Understand 5 1 1018 Cm

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Volume resistivity (Ω x cm) 1015 1012 – 1015 1017 1012 1018 1018 Tensile strength in N/mm2 (MPa) 5 – 20 10 – 25 15 – 30 15 – 45 15 – 40 20 – 25 Elongation at break in% 400 – 600 150 – 400 400 – 800 300 – 600 240 – 400 250 – 350 Water absorption (20°C) in% 1 – 2 0.4 0.1 1.5 0.01 0.01.

• #1010 5/8″ (1,6 cm) $ 3.91 exc. Sold #1012 1/2″ (1,3 cm) $ 3.98 exc. #1018 1/2″ (1,3 cm) (Non Metal Based) $ 2.
• Definition: A centimeter (symbol: cm) is a unit of length in the International System of Units (SI), the current form of the metric system. It is defined as 1/100 meters. History/origin: A centimeter is based on the SI unit meter, and as the prefix 'centi' indicates, is equal to one hundredth of a meter.

Density Calculations
Problems 11-25

Twenty Examples

Probs #1-10

Probs #26-50

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Problem #11: A cylindrical glass tube of length 27.75 cm and the radius 2.00 cm is filled with argon gas. The empty tube weighs 188.250 g. and the tube filled with argon weighs 188.870 g. Use the data to calculate the density of argon gas.

Solution:

1) Volume of a cylinder:

V = πr²h

V = (3.14159) (2.00 cm)² (27.75 cm)

V = 348.7165 cm³

2) Mass:

188.870 g - 188.250 g = 0.620 g

3) Density:

0.620 g / 0.3487165 L = 1.78 g/L

Note that 348.7165 cm³ became 0.3487165 liters. Gas density is typically measured in g/L as opposed to g/cm³ or g/mL. Reminders: 1 cm³ = 1 mL and 1000 mL = 1 L.

Problem #12: The density of silver is 10.50 g/cm³ and the density of benzene is 0.8786 g/cm³. What mass of silver will have the same volume as 15.55 grams of benzene?

Solution:

1) Determine the volume of benzene:

mass / density = volume

15.55 g / 0.8786 g/cm³ = 17.6986 cm³ Instaradio 8 0 1.

2) Determine the mass of silver:

density times volume = mass

(10.50 g/cm³) (17.6986 cm³) = 185.8 g

Problem #13: Calculate the mass of copper in grams (density = 8.96 g/cm³) with the same volume as 100.0 grams of gold (density = 19.31 g/cm³)

Solution:

1) Volume of gold:

100.0 g ÷ 19.31 g/cm³ = 5.17866 cm³

2) Mass of copper:

(8.96 g/cm³) (5.17866 cm³) = 46.4 g

3) Setting up the problem in dimensional analysis style:

1 cm3	8.96 g
100.0 g	x	–––––––	x	–––––––	= 46.4 g
19.31 g	1 cm3

Problem #14: Calculate the mass of zinc in grams (density = 7.14 g/cm³) with the same volume as 100.0 grams of aluminum (density = 2.70 g/cm³)

Solution:

1) Volume of aluminum:

100.0 g ÷ (2.70 g/cm³) = 37.037 cm³

2) Mass of zinc:

(7.14 g/cm³) (37.037 cm³) = 264 g

3) Dimensional analysis:

1 cm3	7.14 g
100.0 g	x	–––––––	x	–––––––	= 264 g
2.70 g	1 cm3

Problem #15: A spherical ball bearing has a radius of 8.50 mm and a mass of 2.315 g. Determine the density of the ball bearing in g/cm³.

Solution:

1) Convert mm to cm:

1 cm
8.50 mm	x	–––––––	= 0.850 cm
10 mm

2) Determine volume of sphere:

V = (4/3)πr³

(4/3) (3.14159) (0.850 cm)³ = 2.57 cm³

3) Calculate density:

2.315 g / 2.57 cm³ = 0.900 g/cm³

Problem #16: 57.0 kg of copper is drawn into a wire with a diameter of 9.50 mm. What is the length of wire in meters? Cu density = 8.96 g/cm³.

Solution:

1) Convert kg to grams:

1000 g
57.0 kg	x	–––––––	= 5.70 x 104 g
1 kg

2) Determine volume of the copper wire:

(5.70 x 10⁴ g) ÷ (8.96 g/cm³) = 6361.607 cm³

3) Convert mm to cm:

1 cm
9.50 mm	x	–––––––	= 0.950 cm
10 mm

3) Determine length of wire:

V = πr²h

6361.607 cm³ = (3.14159) (0.475 cm)² h

h = 8974.91 cm

To three sig figs and in meters, 89.7 m

Problem #17: In the United States, 'copper' pennies made since 1983 actually contain very little copper. If a penny contains 93.975% of its total volume zinc and 6.025% of its total volume copper, what is its apparent density? (density of Cu = 8.96 g/cm³; density of Zn = 7.14 g/cm³.)

Solution:

1) Assume the penny occupies 1.00 cm³. This means:

copper occupies 0.06025 cm³ and zinc occupies 0.93975 cm³.

2) Calculate mass of copper:

(0.06025 cm³) (8.96 g/cm³) = 0.53984 g

3) Calculate mass of zinc:

(0.93975 cm³) (7.14 g/cm³) = 6.709815 g

4) Determine apparent density:

0.53984 g + 6.709815 g = 7.249655 g

since this mass is in 1.00 cm³, the answer is 7.25 g/cm³

Problem #18: Antarctica has an ice sheet covering 1.42 x 10¹⁸ cm² and averaging 1.61 x 10⁵ cm deep. Calculate the total mass if ice has a density of 0.92 g/cm³.

Solution:

1) Calculate volume of ice:

(1.42 x 10¹⁸ cm²) (1.61 x 10⁵ cm) = 2.2862 x 10²³ cm³

2) Calculate mass of ice:

(2.2862 x 10²³ cm³) (0.92 g/cm³) = 2.1 x 10²³ g

Problem #19: Object A is less dense than object B. If both objects are the same mass, what can be said about the volume of A as compared to the volume of B?

Solution:

Object A has a larger volume than Object B.

Problem #20: An ice cube with a volume of 45.0 mL and a density of 0.900 g/cm³ floats in a liquid with a density of 1.36 g/mL. What volume of the cube is submerged in the liquid?

Solution:

The solution to this problem involves the concept of buoyancy.

1) Determine the mass of the cube:

(45.0 mL) (0.900 g/cm³) = 40.5 g

2) The cube will float when 40.5 g of liquid is displaced. We need to know what volume of the liquid weighs 40.5 g.

5 Foot 1

40.5 g) ÷ (1.36 g/mL) = 29.8 mL

This means that 29.8 mL of the cube is submerged (this is the answer to the question), displacing 40.5 g of the liquid. The rest of the cube (45.0 − 29.8) is above the surface of the liquid.

Problem #21: Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = 6.304 x 10¯³ inches) can be produced from the copper that is in 5.88 pounds of covellite, an ore of copper that is 66% copper by mass? (Hint: treat the wire as a cylinder. The density of copper is 8.96 g cm¯³; one kg weighs 2.2046 lb; 1 inch is 2.54 cm and the volume of a cylinder is πr²h)

Solution:

a) Determine pounds of pure copper in 5.88 lbs of covellite:

5.88 lbs x 0.66 = 3.8808 lbs

b) Convert pounds to kilograms:

3.8808 lbs ÷ 2.6046 lbs kg¯¹ = 1.489979 kg (I'm keeping a few guard digits)

1.489979 kg = 1489.979 g

c) Determine volume this amount of copper occupies:

8.96 g cm¯³ = 1489.979 g / x

x = 166.292 cm³

Note: this is the volume of the cylinder.

d) Convert the diameter in inches to a radius in centimeters:

dia = 6.304 x 10¯³ in; radius = 3.152 x 10¯³ in

(3.152 x 10¯³ inch) (2.54 cm / inch) = 8.00 x 10¯³ cm

e) Determine h in the volume of a cylinder:

166.292 cm³ = (3.14159) (8.00 x 10¯³ cm)² h

h = 8.27 x 10⁵ cm = 8.27 x 10³ m

Problem #22: A copper ingot has a mass of 2.15 kg. If the copper is drawn into wire whose diameter is 2.27 mm, how many inches of copper wire can be obtained from the ingot?

Solution:

a) Determine volume of copper:

8.96 g cm¯³ = 2150 g / x

x = 239.955 cm³

Note: this is the volume of the wire.

b) Determine h in the volume of a cylinder (i.e., the wire):

239.955 cm³ = (3.14159) (0.1135 cm)² h

h = 5929.097 cm

Note: 0.1135 cm is the radius

c) Convert cm to inch:

5929.097 cm divided by 2.54 cm/in = 2334.29 in

To three sig figs, the answer is 2330 in

Problem #23: If the copper is drawn into wire whose diameter is 8.06 mm, how many feet of copper can be obtained from a 200.0 pound ingot?

Solution:

a) Convert pounds to grams: Metamovie 2 3 1 – add metadata to your videos.

200.0 lb x (453.59 g/lb) = 9.0718 x 10⁴ g

b) Determine what volume is occupied by this many grams of copper:

8.96 g/cm³ = 9.0718 x 10⁴ g divided by x

x = 1.01248 x 10⁴ cm³

c) Determine the height of the cylinder (Volume of a cylinder = πr²h):

1.01248 x 10⁴ cm³ = (3.14159) (0.0403 cm)² h

h = 1.9844 x 10⁶ cm

d) Convert cm to inches, then to feet:

1.9822 x 10⁶ cm x (1 in/2.54 cm) = 7.8126 x 10⁵ in

7.8126 x 10⁵ in x (1 ft / 12 in) = 6.51 x 10⁴ ft (to three sig figs)

Problem #24: A cube of copper was found to have a mass of 0.630 kg. What are the dimensions of the cube? (The density of copper is 8.96 g/cm³.)

Solution:

a) Determine the volume of the cube (note that kg have been converted to g):

8.96 g/cm³ = 630. g / volume

volume = 70.3125 cm³

b) Each side of a cube is equal in length, so take cube root of the volume for length of cube side:

70.30125 cm³3 = 4.13 cm (to three sig figs)

Problem #25: Calculate the volume (in m³) of a 5,020 tonne iceberg. (1 tonne = 1,000 kg, the density of ice = 0.92 g/cm³)

Solution:

a) Convert tonnes to grams:

5,020 tonne x (1,000 kg / tonne) = 5.02 x 10⁶ kg

5.02 x 10⁶ kg x (1000 g / kg) = 5.02 x 10⁹ g

b) Determine volume in cubic centimeters:

0.92 g/cm³ = 5.02 x 10⁹ g / volume

volume = 5.4565 x 10⁹ cm³

c) Convert cubic centimeters to cubic meters:

5.4565 x 10⁹ cm³ x (1 m³ / 10⁶ cm³) = 5.46 x 10³ m³ (rounded to 3 significant figures)

Bonus Problem: A room contains 11.5 kg of argon. The air in the room consists of 0.225% argon. The density of air in the room is 1.70 g/L. What is the volume of the room in m³?

Solution #1:

a) 11.5 kg is 0.225% of the total mass in the room. Determine total mass (in grams) of air in room:

11.5 kg / 0.00225 = 5111 kg

5111 kg x (1000 g/kg) = 5.111 x 10⁶ g

b) Using density, determine volume of room in liters:

5.111 x 10⁶ g ÷ 1.70 g/L = 3.00 x 10⁶ L

c) Convert liters to cubic meters:

3.00 x 10⁶ L x (1 m³ / 1000 L) = 3.00 x 10³ m³

Solution #2:

5+1 News

a) Determine the liters that 11.5 kg of argon occupies:

11.5 kg x (1000 g/kg) = 11,500 g

11,500 g / 1.70 g/L = 6.7647 x 10³ L

b) The volume the Ar occupies represents 0.225% of the entire room. Determine the total volume of the room:

6.7647 x 10³ L / 0.00225 = 3.00 x 10⁶ L

c) Convert liters to cubic meters:

See step (c) in solution #1.

Twenty Examples

Probs #1-10

Probs #26-50

All the examples & problems, no solutions

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Please provide values below to convert inch [in] to centimeter [cm], or vice versa.

Inch

Definition: An inch (symbol: in) is a unit of length in the imperial and US customary systems of measurement. An inch was defined to be equivalent to exactly 25.4 millimeters in 1959. There are 12 inches in a foot and 36 inches in a yard.

History/origin: The term 'inch' was derived from the Latin unit 'uncia' which equated to 'one-twelfth' of a Roman foot.

5 1 Height Cm

There have been a number of different standards for the inch in the past, with the current definition being based on the international yard. One of the earliest definitions of the inch was based on barleycorns, where an inch was equal to the length of three grains of dry, round barley placed end-to-end. Another version of the inch is also believed to have been derived from the width of a human thumb, where the length was obtained from averaging the width of three thumbs: a small, a medium, and a large one.

Current use: The inch is mostly used in the United States, Canada, and the United Kingdom. It is also sometimes used in Japan (as well as other countries) in relation to electronic parts, like the size of display screens.

Centimeter

Definition: A centimeter (symbol: cm) is a unit of length in the International System of Units (SI), the current form of the metric system. It is defined as 1/100 meters.

History/origin: A centimeter is based on the SI unit meter, and as the prefix 'centi' indicates, is equal to one hundredth of a meter. Metric prefixes range from factors of 10^-18 to 10¹⁸ based on a decimal system, with the base (in this case the meter) having no prefix and having a factor of 1. Learning some of the more commonly used metric prefixes, such as kilo-, mega-, giga-, tera-, centi-, milli-, micro-, and nano-, can be helpful for quickly navigating metric units.

Current use: The centimeter, like the meter, is used in all sorts of applications worldwide (in countries that have undergone metrication) in instances where a smaller denomination of the meter is required. Height is commonly measured in centimeters outside of countries like the United States.

Inch to Centimeter Conversion Table

Inch [in]	Centimeter [cm]
0.01 in	0.0254 cm
0.1 in	0.254 cm
1 in	2.54 cm
2 in	5.08 cm
3 in	7.62 cm
5 in	12.7 cm
10 in	25.4 cm
20 in	50.8 cm
50 in	127 cm
100 in	254 cm
1000 in	2540 cm

How to Convert Inch to Centimeter

Understand 5 1 1018 Cm -

1 in = 2.54 cm
1 cm = 0.3937007874 in

Example: convert 15 in to cm:
15 in = 15 × 2.54 cm = 38.1 cm

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