Recent content by badreligion

I can't see the image try uploading it to imageshack or something then posting the link, thanks.

Try J=F*Δt J=Impulse aka the answer to the first question F=force Δt= Change in time You have Δt but still need force. F=ma m=mass a=acceleration You have mass but still need acceleration a=Δv/t Δv=change in velocity t=time

I had the same feeling not only does the frequency make more since, but so does the answer when it is 8.0 x 109 Hz

Think of Newton's third law "To every action there is an equal and opposite reaction."

There are three forces I can think of off the top of my head. The question doesn't get to specific so it could be 2 or 3 forces and so on. Gravity Centripetal Force Tension Force in the rope The tension one is a most likely but there could not be one. I'll leave you to figure out the rest.

First hint look into conservation of energy. Here are some formulas to look into PE=.5kx^2 k=spring constant x=position KE=.5mv^2 Combine them to form conservation of energy E=.5mv^2+.5kx^2

Magnitude of the change in the dog's momentum basically means an Impulse. Which can be figured out using the formula J=FΔt. J=Impulse aka the answer to the first question F=force Δt= Change in time You have Δt but still need force. F=ma m=mass a=acceleration You have mass but...

Here is the formula I use, it is the same thing as yours just simplified. f'=(1+Vobs/Vsnd)f First Doppler Shift f'=(1+29m/s / 343m/s)*872Hz Second Doppler Shift f''=f'/(1+Vsource/Vsnd) 872Hz-318Hz=f'(1-n/343m/s) 554Hz=f'(1-n/343m/s) 554Hz/f'=1-n/343m/s)...

You will need to use the doppler equation twice, once for the emitted sound wave and again for the return sound wave. If you want me to solve it and show you how you will just need to clarify when you say "8.0x109 Hz" do you really mean 8.0x109 Hz or do you mean 8.0x10^9 (8 times 10 to the...

No problem just curious is this high school physics?

check my edit

Your s=distance? also since when was work a vector quantity? and theta should be zero since the problem never mentions any angles.

Sorry I edited this because I think i made a mistake don't worry I didn't give up on you. I'm sorry again I keep getting an answer very close and is probably just off because of my rounding. Though it's been a while since I have done projectile motion and I would hate to show you the wrong...

P1 +m/V)gy1= P2 +(m/V)gy2 Then move things around so it's P1-P2=(m/V)gy2-(m/V)gy1 Then that changes to Delta P = (m/V)g(y2-y1) Delta P = (m/V)*g*Delta h Delta P is the pressure of the water aka what you are looking for, and Delta h is the different heights of the holes 3cm,13cm,23cm or...

You can get rid of the 1/2(m/V)v's in your Bernoulli's principle equation since that is irrelevant. You need to use the equation without 1/2(m/V)v's to figure out the pressure of the water at each height and then add the air pressure to get the absolute pressure at each height.