So I multiply 7.29E-05 by sin 62 then multiply that by 6370km.
and I get 0.41 km/s or 4.10E+02 m/s
since I have this info do I find the objects acceleration by a= v2/r. r=6.37E+3 km
undefinedundefinedI got it. :biggrin:
we know height so we need to use this formula:
(5Vo)^2=Vo^2+ g2h
>>> Subtract out the Vo^2
>>>>24Vo^2=2gh
>>>>>Vo^2=2gh /2
then, Vo=3.6m/s :laughing:
Assume the Earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the Earth to the object makes an angle of 62.0 degrees with the axis of rotation. Radius of the Earth = 6.37·103 km
I know that...
Sorry about not posting the rest of the problem.
But what you told me helped because I got them right after you explained it.
Thankyou For teaching me something I did not know before.
-Shaun
That is what I thought but the question does not give the direction of V2 which really confused me.
Examples in the book gave directions w/the vectors.
It also asked these True/False questions along with it:
(T/F) The magnitude of V3 can be 2.0 m/s
(T/F) The magnitude of V3 can be...
The velocity vector V1 has a magnitude of 3.0 m/s and is directed along the +x-axis. The velocity vector V2 has a magnitude of 2.0 m/s. The sum of the two is V3, so that V3 = V1+V2
I was never shown how to add vectors and this question came up in class discussion and I was clueless.
??
I do not know if this right but I found:
Vf=17.58m/s
by using Vf^2=2ad.
Than how do I use this to find initial velocity horizontally. Or is that it?
A ball is thrown horizontally from a height of 15.75 m and hits the ground with a speed that is 5.0 times its initial speed. What was the initial speed?
I thought you might have to find time and I got t=1.79s, but after that I cannot seem find my way to the answer
Thanks to whom ever can...