Recent content by Biosyn

Homework Statement Has anyone here read The Mathematics of Juggling by Burkard Polster? I am having a hard time understanding how the inverse of a juggling sequence is calculated on page 27. For example, the table on the page has 7 rows, and in the fourth row, I'm not sure if that symbol...

Would the mean be 10.11 and 29.72? Fall CBM WIF score: 10.11 - 0.85(9.26) Spring CBM WIF score: 29.72 - 1.24(19.96)

Homework Statement On p. 13, they present and discuss Table 1, summary scores for the test periods. At the bottom of the first column, they say that the "fall CBM WIF is 0.85 standard deviations below the mean." What mean are they referring to? What is the value of the mean? Homework...

Oh, okay. Thank you for the help! :)

It's the radius of Earth + the distance of the satellite from the surface of Earth. This is what I used though... Edit: I used 22461010 meters. 6.3675*10^6 + 1.609*10^7 = 22457500 meters ^Using this radius, I still get around 9 hours.

Oh, right. Well, the answer says it takes the satellite 6 hours to complete one orbit around Earth which is about 3 hours shorter than what I came up with (9 hours). Am I wrong?

Actually there's one more question I have which is finding the time it takes the satellite to orbit the Earth once. Can you please verify this? T = 2*pi*sqrt((r^3)/(G*m)) T = 2*pi* Sqrt(\frac{22461010^3}{G*m_E})T = 33627.87 seconds T = 9 hours 20 minutes

Okay. 10,000 miles = 1.609*10^7 meters R = (6.37101*10^6 + 1.609*10^7) R= (22461010) R^2 = 5.044*10^14m g_2 = (6.6742*10^-11)*(5.044*10^14) g_2 = 0.790424 v^2 = 0.790424 * 22461010 v = 4213.51 m/s Edit: Which is approximately 9400 mph. Thanks for helping me! :)

Okay, g_2 = (6.6742*10^(-11))*((5.9736*10^24)/(16371011.61)) g_2 = 1.4875 m*s^-2

Okay, I'm getting confused. In the beginning I converted everything to meters. altitude = 1.609*10^7m + 6.3675*10^6m radius of Earth = 6.3675*10^6m gravitational acceleration = 9.8m/s^2 v = ? v^2 = g*r v = sqrt(g*r) v = sqrt( 9.8 * (1.609*10^7m + 6.3675*10^6m)) v = 14835.2m/s But this...

Okay sorry, I meant 22 mi/h/s... and g_2 = 1.7959 mi/h/s at 10,000 miles above Earth's surface.

g_2 = 1.7959 mph (1.795mph)*(14000mi) = v^2 v = 158.56 mi/√hEdit: For g_1 I used 22mph because the other units are in miles.

I used Sn = \frac{a_1*(1-r^n)}{1-r} Sn = \frac{5*(1-(-5)^98)}{1-(-5)} = \frac{5*(1-(-5)^98)}{6} = (5/6)*(1-(-5)^98) Thanks for your help!

\frac{g_1}{g_2} = (\frac{r_2}{r_1})^2 \frac{22}{g_2} = (\frac{14000mi}{4000mi})^2 g_2 = 1.796mi/h/s v^2 = (1.796)(14000) v = 158.56 mi/h ?

Homework Statement How fast must a satellite 10,000 miles above Earth's surface travel, and how long does it take to complete one orbit of Earth? Homework Equations v^2 = g*r The Attempt at a Solution 10,000 miles + radius of Earth = 14,000mi = v^2 = (9.8)(2.253*10^7m)...