If you have the cross section of a cylinder, then you can find the radius
A=\pi r^2 => r=\sqrt{A/\pi}
From the radius, you could find the volume if you were dealing with a sphere, or if you are dealing with a cylinder, you would need a length/height.
V_{cylinder} = \pi r^2 l
\\
V_{sphere}...
Check again on units. When you have units like cm^3, if you want to go to km^3, the conversion is not the same as going from cm to km (you can check this by typing "cm^3 to km^3" vs. "cm to km" in google).
You need to find the amount of work it takes to put one book on top of another book! You know that the work needed to move one book onto the other one is:
(the mass of the book) * (the acceleration due to gravity) * (the height the book must be lifted).
Think about what the height will be that...
The initial momentum of the ball is p = mv = (.5)(8)=4 and this is at a fifty degree angle with the vertical. Therefore, the y-component is psin(\theta)=4sin(50) and the x-component is pcos(\theta)=4cos(50) When the ball bounces off the wall, the y-component of momentum is staying the same...
Yes. This is how Hooke's law is typically expressed F=-kx
Where x is the displacement from the equilibrium position and k is the spring constant. There are a number of different and non-intuitive ways to come to this conclusion (you will cover springs in differential equations and a course on...
In typical elementary applications, tension is used to describe the force exerted by a rope on an object in order to keep that object in static equilibrium (normally with the force of gravity). It is normally abbreviated as just "T", and often subscripted (T1, T2, etc.).
Your formula for the magnetic field at the center of a loop seems off. Consider the formula for a differential magnetic field from a differential length ds:
dB = \frac{ \mu_0}{4\pi } \frac{ids\times \widehat{r}} {r^{2}}
Where \mu_0 is the magnetic constant. Remember that ids\times...