Recent content by Amanda567

Ok - so if the ball starts at rest, the acceleration is zero, and the distance is zero because it is still in the pitchers hand. I think?

thanks so much!

w=2pi/86400sec w=0.00007 0.00007(6370km)=0.46324km/sec vt=0.46324km^2/6370 vt=0.0000336 Thats what i get, what am i doing wrong?

What is E-2?

Angular velocity is what represents speed? by: F=d/dt (mv)

Would that be the period T, where that is the time to complete one revolution? T=2pi(r)/v

Sorry! I know that gravitational force is Fg=mg And Vxf=Vxi + axt for constant a

Sorry! I know that centripetal acceleration is v^2/r The diameter of the Earth is 12740 km, and half of that is the radius of 6370km. How would I find v?

Knowing that the diameter of the Earth at the equator is 12 740 km, compute the radial acceleration of a point on the surface of the Earth at the equator, due to the rotation of the Earth about its axis.

The gravitational force on a baseball is −Fgjˆ. A pitcher throws the baseball with velocity viˆ by uniformly accelerating it straight forward horizontally for a time interval ∆t =t–0 = t. If the ball starts from rest, (a) through what distance does it accelerate before its release? (b) What...