At the end of the eight laps the person is where they started, so the final displacement is 0. Since velocity is defined as displacement divided by time, time is irrelevant, and the average velocity is 0 also.
You're correct, you've calculated the average speed.
a) is, as you said, the solution for t when
\frac{dm}{dt}=0.
However, the solution to
t^{0.2}=\left[\frac{4.8}{3.35}\right]
is not 2.3375, check your algebra.
b), as you said is m(t) for the solution above
c) and d) can both be found by substituting the times into the expression for...
Try an algebraic approach if you can't get it by thinking about it.
3a = 3\left[\frac{4\pi^2R}{T^2}\right]
Now get rid of the 3 on the RHS by determining a coefficient for T within the square.
For part b), you can probably see there's a relationship involving the angle. As the square is rotated throughout 90 degrees, the flux that passes through the square gradually decreases to 0. Thus, the trigonometric function you're looking for is cos:
63\cos{40} = 48
For c), the only...
On this site it seems people prefer to encourage independent thought rather than just give the solution, but I'm not sure there's much else to say here. The relations relevant at this level are:
V = IR
P = IV
You need to find the time at which the ball reaches the position of the net. What you've written as g_x is actually just acceleration in the x-direction, g is simply used to denote gravitational acceleration (which only occurs in what's defined as the y-direction in this problem). If air...
Uh... not off the top of my head. Gamma is electromagnetic radiation, and is therefore massless and chargeless. Neutrons are almost equivalent in mass to protons (with negligible difference), and therefore have mass 1, and (as the name suggests) electrically neutral, thus no charge.
I just reworked the problem using the plane of the incline as the x-axis and managed to find for x:
x = \frac{2v_{0}^2\cos^2{\phi}\left[\tan{\theta}+\tan{\phi}\right]}{g}
Which I assume is what Mindscrape found...
Which makes:
d =...
I can't follow that at all..
Firstly shouldn't it be:
\cos{\theta}=\frac{AB}{AC}
The next equation is just the calculus-derived expression for x, but I don't see how you're getting to the next equation, nor how you're "simplifying" to get the proceeding one..
Yeah, the solution is quoted in the back of the book, pt a) is correct. The solution to pt b) is \frac{\pi}{4}-\frac{\theta}{2}. I'd like to know how it was solved though. Since the expression for d I found is the solution quoted for pt a) by the text, it seems logical to assume that the...
Homework Statement
A baseball is given an initial velocity with magnitude v_{0} at an angle of \phi above the surface of an incline, which is in turn inclined at an angle \theta above the horizontal.
a) Calculate the distance measured along the incline from the launch point to where the...
This is starting to get extremely frustrating...
I find the same solution approaching the problem a different way.
\vec{r}(t) = (v_{0}\cos{\alpha_{0}}t)\mathbf{\hat{i}} + (v_{0}\sin{\alpha_{0}}t-\frac{1}{2}gt^2)\mathbf{\hat{j}}
\vec{v}(t) = (v_{0}\cos{\alpha_{0}})\mathbf{\hat{i}} +...
Its amazing how many mistakes you can make typing something like this out in TeX. This question is from a text I use all the time, and it always explicitly states 'in terms of var1, var2...' if the solution is general. It's looking for a numerical angle :/ This leads me to believe I'm looking...
Homework Statement
A projectile is thrown from a point P. It moves in such a way that its distance from P is always increasing. Find the maximum angle above the horizontal with which the particle could have been thrown. Ignore air resistance.
Homework Equations
y =...