Recent content by alexpratt

i realize that now, I am also doing a statics course where we assume everything is in equilibrium so that's what i have been doing for most of the semester. Basically i forget how to do these problems.

i was trying to avoid writing it out, but if you want to see it i will i know what I've done actually, i was looking at it as if they were in equilibrium, so this is what i had: T-m2g = Tsin37 + m1gsin37 - m1g - um1gsin37 so T = (m1gsin37 - m1g - um1gsin37 + m2g)/sin37 so i think i know...

Homework Statement two blocks of mass m1 50kg and m2 100kg are connected by a string of negligible mass. The pulley is massless and frictionless. The coefficient of kinetic friction between the 50kg block and the incline is 0.250 the 100kg mass is the hanging mass. theta is 37.0 deg...

ok, that makes sense thank you! i had some other examples that i had the answer to and i noticed some of my answers were half of the right answer but i never made that connection. Unfortunately that question was on the exam :( regardless thank you for your help, it'll come in handy next...

Homework Statement Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the equation: 2 Na(s) + 2 H2O(l) ? 2 NaOH(aq) + H2(g) When 0.030 mol of Na is added to water, the temperature of the calorimeter rises from 25.00C to 37.90C. If the heat...

ahh, thank you! Got the answer to be 0.496 m/s, i almost ran out of tries too!

ok, so its f*mg*d? when i do that i get 0.737, last time i checked anyways. So maybe my math is off.

Homework Statement A block with mass 1.50 kg is attached as shown to a spring with a force constant of 456.0 N/m. The coefficient of kinetic friction between the block and the surface on which it slides is 0.210. The block is pulled 3.60 cm to the right of its equilibrium position and then...

i figured it out on my calculator, but i can't use it on my exams. though I'm assuming with a linear function i can just plug in the x value to get the y value and then make a triangle and solve the area with bh/2 if it was any other function id be lost haha, thank you though, i got the right...

i haven't done integration yet unless i just wasn't paying attention... is that the only way to do it?

Homework Statement A force F = (4.37xi + 3.01yj) N acts on an object as it moves in the x direction from the origin to x = 5.06 m. Calculate the work done on the object by the force. Homework Equations dot product The Attempt at a Solution i took the dot product, FxDx+FyDy since there is...

i figured it out, i was rationalizing the left side but i forgot to get a common denominator afterwards.

I keep getting -1/4 I know you guys are right though, i graphed it to make sure, not that i thought you were wrong haha. I have no clue where I am going wrong though. thanks for your help though

-1/4?and just so i know in the future, how did you get the square root sign?

yes, it is as x approaches 4, sorry about that. but when i rationalize it, wouldn't i just get x-4 in the denominator which would still be division by zero? i'll see if i can figure it out considering you said the limit exists, thank you