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Natasha2012 [34]
3 years ago
9

IF SOME ONE SENDS U A LINK DO NOT PUSH ON IT CUZ THEY WILL HACK UR STUFF .................. BE CAREFUL PLZ

Physics
1 answer:
Karolina [17]3 years ago
7 0

i know that please spread the word around so everyone can know!

it will help us a very much lot,

thanks.

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A solid ball has a radius of 2cm and a length of 7cm. It has a density of 3.1g/cm3
Iteru [2.4K]
Density is mass divided by volume, you would have to solve for the volume of the ball and rearrange the equation to density divided by volume equals mass
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3 years ago
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Answer:

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2 years ago
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10. How do you use a triple-beam balance? Fill in the blanks.
wolverine [178]

Answer:

1st – Place the film canister on the <u>scale</u>.

2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.

3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.

4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.

5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.

Explanation:

The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.

The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.

5 0
3 years ago
A heat engine with a thermal efficiency of 45 percent rejects 500 kj/kg of heat. how much heat does it receive
adell [148]
Alot as far as i know unless you need it in formal terms.
8 0
3 years ago
One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p
Naddik [55]

Answer:

D. the proper replacement unit for one joule per second

Explanation:

When energy is divided by the time the energy was used we get power

P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{mv^2}{t}\\\Rightarrow P=\dfrac{kg(\dfrac{m^2}{s^2})}{s}\\\Rightarrow P=kg(\dfrac{m^2}{s^2})}\times \dfrac{1}{s}\\\Rightarrow P=\dfrac{kgm^2}{s^3}

kg\dfrac{m^2}{s^2}=Joule

P=\dfrac{kgm^2}{s^3}

So, the answer is D. the proper replacement unit for one joule per second

8 0
3 years ago
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