Recent content by bbraves7

Sorry, it was very late last night when I posted this problem. I meant to put dy/dt=v, dv/dt=a. The depth of the lake is 30m. v0= 25ft/s. Sorry again about that. I had been working on this for about two hours so my brain was fried.

Homework Statement A bowling ball is dropped from a boat so that it strikes the surface of a lake with a speed of 25 ft/s. Assuming the ball experiences a downward acceleration of a=10-.9v2 when in the water, determine the velocity of the ball when it strikes the bottom of the lake...

Yeah, I just write it like that as a kind of short hand. Anyways, I figured out what I was doing wrong. The equation should only be used to calculate the magnitude of the force, so I shouldn't put the negative in the equation for the -4 micro Coulomb charge. That takes care of it. Thanks...

Homework Statement Where on the x-axis would the net force on a (-2) micro Coulomb charge be zero? *picture attached* Homework Equations F = k*(q1q2)/d^2 The Attempt at a Solution I think the answer is d= 6, making the position on the x-axis = 7. However, when solving my equation, I have...

That makes sense. Thank you so much for your help, it's actually been quite fun!

Wonderful! I have a quick question. What is the difference between approaching this problem from an average density direction, as opposed to a force equation, where: Force(down) = Force(buoyant) Are they essentially the same thing, or will you get substantially different answers?

So the (base) in terms of (h) would be tan(15) = b/h b = (h)(tan15) 2b = 2(h)(tan15) 1000 = [7500*(.15*30*2)] / 2[(1/2*((2htan15))*h)] 1000 = 67500 / 2(h^2)(tan15) 1000 = 67500 / 2(h^2).27 solving for h h = 11.2

Oh sorry. That's the base of the triangle. 30*sin(15) = 7.76 7.76*2 = 15.52

Awesome, I really think I have it now. So basically what we are doing by taking the average density is evenly distributing the density of the air(negligible) and hull over the area of the triangle (the ship). And the variable (h) that we solve for is going to give us the height where the...

Now I'm really confused! Since I only know the density of the hull, I would have to multiply it by the volume in order to get the mass, right? I thought (.15*30*2) would give me the area of a cross section of the hull?

Ok, then would this be correct:1000kg/m^3 = (7500kg/m^3 * 1/2*base*h*L) / [(1/2*base*height*L) + (.15*30*2*L)] canceling the L leaves 1000kg/m^3 = (7500kg/m^3 * 1/2*base*h) / [(1/2*base*height) + (.15*30*2)] plug in the numbers 1000kg/m^3 = (7500kg/m^3 * 1/2*15.52*h) / [(1/2*15.52*28.98) +...

So how about this: if (h) = the distance the ship sits in the water and (L) = the length of the ship Also, since I want to find the volume of the hull, I think I should be using the formula for finding the volume of a pyramid V = 1/3*base*height*length 1000kg/m^3 = (7500kg/m^3 *...

Thanks for the reply! So here is what I've got after your help: density(water) = average density(ship) so, 1000kg/m^3 = (mass of hull)/(volume of air plus hull) then, 1000kg/m^3 = (density of hull * volume of hull)/(volume of air + volume of hull) 1000kg/m^3 = ((7500kg/m^3)...

Homework Statement Consider a large cargo ship, with a steel (density = 7.5g/cc, 7500kg/m^3) hull. If I ignore the mass of the front and back of the ship, how far into the water (height) will the ship sit? Homework Equations density = m/v Force(buoyancy) = (density of fluid)(volume...