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3 years ago

Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

Answer:

Explanation:

If you subtract the 5 m/s that the train is already going from 45 m/s (giving you 40 m/s) and then divide that by 8 you’d get B. 5 m/s

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Explain where most of the mass of an atom is located. Also, explain why some particles that make up the atom do not contribute m

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Within an atom, there are three elementary particles: the proton, neutron, and electron. Most of the mass of an atom is situated within the nucleus, which is the central part of the atom. It is made up of protons and neutrons, which are the heaviest subatomic particles. The electrons within the atom, orbit around the nucleus at a very far distance. Electrons are also a part of the lightest group of subatomic particles called leptons. That is why these electrons don't contribute much to the majority of an atoms mass. They are very light and they orbit at very far distances.

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3 years ago

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An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute the values then we get

$-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}$

$\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}$

$v=-\frac{20}{3}=-6.7 cm$

Image distance of apple=-6.7 cm

Magnification=m= $\frac{v}{u}=\frac{-\frac{20}{3}}{-20}$

Magnification of apple= $\frac{1}{3}=0.33$

Hence, the magnification of apple=0.33

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3 years ago

WILL MARK BRAINLIEST!!!!

zloy xaker [14]

Answer:

Wave model

Explanation:

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shepuryov [24]

Answer:

I am pretty sure it is B.

Explanation:

I hope this helped if it didn't I am truly sorry

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3 years ago

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In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

Ratling [72]

Answer:

Electric force, $F=-3.59\times 10^6\ N$

Explanation:

Given that,

Electric charge 1, $q_1=+40\ C$

Electric charge 2, $q_2=-40\ C$

Distance, $d=2\ km=2\times 10^3\ m$

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

$F=k\dfrac{q_1q_2}{d^2}$

k is the electrostatic constant

$F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}$

$F=-3.59\times 10^6\ N$

So, the electric force between these two sets of charges is $-3.59\times 10^6\ N$ .

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3 years ago