# Math Help - non linear equations question

1. ## non linear equations question

q1)the period (T) of a pendulum swing is the time taken in seconds for the pendulum to complete one swing. The period is dependant upon the length of the pendulum according to the formula
T=2pie (square root of L divided by 9.8) L= the length of the pendulum in metres.
Determine the time (T) if the clocks pendulum to complete one swing when the pendulum (L) is .85 metres long?

q2) a person pushes her daughter on a swing ands finds that it takes the girl on the swing 7 seconds from where she is let go to when she returns.How long is the rope of the swing to the nearest metre?

Q1) I=_E_
R+r

I is amps E is emf R is resistance ohms r is internal resistance ohms

an unknown battery is connected to a 4 ohm resistor which causes a current of 2 amps to flow in the circuit.When the battery is connected to a 2 ohm resistor a current of 3 amp flows.
show the following 2 equations are formed from this data.
E-2r=8 E-3r=6

solve this equations simultaneously to find the emf (E in volts) and the internal resistance (r in ohms) of the battery.

2. Originally Posted by mat
q1)the period (T) of a pendulum swing is the time taken in seconds for the pendulum to complete one swing. The period is dependant upon the length of the pendulum according to the formula
T=2pie (square root of L divided by 9.8) L= the length of the pendulum in metres.
Determine the time (T) if the clocks pendulum to complete one swing when the pendulum (L) is .85 metres long?

q2) a person pushes her daughter on a swing ands finds that it takes the girl on the swing 7 seconds from where she is let go to when she returns.How long is the rope of the swing to the nearest metre?
Both of these are based on the given equation
$T = 2 \pi \sqrt{ \frac{L}{g}}$

For the first one let L = 0.85 and calculate T. I don't see what the problem is here.

For the second one:
$T = 2 \pi \sqrt{ \frac{L}{g}}$

$\sqrt{ \frac{L}{g}} = \frac{T}{2 \pi}$

$\frac{L}{g} = \left ( \frac{T}{2 \pi} \right ) ^2 = \frac{T^2}{4 \pi^2}$

$L = \frac{gT^2}{4 \pi^2}$

Plug in T = 7 s.

-Dan

3. Originally Posted by mat
Unrelated questions really should go in different threads.

Q1) I=_E_
R+r

I is amps E is emf R is resistance ohms r is internal resistance ohms

an unknown battery is connected to a 4 ohm resistor which causes a current of 2 amps to flow in the circuit.When the battery is connected to a 2 ohm resistor a current of 3 amp flows.
show the following 2 equations are formed from this data.
E-2r=8 E-3r=6

solve this equations simultaneously to find the emf (E in volts) and the internal resistance (r in ohms) of the battery.
$E - 2r = 8$

$E - 3r = 6$

Using the substitution method:
Solve the second equation for, say, E:
$E = 3r + 6$

and insert that value of E into the first equation:
$(3r + 6) - 2r = 8$

$r + 6 = 8$

$r = 2$

Now plug that value of r into either of the original equations (or the one we solved for E):
$E = 3(2) + 6 = 12$

(Don't forget to put in the appropriate units when giving your answer.)

-Dan

4. thankyou for your help topsquark.