on the surface of the second?
Answer
For a sand pile we know that the volume of the sand pile is directly proportional to the cube of the height of the cone, thus we have V = a( h^3), where a is a proportionality constant.
Also we know density = mass / volume, and since density remains the same and thus mass is directly proportional to the volume. Thus mass will be directly proportional to the cube of the height.
Thus m = k(h^3) where k is the proportionality constant , m is the mass and h is the height.
We know that for h = 2.1 meters we have mass of 5000.939 kg, thus
5000.939 = k ( 2.1 ^ 3 ) = k * 9.261
Thus k = 539.99989 kg / cubic meter which is approximately k = 540 kg / cubic meter.
Thus in order to calculate mass for a sand pile of 6.1 meters high
We know k and h = 6.1 meters
Thus m = 540 kg / cubic meter * ( 6.1 ^ 3 cubic meter ) = 540 kg/ cubic meter * 226.981 cubic meter = 122569.74 kg is the mass of the sand pile 6.1 meters high.
Grains of sand exposed on the surface of first sand pile = 1.323 billion
We saw that the volume was proportional to the cube of height, the surface area of first sand pile will be proportional to the square of the height.
The surface area will also be directly proportional to the number of grains exposed thus the number will be directly proportional to the square of height.
Let number of grains exposed be N
Thus we have N = k1 * ( h^2 ) , where k1 is the number of grains exposed function proportionality.
Thus for h = 2.1 meters we have N = 1.323 billion.
Thus 1.323 billion = k1 * ( 2.1 * 2.1 square meter )
k1 = 0.3 billion / square meter
next for height h = 6.1 meters to calculate grains of sand exposed we use the value of k1
thus we have N(6.1) = 0.3 billion / square meter * ( 6.1 * 6.1 square meter )
thus N(6.1) = 11.163 billion
thus we would expect 11.163 billion sand grains exposed on the surface of the second sand pile.
"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
#*&!#*&!
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