How is radioactivity produced

A ball is rolling along at speed v without slipping on a horizontal surface when it comes to a hill that rises at a constant ang

MissTica

Answer:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

Explanation:

The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.

This is because:

If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.

Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.

Therefore, static friction is infact necessary for a ball to begin rolling.

Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.

So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.

6 0

3 years ago

While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?

timurjin [86]

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).

5 0

3 years ago

Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$