Lead Shield Cone - Revision history

Vanwdani: /* Effects of DC/Tracker on Moller particles */

2018-01-23T22:30:46Z

Effects of DC/Tracker on Moller particles

2018-01-23T22:19:54Z

Effects of DC/Tracker on Moller particles

2018-01-23T22:11:53Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:15:54Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:14:34Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:12:26Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:09:27Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:07:48Z

Effects of DC/Tracker on Moller particles

2018-01-23T17:04:10Z

Effects of DC/Tracker on Moller particles

2018-01-23T16:59:47Z

Effects of DC/Tracker on Moller particles

@@ Line 184: / Line 184: @@
-<math>R_{\rm specific}=\frac {8.314 J / mol. K.}{40.354 g/mol}=335.505 \frac{J}{g \cdot K}</math>
+<math>R_{\rm specific}=\frac {8.314 J / mol. K.}{40.354 g/mol}=0.2060 \frac{J}{g \cdot K}</math>
@@ Line 192: / Line 192: @@
-<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>
+<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{0.2060 \frac{J}{g \cdot K} 293.15 K} = \frac{101325kg⋅m^{−1}⋅s^{−2}}{60.40 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>
-<math>\rho =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=0.0013mg/cm^3</math>
+<math>\rho =\frac{1678g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=1.678mg/cm^3</math>

← Older revision		Revision as of 22:19, 23 January 2018
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−	<math>\rho~~= \frac{1.01325000 g}{98353291 m^3}~~ =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=	+	<math>\rho =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=0.0013mg/cm^3</math>

← Older revision		Revision as of 22:11, 23 January 2018
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−	<math>\rho = \frac{~~101325000 g⋅m~~^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{~~101325000 g⋅m~~^{−1}⋅s^{−2}}{~~98353291~~ \frac{~~g⋅m~~^2⋅s^{−2}}{g}} </math>	+	<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>


−	<math>\rho= \frac{~~101325000~~ g}{98353291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=	+	<math>\rho= \frac{1.01325000 g}{98353291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=

← Older revision		Revision as of 17:15, 23 January 2018
Line 195:		Line 195:


−	<math>\rho= \frac{101325000 g}{98353291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}	+	<math>\rho= \frac{101325000 g}{98353291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}\frac{1000mg}{1g}=

← Older revision		Revision as of 17:14, 23 January 2018
Line 192:		Line 192:


−	<math>\rho = \frac{~~101325 kg⋅m~~^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{~~101325 kg⋅m~~^{−1}⋅s^{−2}}{~~98353.291~~ \frac{~~kg⋅m~~^2⋅s^{−2}}{g}} </math>	+	<math>\rho = \frac{101325000 g⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325000 g⋅m^{−1}⋅s^{−2}}{98353291 \frac{g⋅m^2⋅s^{−2}}{g}} </math>


−	<math>\rho= \frac{~~101325~~ g}{~~98353.291~~ m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}	+	<math>\rho= \frac{101325000 g}{98353291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}

← Older revision		Revision as of 17:12, 23 January 2018
Line 195:		Line 195:


−	= \frac{101325 g}{98353.291 m^3}	+	<math>\rho= \frac{101325 g}{98353.291 m^3} =\frac{1.03g}{m^3}\frac{1m^3}{1000000cm^3}

← Older revision		Revision as of 17:09, 23 January 2018
Line 193:		Line 193:

	<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>		<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>
		+
		+
		+	= \frac{101325 g}{98353.291 m^3}

@@ Line 189: / Line 189: @@
 <math>\rho = \frac{p}{R_{\rm specific} T} </math>
-Using the normal temperature and pressure of 20 °C (293.15 K, 68 °F) and an absolute pressure of 1 atm (14.696 psi, 101.325 kPa) 1Pa=1 kg⋅m−1⋅s−2
+Using the normal temperature and pressure of 20 °C (293.15 K, 68 °F) and an absolute pressure of 1 atm (14.696 psi, 101.325 kPa) 1Pa=1 kg⋅m−1⋅s−2   1 J=kg⋅m2⋅s−2
-<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{J}{g}} </math>
+<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{kg⋅m^2⋅s^{−2}}{g}} </math>

← Older revision		Revision as of 17:04, 23 January 2018
Line 192:		Line 192:


−	<math>\rho = \frac{101325 ~~kg⋅m−1⋅s−2~~}{335.505 \frac{J}{g \cdot K} 293.15 K} </math>	+	<math>\rho = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{335.505 \frac{J}{g \cdot K} 293.15 K} = \frac{101325 kg⋅m^{−1}⋅s^{−2}}{98353.291 \frac{J}{g}} </math>