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olya-2409 [2.1K]
2 years ago
7

The volume of the water in the graduated cylinder rose as some of the water was displaced by the table tennis ball. Find the vol

ume of the ball using the formula
Vb = Vf – Vi
where Vb is the volume of the ball, Vf is the final volume of the water, and Vi is the initial volume of the water. Record the volume of the ball in Table A of your Student Guide.

What is the volume of the table tennis ball?

cm3
Physics
1 answer:
Leokris [45]2 years ago
6 0

The approximate volume of table tennis ball is  80 cm³

<h3>What is volume?</h3>

Volume is defined as the amount of space occupied by the three dimensional object. S I unit of volume is m³ or cm³.

To find the volume of tennis ball using graduated cylinder.

Step 1 - Fill the graduated cylinder half or full.

Step 2 - Mark the initial volume of the water i.e. 100 cm³ (Vi)

Step 3  - Put the tennis ball in the graduated cylinder. Some of the water was displaced by the table tennis ball.

Step 4 - Mark the Final volume of the water (Vf) i.e. 180 cm³

Step 5 = Calculate the volume by using Formula

Vb = Vf – Vi = 180 cm³ - 100 cm³ = 80 cm³

Hence the volume of tennis ball (Vb) is 80 cm³

For more Volume related question visit here:

brainly.com/question/14996332

#SPJ1

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1) 293 ^{\circ}C

2) 859^{\circ}C

Explanation:

1)

The average kinetic energy of the molecules of an ideal gas is directly related to the Kelvin temperature of the gas, by the formula

KE=\frac{3}{2}kT

where

KE is the kinetic energy

k is the Boltzmann constant

T is the Kelvin temperature

We can say  therefore that the average kinetic energy of the particles is directly proportional to the absolute temperature of the gas; so, we can write:

KE\propto T

And therefore

\frac{KE_1}{KE_2}=\frac{T_1}{T_2} (1)

In this problem, we have:

KE_1 = K_{10} is the initial kinetic energy of the molecules when the temperature of the gas is

T_1=10^{\circ}+273=283 K

Here we want to find the temperature T_2 at which the average kinetic energy of the particles is

KE_2=2K_{10}

So, twice the initial value. Substituting into eq.(1) and solving for T2, we find:

T_2=\frac{T_1 KE_2}{KE_1}=\frac{(283)(2K_{10})}{K_{10}}=566 K

Converting into Celsius degrees,

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2)

The root-mean-square (rms) speed of the molecules in a gas is given by the equation

v=\sqrt{\frac{3kT}{m}}

where

k is the Boltzmann constant

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m is the mass of each molecule

Therefore, from the equation we can say that the rms speed is proportional to the square root of the temperature:

v\propto \sqrt{T}

So we can write:

\frac{v_1}{v_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}} (2)

where in this problem:

v_1 = v_{rms} is the rms speed of the molecules when the temperature is

T_1=10^{\circ}C+273=283 K

v_2=2v_{rms} is the final rms speed of the molecules

Solving eq.(2), we find the temperature at which the rms speed is twice the initial value:

T_2=T_1 (\frac{v_2}{v_1})^2=(283)(\frac{2v_{rms}}{v_{rms}})^2=1132 K

Converting into Celsius degrees,

T_2=1132-273=859^{\circ}C

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Answer:

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