Recent content by shawpeez

no there is no angle given

A banked circular highway curve is designed for traffic moving at 60km/h. The radius of the curve is 200m. Traffic is moving along the highway at 40km/h on a rainy day. What is the minimum coeficient of friction between tires and road that will alow cars to take the turn without sliding off...

I've been looking at this question for the last two hours, here's the question and what I've done so far The picture shows 3 blocks, A B and C, block C is ontop of block A. Block B is connected to block A by a rope, and is hanging over the edge. here it is, Blocks A and B have weights...

thanks for the help:smile:

My first instinct was that the upward acceleration of the cab was slowing down, this is what i did Fnet= T - mg = ma T - 2000(9.8) = 2000(-1.8) T - 19600 = -3600 T= 19600 - 3600 T = 16000 N Is this correct ?

here's the question, An elevator cab is pulled upward by a cable. The cab and its single occupant have a combined mass of 2000kg. When that occupant drops a coin, its acceleration relative to the cab is 8.00m/s^2 downward. What is the tension in the cable? could someone explane to me...

Your calculations look correct to me.:smile:

from this i took the position r=(36.75m)i + (30m)j then i found the angle relative to the x axis Tan(theta)= 30/36.75 theta = arctan .82 = 39.4 90 - 39.4 = 50.6 which would be the angle from the y-axis can someone tell me if I am doing this right i would appreciate it.:smile:

still pretty much stuck, could anyone help

Ok this is what I've tried so far Particle B Y-axis displacement= 30m 30 = V0(t) + 1/2a(t^2) t= 12.25s v = v0 + at v = (.40)12.25 v= 4.9m/s In 12.25s particle A travels 3.0m/s * 12.25s = 36.75m X-axis displacement = 36.75m 36.75 = vo + 1/2a(t^2) 36.75 =(.2)(t^2) t= 13.56s...

Wouldn't x - axis be a: acos(angle) and y -axis be a: asin(angle)

Could someone point me in the right direction, I am totally lost particle A moves along the line y=30m with a constant velocity v of magnitude 3.0m/s and parallel to the x axis. At the instant particle A leaves the y axis, particle B leaves the origin with zero initial speed and constant...

so 3/4 of the trip was 18m , then the total circumference would be 24m , the diameter 7.6 and radius 3.8 m. Then the centre of the circle would be x = 8.8 y = 6

from the diagram i see that at the first point its in the 2nd quarter and at the second point its in the 4th quarter( moving in a clockwise direction). Would the radius be the square root of (5^2 + 6^2). I still can't understand how this would relate to an xy centre coordinate(I thought the...

thanks for the post, but I'm still lost?