Recent content by samantha.

Thank you, Sir. :)

88kPa is the value given. If it was a typo it wasn't on my part. I noticed the water floating on top of oil.. strange yes. Haha. I spent half of my night panicking over my exam tomorrow, trying to figure out where I was going wrong in all of these problems that my professor gave me when in...

Homework Statement Consider a manometer with a glass bulb (Pgas=88kPa) at one end and which is open to the atmosphere (P =101 kPa) at the other. The manometer has two liquids oil and water in it, as shown. If the oil has a density of 823 kg/m3 and water a density of 1.00x10^3 kg/m3 find the...

Okay I re did it and fixed the signs 4.906*t^2-36.53*t+33=0 After doing the quadratic equation, I ended up with the same answer that I did my second attempt :/

Homework Statement A ball is launched from a 35.0m high cliff with an initial velocity of 47.0m/s at an angle of 39.0 degrees above the horizontal. On the way down it just clears a 2m high wall. A) How far away is the wall? Homework Equations Delta D=Vx*T + 1/2*a*t^2 The Attempt at...

I arrived at the answer by chance but now that it's done and I can see my procedure, it makes logical sense.

I actually managed to get the right answer. Thanks though! :)

Sorry, Is that a intuitive question for me or is that information you need to know?

My bad.. I didn't copy it properly. The question is; Three blocks on a frictionless surface are connected by massless strings, with M1 = 1.60kg, M2 = 2.10kg, and M3 = 3.70kg. Due to the force F acting on M3, the system accelerates to the right. Given that T1 is 4.10Newtons, calculate T2. I'm...

Homework Statement A 4.98-kg block is placed on top of a 11.4-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.500. What is the maximum horizontal force that can be applied before the 4.98-kg block begins to slip relative to the...

Ohhh you're very right. Thanks so much again!

Woops. I used the wrong value. But doing so, I fixed it to the correct one but I was still wrong.

Thanks! To calculate the component of the applied force that was parallel to the incline I did this; Fapp(x)=573*cos(20.1) = 538.1N Applying this to the equation I used previously, 2063.6N*U-538.1N = 0 0.261. I confidently entered this answer and I was once again rejected. :(

Thanks for your quick reply. I'm not really sure how to incorporate the applied force into the normal force since it's purely a horizontal force.. Sorry if I'm a bit slow. I'm a chemist.

Homework Statement A 224-kg crate rests on a surface that is inclined above the horizontal at an angle of 20.1°. A horizontal force (magnitude = 543 N and parallel to the ground, not the incline) is required to start the crate moving down the incline. What is the coefficient of static friction...