Can someone please help

A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what

kondor19780726 [428]

The kinetic energy of the tomato is :

K.E = 1/2 mv^2

K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

Hope this helps

7 0

3 years ago

A machine is applying a torque to rotationally accelerate a metal disk during a manufacturing process. An engineer is using a gr

Viefleur [7K]

Explanation:

Question 9 A machine is applying a torque to rotationally accelerate a metal disk during a manufacturing process. An engineer is using a graph of torque as a function of time to determine how much the disk's angular speed increases during the process The graph of torque as a function of time starts at an initial torque value and is a straight line with positive slope. What aspect of the graph and possibly other quantities must be used to calculate how much the disk's angular speed increases during the process? The slope of the graph multiplied by the disk's radius will equal the change in angular speed The area under the graph multiplied by the disk's radius will equal the change in angular speed. The slope of the graph divided by the disk's rotational inertia will equal the change in angular speed. The area under the graph divided by the disk's rotational inertia will equal the change in angular speed. The area under the graph multiplied by the disk's rotational inertia will equal the change in angular speed E

8 0

3 years ago

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A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis

Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

$M_{before} = M_{after}$