What is the net force on a ball that falls downward?

1 answer:

6 0

There's no way to tell without knowing the downward acceleration of the ball.

For example, if it's falling at a constant speed, then the net vertical force on it is zero, and we still don't know anything about the net horizontal force.

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How does the electrostatic force compare with the strong nuclear force in the

diamong [38]

Answer:

Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force

Explanation:

There are mainly two forces acting between protons and neutrons in the nucleus:

- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by

$F_E=\frac{kq_1 q_2}{r^2}$

where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.

The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.

- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.

At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.

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2 years ago

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Which tool is a wheel and axle?

Anton [14]

A steering wheel, a wrench, a screwdriver, and the back wheel of a bike are all examples of tools with a wheel and axle.

5 0

2 years ago

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The image above was taken by the spacecraft Messenger as it flew by the planet Mercury. The terrain shown in this image is typic

lorasvet [3.4K]

The answer is

C

Hope this helps

7 0

3 years ago

You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice

Galina-37 [17]

The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

$\sum F =ma$

where

$\sum F = F_f = -0.14 N$ is the net force acting on the puck (the force of friction, negative because it acts in a direction opposite to the direction of motion)