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Tresset [83]
3 years ago
15

For an object that floats in a liquid, which of the following statements is true? Group of answer choices The specific gravity o

f the object is greater than or equal to the specific gravity of the liquid. The mass of the object is equal to the mass of liquid displaced.
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Explanation:

Object is Floating in a liquid

weight of object is equal to weight of displaced fluid

i.e.

\rho _oVg=\rho _mV'g

i.e.\rho _oV=mass\ of\ object

\rho _mV'=mass\ of\ liquid\ displaced

where V and V' is the volume of object and volume of object inside the liquid.

therefore mass of liquid displaced must be equal to mass of object                          

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a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100
Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

       

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = \frac{4}{3} \ \pi r^{3}

          V = \frac{4}{3} π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = \frac{0.010}{4.189 \ 10^{-3} }

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

             

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm

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3 years ago
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The properties of minerals that might help Jimmy and his partners to identify the sample other than Hardness (scratch test) are:

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3 years ago
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How much work does it take to lift 345 boxes to a height of 6.00 m of each box has a mass of 7.89 kg
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Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

W = F × d

Where F is force applied or Weight and d is distance

Also Force = Weight = mass × acceleration due to gravity.

Since gravity is acting on the boxes as it been lift

W = Weight × height from ground level

W = mg × d

Where m is mass of the boxes, g is accelration due to gravity( g = 9.8m/s² ) and d is distance from ground level.

Given the data in the question;

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To determine the work done, we substitute our values into the expression above.

W = mg × d

W = 2722.05kg × 9.8m/s² × 6.0m

W = 160056.5kgm²/s²

W = 160056.5J

W = 1.6 × 10⁵J

Therefore,  Given the value of the mass of each boxes, the work done in lifting the boxes to the given height is 1.6 × 10⁵J.

Learn more about work done here: brainly.com/question/26115962

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