Recent content by 7starmantis

So basically since\rhogv = 2N Volume = 2N / \rhog Volume = 2.0X10^-4 Then the weight in air is 15N / 9.8 = 1.53kg \rho object = m/v = 1.53kg/2.0X10^-4 = 7650 That make sense? I hope so, cause it does to me... :)

We did a lab like this, but my teacher hasn't returned our lab reports yet and it has all my data in it. I just can't for the life of me understand how to setup the problem.

Thats ok, I just wanted to understand the concept and make sure I was setting it up right.(although it does sound like a big weight) Thank you very much for your help, you explained it in a way I could understand. I really appreciate it! a lot!

Ok, so Vol of water displaced is: vol = a*h = (15.24X6.10)*(.0367) = 3.41m^3 Then: m = v*\rho = (3.41)*(1000) = 3410kg Wt = 3410kg * 9.8 = 33418N So that is the weight of the water displaced which is also the weight of the truck. That makes sense to me. Why couldn't I figure that out...

I'm so frustrated by this problem I can't even think straight anymore. I don't understand what you mean by it being an additive term, isn't it multiplied in the equation? But at this point I can't even figure out how to solve the equation with or without it.

Could I not find the pressure at that depth by \rhogh (1000)(9.8)(.0367) = 359.7pa F = PA = (359.7)(15.24m X 6.10m) = 33439.2 / 9.8 = 3412.16 kg ?

This is making me feel really stupid. I understand the concept that the water displaced will equal the weight of the truck, but I can't seem to find an equation to find that mass of the water. I can't determine the volume or weight of the ferry.

I didn't get that he said to set them equal to each other, but I also don't know the density of the air. So: m*g - volume of the object* density of the air* g = m*g - volume of the object* density of water*g I have everything except density of air.

So we are looking at something like: Buoyant Force only changes once the truck drives onto the ferry. So the displaced water equals the weight of the truck? That would mean... Buoyant force (wt of truck) = P2A - P1A P1 - ATM pressure P2 = \rhogh A = length X width so... =...

Ah, then I just questioned myself for no reason. I already had the correct answer. Correct?

I'm confused, that still leaves me with two separate values for mass. Which value would I use in the density equation?

Yeah I forgot to complete the area units, sorry. I didn't think area = volume / length was the right equation. If it is then I have the right answer already, right?

How do I go about finding the density? Don't I need to find volume first then use volume in the density equation?

[SOLVED] Radius Problem A brass rod (density = 8470 kg/m3) is 12cm long and weighs 20N. What is the radius of the rod? I'm confused when finding area. I don't think I'm using the right variables. volume = mass/density v = 2.04 kg / 8470 kg/m3 = 2.4X10^-4 Area = volume/length? A = 2.4X10^-4 /...

An object weighs 15N in air and 13N when submerged in water. determine the density of the object. Am I setting this up correctly, I'm confused about the two masses. Volume = W(air) - W(water) / pg 15-13 / (1000)(9.8) = 2.0x10^4 p(density) = m/V What mass do I use here though, the one in air or...