Recent content by kieslingrc

Thanks for the reply and for pointing out my error on the units. I knew there was something in common with the J and kg m2/s2, I just couldn't find it in my notes. Otherwise, I believe the math to be correct? Thanks for the speedy reply. I really love math, but it's so hard to pick up again...

1. A lead bullet with a mass of 0.03 kg traveling at 198 m/s strikes an armor plate and comes to a stop. If all the energy is converted to heat and absorbed by the bullet, what is the temperature change? 2. Q=KE=1/2mv^2; Q=CmΔT 3. Heat capacity (C) of lead = 130J/kg-°C, mass of...

Got it! 352.8cm^3 = .000353m^3. Wd of water is 1000kg/m^3. 1000kg/m^3*.000353m^3=.353kg 1kg = 9.8 N so 9.8*.353=3.459 Thank you. You got me pointed in the right direction.

No worries, I appreciate your efforts.

Well, that makes a lot more sense now that I see what you mean by g. Not g as in gravity but g as in gram. I still don't see the Wd of water in g/cm^3 though. Maybe I am too tired or just so overwhelmed with this.

Dw is the weight density of water, straight out of my book and it is represented as 62.4ft^3. What does 1 cm^3 = 1 g have to do with this? I seriously need some guidance before I lose my mind. I am taking an online class with a non existent instructor to answer questions while working full...

1. An object with a mass of 56 g displaces 352.8 ml of water when it is completely immersed. What is the buoyant force on the mass? Notes: * Report your answer in Newtons and use N as the unit. * Report your answer to one decimal place. * Acceleration due to gravity = 9.8...

Yeah, never mind. I was totally off on a tangent here with some crazy formula in my text. It is as simple as calculating the weight = Dw of the wood * V = 35 * .35 = 12.25lb. Then buoyant force = Dw of the water * V = 62.4 * .35 = 21.8lb. Fnet = Fb - W = 21.8 - 12.25 = 9.55 lb. Easy once it's...

1. A juniper wood plank measuring 7 ft by 0.5 ft by 0.1 ft is totally immersed in water. What is the net force acting on the plank? () 2. a) F top = P top * A top(l x w) ; b) P bottom = P top + (Dw * h); c) F bottom = (F top) + (Dw*l*w*h); d) Fb = F bottom - F top e) Weight = Dw * l * w...

Thanks Lawrence for helping me understand what I was doing wrong.

OK, I am totally confused. Let me start over. The pascal conversion was fine, no need to work with that anymore. 38.8 lb/in^2 Area: [1m^2 = 10.8ft^2]; [10.8ft^2 = 1555.2 in^2] Convert the area 1555.2 * 0.8 = 1244.2in^2. The force is 38.8 * 1244.2 = 48,275lbs

So that leave me 48,275 psi, or 335.2 ft lb^2. The force is 112,359 lbs.

gotcha, I needed to multiply by 144, not 12, right? But in that case, my numbers seem way too high.

We are to use all units in the book and to 1 decimal place rounded up so that's where I came up with the Pa conversion. Hmmm, in my book, the conversion is 1m^2 = 10.8ft^2, so I converted the m - ft by multiplying it leaving 8.6ft^2. I had to get the units to be the same so converted ft^2 to...

1. The water in a tank is at a gauge pressure of 267,018 Pa. If the bottom of the tank that has an area of 0.8 m2, what is the force on the bottom of the tank? 2. P=F/A; F=P*A 3. Converted Pa into 38.8psi for P (267,018/6890); Converted the Area into 103.7 in^2 (.8*10.8*12)...