Well I have been trying to figuring out these two problems for a couple of days now, and no one has been able to help me. So I figured I would throw them up on here and maybe there is a bright mind out there somewhere who could better explain these to me. These are physics problems.. here they are..
1.) ----- T= 2pi* sqrt m divided by k

Consider the case of a student wishing to determine the spring constant for a spring by measuring the period of oscillation of a mass on the spring. The student knows that the relationship between the spring Constant (k), the mass (m) and the period of oscillation (T) is given by the above equation. This is not a linear equation. The equation can be made linear by squaring both sides of the equation above. Nest try to put the equation into y=mx+b form( do not confuse m for slope).

So my question is how do I turn this equation into y=mx+b? I really have no clue, it also says that once I graph the line I can use the slop to determine the constant(K). How would I go about doing that?

- It also asks me to graph T^2 after i square both sides, I am assuming the graph of t^2 would be an exponential graph, but the instructions lead me to believe the graph is suppose to be linear. If the graph is exponential how do I find the slope?

2.)
A student wants to determine the Stefan-Boltzmann constant, o. The student then measures the luminosity, surface area, and temperature of an object. The temperature of the object is changed several times. The student measures the new temperature and luminosity for each of the new cases. The surface area is assumed to be the same in each case. The data is then plotted to determine the Stefan-Boltzmann constant. The student knows that the relationship between these quantities given bu the Stefan-Boltzmann Law:

L=oAT^4

-How should the data be plotted so that the Stefan-Boltzmann Constant can be determined? What is the slope of the line when the data is so plotted? What are the labels for the Vertical and horizontal axis

- I am assuming that because T^4 this graph will be an exponential graph most likely with a y intercept of 0, but how would I find the slope?

Sorry mate, i can barely work out 2+2 lol

laughing at above comment!!

1) since you are trying to determine the spring constant, k becomes equivalent to y in the y = mx+b form - so try to isolate k, getting an equation k = ...

if you could do that bit then I'll help you with the second part about graphing T^2

2) if you plot the graph with L on the y-axis and T^4 on the x-axis you should get a straight line, and the gradient of that will be oA.

...I think

oohhh clever!!

shall we name this the 'Official Homework Help Thread'??

Haha maybe see how it goes first flumps, might change their mind about wanting help pretty quick

here we go, my professor finally emailed me back and now i get it! here is how you do it in case anyone is curious lol

First, as the problem states that T=.... (18) is not a linear equation, you can do that by squaring both sides.
Doing so you will have T^2= 4pi^2 m/K
However, you still not out of the woods, to make it linear you have to consider (y=T^2), so you can satisfy the form y=mx+b where y=T^2, m= 4pi^2/k (the slope) and x=m(the mass). Now b=0. x= could have been x=1/k, but because they asked you to plot T^2 verses mass(m) you have to follow.
Now based on the table below in the paper (case 3), you have to calculate T^2 of the given data. After doing so, draw a coordinate system with x & y axes. y=T^2, x=m. after connecting the dots you should get a straight line. The slope of this line is m=4pi^2/K, To get K value you need to find the value of m (the slope) first by m= delta y/ delta x or m=y2-y1/x2-x1.
when you calculate m (the slope) then k=4pi^2/m.

when you want to do it on excel, you just give the first column the T^2 values and the second column the masses. Then the slope is not K rather is again (4pi^2/k).

great

Should have come to me, i've just finished A2 physics

yea with a C

Here is an easier one that i cant seem to get the right answer to...

A sports car moving at constant speed travels 120 m in 5.1 s.

a.)If it then brakes and comes to a stop in 4.5 s, what is its acceleration in m/s^2?

b.)Express the answer in terms of "g's," where 1.00\,g = 9.80\m/s^2.

If you guys could please help me figure this one out I would greatly appreciate it. I cant seem to do it myself lol.

Here is what i did but the answer is wrong.
figured the velocity by dividing distance/time 120/5.1 and came up with 23.5 m/s. then I used that to figure out the distance the car traveled in 4.1 seconds (where it stops) i did 23.5* 4.1 and came up with 105m. then i tried to find the velocity using what i just found and did 23.5^2 =0^2+2a(105) and came up with an acceleration of 2.63 m/s which would have to be negative since you are coming to a stop. What did I do wrong?
a

a)
acceleration = (final velocity - initial velocity) / time
a = (v - u) / t
v = 0, u = (120m / 5.1s) = 23.53
t = 4.5

=> a = -23.53/4.5 = -5.23 m/s^2

b)
-5.23/9.8 = (-).534g

...I hope those are right