Which force, in real life, will have the least effect on a bowling ball rolling down a lane toward bowling pins?

A) magnetism

B) air resistance

C) gravity

D) friction

Answers

Answer 1

Answer:

Its Friction

Explanation:

the pins are not floating and they are not a magnet and not involved with air

Answer 2

The force, in real life, that will have the least effect on a bowling ball rolling down a lane toward bowling pins is magnetism. The correct option is A.

What is magnetism?

Magnetism is basically the force which indeed magnets exert when they attract or even repel one another. The movement of electric charges resulting in magnetism.

Every substance is composed of tiny units referred to as atoms. Each atom contains electrons, which are charged particles.

To increase stability, the pins themselves have a low center of gravity. Because they are spherical in shape, they can roll and strike other pins in a variety of directions.

The force acting on the bowling ball is friction and air resistance. The friction force is equal to the friction coefficient multiplied by the normal force, and thus mass times acceleration.

Thus, the correct option is A.

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Related Questions

Option B.
Consider a setup in which two springs are attached to a mass in parallel.
Convince yourself that in this setup, the compression of each spring must be the same. Using
this fact, derive the effective spring constant for springs in parallel
This is asking, "ll1 replace the two springs by a single imaginary spring, what would its spring
constant be such that the force stays the same?" Your answer should only depend on k, and k

Answers

Answer:

it would be...

Explanation:

Vesta is a minor planet (asteroid) that takes 3.63 years to orbit the Sun.
Calculate the average sun -Vesta distance

Answers

Using Kepler's third law, the average sun -Vesta distance is 2.36 AU.

According to Kepler's laws, the square of the period of revolution of planets are proportional to the cube of their average distances from the sun. Hence, we can write; [tex]T^{2} =r^{3}[/tex]

Where;

T = period of the planet

r = average distance of the planet

When;

T =  3.63 years

r = [tex]\sqrt[3]{T^2}[/tex]

r = [tex]\sqrt[3]{(3.63)^2}[/tex]

r = 2.36 AU

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