1- yea i wasnt in radian mode ...THANKS
2- I tried it in radian and i got 1.212 and its also wrong :S
3- i can rearrange the x(t) equation to have it as a function of t but i have two x's...im confused...
4- yea i believe its the rite equation and i did try it and i got the same answer
5- i...
I have several questions i will write all of them and give you my attempt at a solution, hope you can help me figure out what i am doing wrong, Thanks.
Question 1:
An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t=0.00 s. It then...
hmm i must be doing something wrong again, this is how i worked it out
I used impulse = change in momentum so;
F t = m (V2-V1)
F = [(0.64kg) (56m/s)]/ (0.0174s)
F = 2059.77 N
==> this is the wrong answer could u please tell me where I am going wrong...thx
Homework Statement
A 64.0 g tennis ball with an initial speed of 28.0 m/s hits a wall and rebounds with the same speed. The figure below shows the force of the wall on the ball during the collision. What is the value of Fmax, the maximum value of the contact force during the collision, if the...
Homework Statement
Suppose that a population develops according to the logistic equation
dp/dt = 0.03 p - 0.006 p^2
where t is measured in weeks.
-What is the carrying capacity and the value of k?
Homework Equations
dp/dt = kP ( 1 - (p/K)) where K is the carrying capacity
The...
sorry i got one more question, its conceptual and i have a little problem with it,
1. The problem:
Bill and Susan are both standing on identical skateboards (with really good ball bearings) initally at rest. Bill weighs three times as much as susan. Bill pushes horizontally on Susan's back...
well if what i said was correct and fnet(x-direction) = ma = Ff, then
ma = Ukmg right? ohhh ic...lol so a=Ukg... my bad this is where i make another stupid mistake...THANKS
Sled question
I have a physics midterm tonite and i was trying some of the practice tests and i got stuck on this question, ur help would be greatly appreciated :)
1. The problem:
A sled is traveling at 4.00m/s along a horizontal stretch of snow. The coefficeint of kinetic friction is...
yay i got the right answer, i was doing it wrong in terms of calculating time, i had to first calculate the time it took it to reach its max hight and then calculate the time it took from its max hight to reach the floor.
so for my y-direction my v1=6.38m/s, and v2= 0 and i know that a=...
i think so, idk i used my calculator to get the answer, but here's how i set it up:
d=32.8m, v1= 6.38m/s (maybe this number should b negative since speed is down in the y-direction, wat do u think?) and a= -9.8m/s^2
so d=v1 t + .5 a t^2
32.8 = 6.38t + -4.9 t^2
4.9t^2 - 6.38t + 32.8 = 0...