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3 years ago

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

Physics

1 answer:

pav-90 [236]3 years ago

5 0

Answer:

Explanation:

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.

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

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 1000 $cm^{3}$ = 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

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 $cm^{3}$

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 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 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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yuradex [85]

Answer:

9) a = 25 [m/s^2], t = 4 [s]

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Explanation:

To solve this problem we must use kinematics equations. In this way we have:

$v_{f}^{2} = v_{i}^{2}-(2*a*x)\\$

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a = acceleration [m/s^2]

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$v_{f} =v_{i} - (a*t)$

0 = 100 - (25*t)

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10)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

Note: The positive sign of the equation means that the car increases his speed.

5^2 = 2^2 + 2*a*(125 - 5)

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a = 0.0875 [m/s^2]

Now using the following equation:

$v_{f}= v_{i}+a*t\\$

5 = 2 + 0.0875*t

3 = 0.0875*t

t = 34.3 [s]

11)

To solve this problem we must use kinematics equations. In this way we have:

$v_{f} ^{2} = v_{i} ^{2} + 2*a*(x-x_{o})$

10^2 = 2^2 + 2*a*(200 - 10)

100 - 4 = 2*a* (190)

a = 0.25 [m/s^2]

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$v_{f}= v_{i}+a*t\\$

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8 = 0.25*t

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