Speed = (distance) / (time)
Speed = (100 mi/2 hr)
Multiply by a fraction equal to 1 :
Speed = (100 mi/2 hr) x (1 hr/60 min)
Speed = (100 x 1 / 2 x 60) (mi-hr/hr-min)
Speed = (100 / 120) (mi / min)
Speed = 5/6 mi/min
<em>Speed = 0.833... mi/min</em>
Answer:
Explanation:
It is given that the sphere is insulated from ground and a large charge is placed on the sphere. The charge on the hollow sphere will always remain on the outer surface of the sphere and there will be no charge on the inner surface of the sphere.
If a person touches the inner surface of the sphere then he will not be harmed as there is no charge on the inner surface of the sphere.
If a person carries the charge of the opposite sign of the same magnitude then the sphere and person get neutralized upon touching the sphere.
If a person does not touches the sphere then the charge on the outer surface will be zero and there will be a positive charge on the inner surface of the sphere
A ball falling through the air has a mass, a density, a volume...it is facing air resistance and is being acted on by gravity...it is accelerating and gaining velocity...and it is increasing in kinetic energy.
I suppose out of all those the biggest thing the ball has in this case is ENERGY. There are two main types to focus on...
Kinetic Energy - The further the ball fall the more KE it has...until terminal velocity is reach, then KE would become constant.
Potential Energy - Conversely to that of KE, the further the ball falls the less PE it will have.
<em>Heat/Thermal Energy is technically also present due to the friction from the air resistance, but the transfer of energy between the air and ball is quite complex and not necessary important for basic physics.
</em>
The question itself seem kind of vague and open ended, but I could just be viewing it the wrong way.
Comment if you need more help!
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
