hey Doc, thanks for the hint.
If I understand, can I use the mass of the Earth and multiply it with acceleration to five me force? so if I take 5.98x10^24 and multiply it with 9.8, it'll give me 5.86 x 10^25. Is this the appraoch I should be taking?
Hello all,
Need some help with this question:
You stand on the seat of a chair and hop off.
a) During the time you are in flight down the floor, the Earth is lurching up toward you with an acceleration of what order of magnitude? In your solution explain your logic. Model the Earth as a...
k, I think I solved this one. I don't want to bother writing all my steps. I just want to check the answer. If it's wrong, don't tell me the answer, I'll post up the work if the answers are wrong.
So for a) it's 20 degrees (Because complemntary angles with the same intital speed arrive at the...
ok let's see.
So I take 2.85714 and subtract it from 3 to give me 1.4286.
Now to find my displacement in x I use X=(v)(t):
(343)(1.4286) = 49.00
I can use teh kinematic equation to find xf for the y-component:
-40 = vyi - 4.9 (1.4286)
-40/-7.000 =
vyi = 5.71428 m/s...
ooop! I see my error. Just realized I can't have 13.05 as an inital x velocity.
So going back to the begining, I have the two x and y equations:
10 = vxi (cos) (40) t
and
1.05 = vyi (t) - 4.9t^2
the way I see it, I can find time from the y equation, if I assume that vyi is 0 as...
from what I understand from what you're saying. I can use this equation?:
-40= vi(sin)(0)t-1/2(9.8)t^2 =
-40= -4.9 t^2
-40/-4.9 = 4^2
t = 2.85714 sec ?
yeah?
would it be this?
10/0.76604 = 13.05 (Initional velocity)
I than use the intital velocity and plug it into x=(vi)(cos)(angle)(t)
10 = 13.05cos 40 \t
10=(9.9996)t
t = 1.000 sec
=
so now that I have my time, I plug that into my y-component equation:
1.05 = vi sin 40...
k, I think I may have figured out my error. Let me know if this sounds good. :)
4.54) A basketball player who is 2.00 m tall is standing on the floor 10.0 m from the basket, as in Figure P4.54. If he shoots the ball at a 40.0 degree angle with the horizontal, at what initial speed must he...
alrighty, for this problem, I figured out how Jtbell, founf the minimum value for the sound coming from the splash. so I subtracted that value from t = 3.00s and cam up with 2.88. Now, because the ball starts off in a horizontal motion on top of the cliff, I assume that my angle of orientation...
Ooops, I think I made a grave error after I founf my vxi = 13.05 m/s.
I assumed I could take that velocity and plug it into my equation for the y-component, which I now realize is totally wrong. Both x and y components should be solved seperately. I mistook 13.05 to be vi instead of vxi...
Oh Dan! You were absolutley right! I made a typo with the question. here's the question again but with the correction underlined. I must have missed it when I re-read the question!
4.13)******* One strategy in a snowball fight is to throw a snowball at a high angle over level ground. While...
oh I see! I was close though! lol. I will make the necessary corrections and post them soon. Can't express how much I appreciate your help Dan!
Thanks a million!
Neeraj