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vaieri [72.5K]
3 years ago
8

In which situations can you conclude that the object is undergoing a net interaction with one or more other objects? (Select all

that apply.
1)A car travels at constant speed around a circular race track.
2)A book slides across the table and comes to a stop.
3)A hydrogen atom remains at rest in outer space.
4)A proton in a particle accelerator moves faster and faster.
5)A spacecraft travels at a constant speed toward a distant star.
Physics
1 answer:
just olya [345]3 years ago
5 0

Answer:

<em>4, A proton in a particle accelerator moves faster and faster.</em>

Explanation:

A particle accelerator according to Wikipedia.com is a machine that uses electromagnetic fields to propel charged particles to very high speeds and energies, and to contain them in well-defined beams. Large accelerators are used for basic research in particle physics.

As the proton moves around, it collides with other proton and the wall of the accelerator thereby undergoing net interaction.

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As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
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Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

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What are facts about balloons?
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Answer:

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