Question

Help, please.
1) Explain why a particle moving with constant speed along a circular path has an acceleration.
2) The value of radial acceleration is given by v^2/r, where v is the speed and r is the radius of the path. Using base units, show that this expression is dimensionally correct.
3) A 15 year old girl listens to her ipod via a single earphone placed in her right ear. Describe an experiment which you could do in the physics lab to test whether the frequency response of her right ear is now different that that of her left ear.
Thanks in advance. I really need help with these.

Answer 1

because the speed is constant but direction is changing due to which velocity is changing thus it has accleration.

Answer 2

2) \[\frac{ v ^{2} }{ r}=\frac{ \frac{ m ^{2} }{ s ^{2}} }{ m}=\frac{ m }{ s ^{s}}=a\]

Answer 3

"a" stands for acceleration.

Answer 4

Acceleration exixts everytime there is a change of the velocity vector, so if we consider our particle when it is moving on its trajectory at two different times, we have the subsequent drawing:

going from A to B, we have a variation of the velocity vector, by a quantity:
\[\Delta {\mathbf{v}} = {{\mathbf{v}}_{\mathbf{B}}} - {{\mathbf{v}}_{\mathbf{A}}}\]

Answer 5

Thanks everyone! ^_^ I fully understand now, 100%.

Answer 6

@Michele_Laino I've been struggling with this for a while. Can you help me?

The speed of ocean waves in deep water is given by:\[V=\sqrt{\frac{ g \lambda }{ 2\pi }} \]
where g is the acceleration of free fall and \[\lambda \] is the wavelength of the waves. Show that T, the period of the waves, in terms of g and \[\lambda \] is given by:
\[T= \sqrt{\frac{ 2\pi \lambda }{ g }}\]

Answer 7

It is simple, you have to use this formula:
\[V = \frac{\lambda }{T}\]

Answer 8

so we can write:
\[\frac{\lambda }{T} = \sqrt {\frac{{g\lambda }}{{2\pi }}} \]

Answer 9

or, taking the inverse of both sides:
\[\frac{T}{\lambda } = \sqrt {\frac{{2\pi }}{{g\lambda }}} \]
now, please solve for T, what do you get?

Answer 10

I'm on it.

Answer 11

Okay, so I used the first one and I got back the \[T=\sqrt{\frac{ 2\pi \lambda }{ g }}\]
It took me a while but I got it. Thank you very much for your help!