Recent content by bluebear19

Airlines often overbook flights. You are a manager selling 40 tickets on a flight with a capacity of 35. You estimate the probability of a given passenger NOT showing up is 0.2. I already calculated some parts of the problem: 1) how many passengers do you expect to show up = (.8)(40) = 32...

the eqn you got i think only solves for x, but not Xmax y = 1/2 h for max range, i think you can solve for the rest you had the right idea with sqrt(4y(h-y)) take the 4 out and u get 2sqrt(y(h-y)) looking at inside the radical y(h-y). you want that value to be the largest as possible. But...

the velociyty for an obj falling in kinematics is sqrt[2gh] Vf^2 - Vi^2 = 2as where a = acceleration and s = displacement here a = g since its falling and s = h-y = distance the water has fallen up to the point of the hole. and Vi = 0 so Vf^2 = 2*g*(h-y) and solve for v by taking the...

that's what i got for v, what did u get for the max range of x??

oh i see the mass of the pulley is 2.6 so it should be .5(2.6) this is confusing because in my problem the actualy mass of the pulley is 2.4kg, so other than that everything else is right? thank you sooo much!

oh yes, thank you so much! so the answer should be T1 - T1 = 2g - 4a = .5mpa +M/r 2g - 6a = .5(2.4)a + .51/.085 2g - 6a = 1.2a + 6 a = 1.89 t = 1.03s? does this look right?

wouldn't that give the same answer T1 - T2 = (m1-m2)g - (m1-m2)a = (4-2)(g) - (4- 2)a = 2g - 2a is the same thing as 4g - 2g - 4a +2a = 2g - 2a

T1 = m1 g - m1 a T2 = m2 g + m2 a then T1 - T2 = (m1-m2)g - (m1-m2)a doesn't it? i don't see what is wrong, can you please tell me?

the above is what i tried to follow, and by following that this is what i arrived at T1 - T2 = .5mpa +M/r = (m1-m2)g - (m1-m2)a plugging the numbers in i get: .5(2.6)a + .51/.085 = (2)(9.8) - 2a solving for a : a= 4.12 m/s^2 solving for t: t = sqrt[2ah] = sqrt[2(1) /4.12] = 0.696s is this...

First let's get the symbols right: m1 = larger mass, 4 kg m2 = smaller mass, 2 kg T1 = rope tension on the m1 side T2 = rope tension on the m2 side r = radius of the pulley, 0.085 m mp = mass of pulley, 2.6 kg M = moment or torque due to friction at pulley spindle, 0.51 Nm J = polar...

i did in the yahoo link. i replied to the answer with my own workings

but then how come in the yahoo answer, the time was 1.7s?? using this forumla i only get that t = around0.7 s? can u tell me what i am doing wrong please thanks so much

but isn't the Vo = 0 zero since they are dropped from rest so solving for t so that should give you t = sqrt[2h/a], is that right now?

the response on yahoo used that forumla, i don't know what i should use, should it be. thanks for helping

Homework Statement The two blocks are connected by a massless rope that passes over a pulley. The pulley is 17 cm in diameter and has a mass of 2.6 kg . As the pulley turns, friction at the axle exerts a torque of magnitude 0.51 Nm. Homework Equations...