I'm not sure how to start solving this

  • Thread starter Thread starter zekester
  • Start date Start date
AI Thread Summary
To solve the system of equations involving r and c, start by dividing the two equations to eliminate the denominator, allowing you to express c in terms of r. Substitute this expression for c back into one of the original equations to create a single equation in r. Simplify the resulting equations by expanding and rearranging terms to isolate r and c. After deriving expressions for r, substitute back to find corresponding values for c. Finally, ensure to verify solutions by substituting them back into the original equations.
zekester
Messages
30
Reaction score
0
i have two equations -r^2/wc
divided by r^2-(1/(wc)^2)
equals -7903

and r/(wc)^2
divided by r^2-(1/(wc)^2)
equals 3287



w is a known quantity. How do I start to solve for r and c.
 
Physics news on Phys.org
1. Step:
Divide the equations by each other; this gets rid of the difficult denominator, and you may express "c" in terms of "r" (or the other way around, if you prefer that).
2.Step:
Insert the given expression for "c" into one of your ORIGINAL equations.
You have now a single equation in "r" which you may solve.
3.Step
With your "r"-solutions, put these into the relation you derived in 1. between "c" and "r" to determine the corresponding solving values for c.
 


To solve for r and c, we can use the system of equations provided. First, we can rewrite the equations in a more simplified form by multiplying both sides by the denominator of the second equation, which is r^2-(1/(wc)^2). This will eliminate the fractions and give us two equations in terms of r and c:

1) -r^2 = -7903(r^2-(1/(wc)^2))
2) r = 3287(r^2-(1/(wc)^2))

Next, we can simplify these equations by expanding the parentheses and combining like terms:

1) -r^2 = -7903r^2 + 7903/(wc)^2
2) r = 3287r^2 - 3287/(wc)^2

Now, we can rearrange the equations to isolate r and c:

1) -r^2 - 7903r^2 = 7903/(wc)^2
2) r - 3287r^2 = -3287/(wc)^2

We can then combine the like terms and move all the variables to one side:

1) -7904r^2 = 7903/(wc)^2
2) -3287r^2 - r = -3287/(wc)^2

Finally, we can solve for r by dividing both sides by the coefficient of r^2 and taking the square root of both sides:

1) r = √(7903/(7904(wc)^2))
2) r = -√(3287/(3288(wc)^2))

Now, we can plug in the known value of w to solve for c:

1) r = √(7903/(7904(c)^2))
2) c = √(7903/(7904r^2))

Using the values of r and c, we can then solve for the unknown quantity w by plugging them into the original equations. I hope this helps you get started on solving this problem. Remember to always check your solutions by plugging them back into the original equations to make sure they satisfy the given conditions.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Replies
12
Views
2K
Replies
4
Views
2K
Replies
5
Views
3K
Replies
7
Views
2K
Replies
4
Views
7K
Replies
8
Views
1K
Replies
1
Views
10K
Back
Top