Rational function helpp needed immedietly

  • Thread starter Thread starter gamefreak
  • Start date Start date
  • Tags Tags
    Function Rational
AI Thread Summary
Astronauts experience a decrease in weight as they ascend due to reduced gravitational pull, described by the formula W(h) = We/(1+h/6400)^2. To determine the altitude at which an astronaut's weight falls below 10 N, the equation W(h) < 10 must be solved for h. Initial attempts to solve the equation led to confusion, particularly in manipulating the inequality. A suggested approach is to multiply both sides by (1+h/6400)^2 and then take the square root to simplify the problem. This method aims to clarify the steps needed to find the correct altitude range.
gamefreak
Messages
2
Reaction score
0
rational function helpp needed immedietly !

1. when astraunauts go into space, they feel lighter. this is because weight decreases as a person rises above Earth's gravitational pull according to the formula

W(h) = We/(1+h/6400)^2

where We is the person's weight in Newtons, at sea level on Earth and W(h) is the weight at h km above sea level

Q) at what range of altitudes will this astraunaut have a weight of less than 10 N?

Homework Equations



W(h) = We/(1+h/6400)^2

The Attempt at a Solution



i attempted it by trying various methods, it will take too long to type it up here and i already have the answer in the book i tried but i could not get that answer

so please help ! I would really appreciate that . thank you !

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
Physics news on Phys.org


W(h) is the weight of the person at a height of h above sea level. For this person to weigh less than 10 Newtons, W(h) needs to be less than 10. So set W(h)<10 and solve for h.
 


Well you basically want to find when W(h)<10


so just solve for h in

\frac{820}{(1+\frac{h}{6400})^2}&lt;10
 


ok so when i solve that I end up with

82 (h+1)^2 > 0

im lost now?
 


You didn't solve for h correctly. Hint: multiply both sides by (1+h/6400)2 and then take the square root of both sides.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top