Recent content by mg11

Yep... makes sense and works great, thank you!

Homework Statement A disk of radius r meters has a string wrapped around its perimeter. A mass of m kilograms is attached to the end of the string and is allowed to descend freely from a height of h meters, it takes t seconds for the mass to travel the distance. Find the moment of inertia (I)...

Great, Thank you!

Can anyone just give me a yes/no answer to this? Did I solve right? I have to go to sleep soon before the test tomorrow. I would really appreciate some help.

I differentiated the original function v(x) and got dv/dx = 2 * 4.0[1/ms] * x Then I multiplied it by the original v(x) and got 2 * 4.0[1/ms] * X * 4.0[1/ms] * x^2 = 32.0[1/m^2*s^2] * x^3 --------------------- -------------------- dv/dx v(x)

I have an exam tomorrow that includes a question of this type. Could someone please just confirm if I'm right? Thanks

OK, so in that case would the answer be F(x)=m*a(x)=m*v(x)*v`(x) ? I did this and got F(x)=1500[kg]*32.0[1/m^2*s^2]*x^3 Is that right?

dx/dt is the same as v(t) but I don't know v(t), or do I? Thanks!

Homework Statement The movement of an object with a mass of 1500kg is given by v(x)=(4.0 [1/ms]) * x^2 Determine the net force acting on the object as a function of x. Homework Equations F=ma The Attempt at a Solution I know I'm supposed to use the chain rule to solve this but...

I didn’t use the right notation here but it wasn’t because I didn’t understand the meaning of the result. The bottom line here, as we seem to agree, is that the uncertainty is very high (±0.1 x 10^6) when we take the actual uncertainty factor into account. What you’re saying may be true but...

Indeed, significant figures are very far from being exact science. In fact, I think they're pretty far from being science at all. I did some calculations of my own to check the validity of significant figures. I performed similar calculations using sig figs on one side and precise error notation...

If you draw this line on a coordinate system you will get a vertical line passing through all y values when x=-1. So x=-1 regardless of the y value, therefore the equation is x=-1 Let's generalize: The equation of a vertical line passing through (a,b) will be x=a You should also note that a...

Mute, Thank you for taking the time to answer. Your answer makes a lot of sense but I still have some difficulty accepting it. While using method B will always give the same result (on a calculator) it is still sensitive to parsing. Lets bring back this calculation: (5.0+5.0)/11.0 Whether I...

Hi, I am currently taking a college-level Physics course and ran into some troubling questions regarding significant figures. I wasn’t sure which forum I should post this to but the math forum seemed like the most adequate one. So here goes… I have no problem understanding the basic rules for...