Density - Revision history

Dmunagapati at 18:54, 2 December 2025

2025-12-02T18:54:29Z

2025-12-02T18:38:41Z

← Older revision		Revision as of 14:38, 2 December 2025
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	This topic covers Density -- Claimed by ~~Ritvik Khanna~~, ~~M08 Dr. Greco Spring 2016~~		This topic covers Density -- Claimed by Deepthi Munagapati, Fall 2025

2019-06-01T20:26:27Z

Difficult

← Older revision		Revision as of 16:26, 1 June 2019
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	e) Determine h in the volume of a cylinder:		e) Determine h in the volume of a cylinder:

	166.664 ~~cm3~~ = ~~(3.14159)~~ (8.00 ~~x 10¯3 cm~~)2 h		<math>166.664 </math> cm^3 = <math> \pi \cdot (8.00 \cdot 10^{-3})^{2}</math> cm <math>\cdot h</math>
	h ~~= 8.29 x 105 cm~~ = 5.63 ~~x 103~~ m
			<math> h = 5.63 \cdot 10^{3} </math> m

	==Connectedness==		==Connectedness==

2019-06-01T20:23:57Z

Difficult

← Older revision		Revision as of 16:23, 1 June 2019
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	<math>3.8808</math> lbs / <math>2.6046</math> lbs kg^-1 = 1.489979 kg (keep a few guard digits)		<math>3.8808</math> lbs / <math>2.6046</math> lbs kg^-1 = 1.489979 kg (keep a few guard digits)

	1.489979 kg = 1489.979 g		1.489979 kg = 1489.979 g

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	<math> 8.94 </math> g/cm^3 = <math> 1489.979 </math> g / V		<math> 8.94 </math> g/cm^3 = <math> 1489.979 </math> g / V

	V = 166.664 cm^3		V = 166.664 cm^3

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	d) Convert the diameter in inches to a radius in centimeters:		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		diameter = <math>6.304 \cdot 10^{-3}</math> in; radius = <math>3.152 \cdot 10^{-3}</math> in
	(3.152 ~~x 10¯3 inch) (~~2.54 cm / ~~inch)~~ = 8.00 ~~x 10¯3~~ cm
			<math>3.152 \cdot 10^{-3}</math> in <math>\cdot</math> <math> 2.54 </math> cm/in = <math>8.00 \cdot 10^{-3} cm</math>

	e) Determine h in the volume of a cylinder:		e) Determine h in the volume of a cylinder:

2019-06-01T20:21:20Z

Difficult

← Older revision		Revision as of 16:21, 1 June 2019
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	<math>3.8808</math> lbs / <math>2.6046</math> lbs kg^-1 = 1.489979 kg (keep a few guard digits)		<math>3.8808</math> lbs / <math>2.6046</math> lbs kg^-1 = 1.489979 kg (keep a few guard digits)
	1.489979 kg => 1489.979 g		1.489979 kg = 1489.979 g

	c) Determine the volume that this mass of copper occupies:		c) Determine the volume that this mass of copper occupies:

	<math> 8.94 </math> g/cm^3 = <math> 1489.979 <math> g / V		<math> 8.94 </math> g/cm^3 = <math> 1489.979 </math> g / V
	V = 166.664 cm^3		V = 166.664 cm^3

2019-06-01T20:19:32Z

Difficult

← Older revision		Revision as of 16:19, 1 June 2019
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	b) Convert pounds to kilograms:		b) Convert pounds to kilograms:

	3.8808 lbs ÷ 2.6046 lbs ~~kg¯1~~ = 1.489979 kg (~~I'm keeping~~ a few guard digits)		<math>3.8808</math> lbs / <math>2.6046</math> lbs kg^-1 = 1.489979 kg (keep a few guard digits)
	1.489979 kg = 1489.979 g		1.489979 kg => 1489.979 g

	c) Determine volume this ~~amount~~ of copper occupies:		c) Determine the volume that this mass of copper occupies:

	8.94 g ~~cm¯3~~ = 1489.979 g / x		<math> 8.94 </math> g/cm^3 = <math> 1489.979 <math> g / V
	x = 166.664 ~~cm3~~		V = 166.664 cm^3

	Note: this is the volume of the cylinder.		Note: this is the volume of the cylinder.

2019-06-01T20:16:09Z

Difficult

← Older revision		Revision as of 16:16, 1 June 2019
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	=== Difficult ===		=== Difficult ===

	Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = <math> 6.304 \cdot 10^{-3}</math> inches) can be produced from the copper that is in <math>5.88</math> lbs of covellite, an ore of copper that is <math>66%</math> copper by mass? (Hint: treat the wire as a cylinder. The density of copper is <math>8.94</math> <math>\frac{g}{cm^3}; one kg weighs <math>2.2046 </math> lbs. The volume of a cylinder is <math> V_{cyl} = \pi \cdot r^2 \cdot h </math>.		Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = <math> 6.304 \cdot 10^{-3}</math> inches) can be produced from the copper that is in <math>5.88</math> lbs of covellite, an ore of copper that is <math>66%</math> copper by mass? (Hint: treat the wire as a cylinder. The density of copper is <math>8.94</math> <math>\frac{g}{cm^3} </math>; one kg weighs <math>2.2046 </math> lbs. The volume of a cylinder is <math> V_{cyl} = \pi \cdot r^2 \cdot h </math>.

2019-06-01T20:15:37Z

Difficult

← Older revision		Revision as of 16:15, 1 June 2019
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	=== Difficult ===		=== Difficult ===

	Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter =<math>6.304 x 10^ {-3}</math> inches) can be produced from the copper that is in<math>5.88</math> ~~pounds~~ of covellite, an ore of copper that is <math>66%</math> copper by mass? (Hint: treat the wire as a cylinder. The density of copper is <math>8.94</math>g/ cm^3; one kg weighs <math>2.2046 </math> lb. The volume of a cylinder is ~~πr2h~~.		Copper can be drawn into thin wires. How many meters of 34-gauge wire (diameter = <math> 6.304 \cdot 10^{-3}</math> inches) can be produced from the copper that is in <math>5.88</math> lbs of covellite, an ore of copper that is <math>66%</math> copper by mass? (Hint: treat the wire as a cylinder. The density of copper is <math>8.94</math> <math>\frac{g}{cm^3}; one kg weighs <math>2.2046 </math> lbs. The volume of a cylinder is <math> V_{cyl} = \pi \cdot r^2 \cdot h </math>.


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

	5.88 lbs x 0.66 = 3.8808 lbs		<math>5.88 </math> lbs <math>\cdot 0.66 = 3.8808 </math> lbs

	b) Convert pounds to kilograms:		b) Convert pounds to kilograms:

2019-06-01T20:09:51Z

Middling

2019-06-01T20:05:51Z

Middling

← Older revision		Revision as of 16:05, 1 June 2019
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	b) Determine h in the volume of a cylinder (the wire)		b) Determine h in the volume of a cylinder (the wire)

	<math>~~240.49217~~</math> ~~cm^3 =~~ <math>\pi</math> * <math>0.1135</math>2 {h}		This "height" translates to the length of the wire, since the wire can essentially be treated as a long cylinder.

			<math> V = \pi \cdot r^2 \cdot h </math>

			Isolating for height (h):

			<math> h = \frac{V}{\pi \cdot r^2} </math>

			where r = radius

			<math> h = \frac{240.49217 cm^3}{\pi \cdot 0.1135 cm^2} </math>

	{h} = <math>5942.37</math> cm		{h} = <math>5942.37</math> cm