Recent content by Superman123

Yes

What exactly does that prove? Sorry, I have a very slow mind.

Well, It should be the same? because even if B has twice the mass of A, its density is also twice as that of A making the atoms more compact.

1. The task is to figure out if the buoyant force is proportional to the mass of the object that is being sunk in the fluid (the fluid in this case water). I had an answer and that the buoyant force is proportional to the mass, but my answer only took into consideration objects with the same...

Oh, thanks for your time and help!

Is saying " The energy of the fall will be preserved in the rope and because there's no energy loss, the rope will pull up UT with the same energy, getting here to the same she started in." correct, considering physics terms?

Well, the energy produced will be enough to bounce her back where she started, if nothing of it is lost; though I don't know how to present it with formulas to show that my reasoning is enough

1. Ut will bungee jump from the golden gate bridge. The height that she will jump from is 65 meters. The rope pulls her up 8 meters above the water. If UT's weight is 63 kg, how far will the rubber rope pull her upwards, if there's no loss to friction or air resistance? Homework Equations...

Thank you very much, I appreciate your help !:smile: Just a very tiny question, did you use suvat-equations to construct this function for the optimal angle (π-atan(d/h))/2?

Well, if one speed is 5.85 m/s (did the calculation again with 11.6) and the other one is 7.54 then launch speed is √(5.58^2+7.54^2)= 9.3 m/s The speed available from the bow is 179=(1.28 kg *v^2)/2 v=16.703 16.703 > 9.3 , meaning that it doesn't exceed the speed available from the bow...

OOhh, Well, we can assume that it goes over the wall. Then U would be 9=U√(2*12/9.82) U=5.76 m/s; 5.76 < 5.85 What does this tell us, does it just tell us that it requires a lower velocity to get over the wall or that it can't get over the wall from a 9 meters distance?

There is another suvat equation that I could use to solve this, but we just used it and I am unsure if it gives the right answer. Δy=Vy*t+1/2g*t^2 At the top of the height Δy=0 0=Vy(1.537)+1/2(-9.82)(1.537^2) Vy= 7.5466 m/s θ=atan(Vy/Vx) θ=atan(7.54/5.58) θ=53.52 degrees

S= 9 meters s= U* time s= U√(2h/g) u=5.85 m/s Is this correct? the angle would then be θ=atan(Vy/Vx) I know Vx, it's 5.58, but how do I determine Vy?

There is another way that you mentioned about a suvat equation that requires distances and that would help me get the time: s= ut + ½at^2 s = h = ut + ½at^2 = 0t + ½gt^2 making h = ½gt^2 Which makes t= √(2h/g) I am not sure about this way or the other one, but this way give me an answer to...

Sorry, I just thought about what you said earlier about writing the time as some function of v, θ, h and how long it would take to reach the h with the vertical velocity, Vx. I could write. tan θ= Vy/Vx. Vx=Vy/tan θ I could then write the time as t= h/Vx or t= h/(Vy/tan θ), this function has v...