Two people are standing in the middle of a frozen pond that is frictionless. They want to move to the edges of the pond but are
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Answer:
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Explanation:
<u>Question 6:</u>
The density of gold is 19.3 g/cm³
The density of silver is 10.5 g/cm³
- The density of the substance in Crown A;
Density = mass ÷ volume = = 19.3 g/cm³
Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.
- The density of the substance in Crown B;
Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)
Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.
- The density of substance in Crown C;
Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)
<h3><u>The density of the mixture:</u></h3><h3 />For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g
∴ for 1 cm³ of the mixture, its mass equal to = 14.9 g
Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.
* Crown C is not made up of a mixture of gold and silver.
<u>Question 7:</u>
<u />
- An empty masuring cylinder has a mass of 500 g.
- Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.
- The total mass is now 850 g
The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g
Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³
<u>Question 8:</u>
<u />
A tank filled with water has a volume of 0.02 m³
(a) 1 liter = 0.001 m³
How many liters? = 0.02 m³ ?
Cross multiplying gives:
= 20 liters
(b) 1 m³ = 1,000,000 cm³
0.02 m³ = how many cm³ ?
Cross-multiplying gives;
= 20,000 cm³
(c) 1 cm³ = 1 ml
∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml
<u>Question 9:</u>
<u />
Caliper (a) measurement = 3.2 cm
Caliper (b) measurement = 3 cm
<u>Question 10:</u>
<u />
- A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
- in a displacement can
- The mass of liquid which overflow is 20 g
The mass of the liquid which overflow = mass of the stone = 20 g
1 gram of the liquid occupies 1 cm³ of space.
20 g of the liquid will occupy; = 20 cm³
(a) Since the volume of the water displaced is equal to the volume of the stone.
∴ The volume of the stone = 20 cm³
(b) Mass = density × volume
Density of the stone = 5.0 g/cm³
Volume of the stone = 20 cm³
Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g
Answer:
The 5-number summary is
1. Median = 0.93 W/kg
2. Lower quartile = 0.69 W/kg
3. Upper quartile = 1.16 W/kg
4. Minimum value = 0.54 W/kg
5. Maximum value = 1.42 W/kg
Explanation:
We are given the measured radiation absorption rates (in W/kg) corresponding to 11 cell phones.
1.16 0.85 0.69 0.75 0.95 0.93 1.18 1.17 1.42 0.54 0.57
What is 5-number summary?
A 5-number summary refers to a box plot that basically shows 5 statistical characteristics of a data set.
These statistical characteristics are:
1. Median
2. Lower quartile
3. Upper quartile
4. Minimum value
5. Maximum value
1. Median:
Arrange the data in ascending order
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
(n+1)/2 gives the median value of the data set.
(11 + 1)/2 = 6th position
Therefore, 0.93 W/kg is the median of the data set.
2. Lower quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The lower quartile is the median of the lower half of the data set.
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
The median is 6/2 = 3rd position
Therefore, the lower quartile of the data set is 0.69 W/kg
3. Upper quartile:
Divide the data set into two equal halfs (include median in both if n = odd)
Lower half = 0.54 0.57 0.69 0.75 0.85 0.93
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The upper quartile is the median of the lower half of the data set.
Upper half = 0.93 0.95 1.16 1.17 1.18 1.42
The median is 6/2 = 3rd position
Therefore, the upper quartile of the data set is 1.16 W/kg
4. Minimum value:
The minimum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the minimum value of the data set is 0.54 W/kg
5. Maximum value
The maximum value is the least value in the data set.
0.54 0.57 0.69 0.75 0.85 0.93 0.95 1.16 1.17 1.18 1.42
Therefore, the maximum value of the data set is 1.42 W/kg
The box plot is illustrated in the attached diagram.
