How is energy distributed in the ocean?

A Eureka can of mass 60g and cross sectional area of 60 square centimeters is fillied with water of density 1g/ cubic centimeter

Natalka [10]

a.600 cm³

b. 2.5 g/cm³

c. 2.0775 kg

<h3>What is Eureka moment?</h3>

In Physics, this Archimedes' Eureka moment is called the Archimedes Principle, which states that when a body is immersed in a liquid, it experiences an upward buoyant force, which is equal to the weight of the liquid displaced by the body.

According to the question

Mass of Eureka is 60g

Cross sectional area is 60 square centimeters

Mass of steel piece is 20g

Density = 8g /cubic centimeter

a)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.

To find 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.

Here,

60cm² x 10cm = 600 cm³

To find the mass that water has in this case we have to multiply the water's density by the volume,

so we get....

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

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 + 60g = 1060g or 1.06kg

b)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 ÷ $\frac{8}{cm^3}$ = 2.5 g/cm³

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 2.5 overflowed.

So now we the same process as in number a) just with a few adjustments.

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

Hence,

1060g + 997.5g + 20g = 2077.5 g or 2.0775 kg

Learn more about Archimedes Principle here:

brainly.com/question/13106989

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4 0

2 years ago

Calculate the mass of earth

Roman55 [17]

Answer:

$m=6.85\times 10^{24}\ kg$

Explanation:

The radius of Earth is $6.67\times 10^6\ m$

The density of an object is given by :

$d=\dfrac{m}{V}$

The density of Erth is, d = 5515 kg/m³

Where

m is the mass of the Earth

So,

$m=d\times V\\\\m=\dfrac{4}{3}\pi r^3\times d$

Put all the values,

$m=\dfrac{4}{3}\pi \times (6.67\times 10^6)^3\times 5515 \\\\m=6.85\times 10^{24}\ kg$

So, the mass of the Earth is equal to $6.85\times 10^{24}\ kg$ .

6 0

3 years ago

What are the differences between convex lens and convex mirrors ?

MaRussiya [10]

Answer:

The mirror is opaque while the lens is transparent

Explanation:

The lens is transparent throughout (lets light pass through), while the mirror reflects light. You are basically asking for the difference between a lens and a mirror

But how about the the behaviors of the convex mirror and convex lens? The convex lens will let light pass through and focus it on the other side of it. If you stand on one side of the lens, you can see what is on the other side of it. Think about a camera lens that's just curved!The convex mirror will reflect the light on the same side of it you'll basically will just see yourself but you'll look funny.

4 0

3 years ago

If an ideal monatomic gas undergoes an adiabatic expansion, in which the volume increases by a factor of 4.0, by what factor doe

julsineya [31]

Answer:

e) 0.099

Explanation:

For an adiabatic expansion:

$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$