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zimovet [89]
3 years ago
14

Each of two identical objects carries a net charge. The objects are made from conducting material. One object is attracted to a

positively charged ebonite rod, and the other is repelled by the rod. After the objects are touched together, it is found that they are each attracted by the rod. What can be concluded about the initial charges on the objects
Physics
1 answer:
lilavasa [31]3 years ago
4 0

Answer:

Initially one object is positive and one is negative , with the positive charge having a greater magnitude than the negative charge.

Explanation:

The object that is attracted to a positively charged ebonite rod, must have the negative charge as the oppose charges attract. However the object that is repelled by the rod must have the positive charge because the same charges repel that's why object move away from the rod.

Now after the objects are touched together, then there must be a flow of charge from one object to other until they got the equal charge in magnitude. As the both objects repel the rod it means both have now got the positive charge. So there must be initially a positive charge wih higher magnitude.

So we can conclude that,

" Initially one object is positive and one is negative , with the positive charge having a greater magnitude than the negative charge."

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Home Business Cottage and Small Industries Development Committee to organize National Industrial Goods and...

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Cottage and Small Industries Development Committee to organize National Industrial Goods and Technology Expo

By Glocal Khabar - 2368 0

National Industrial Goods and Technology Expo- Glocal Khabar

Kathmandu, February 19, 2018: Cottage and Small Industries Development Committee is set to organize the twenty-ninth edition of National Industrial Goods and Technology Expo from March 23, 2018 at Bhrikutimandap, Kathmandu.

The motto of the five-day event is ‘Let’s use home-made products, move ahead towards prosperity.’ The committee has shared that handicrafts, wool and bamboo products, goods made from handmade papers, different types of pickles and Palpali Dhaka would be put on display in the expo.

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3 years ago
I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......
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Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100cm^{2} x 10cm = 1000cm^{3}

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

\frac{1g}{cm^{3} } x 1000cm^{3} = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

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The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ \frac{8g}{cm^{3} } = 2.5 cm^{3}

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5cm^{3} overflowed. So now we the same process as in number a) just with a few adjustments.

\frac{1g}{cm^{3} } x (1000cm^{3} - 2.5cm^{3}) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

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Explanation:

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