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3 years ago

Difference between MKS system and CGS system

Physics

1 answer:

ivolga24 [154]3 years ago

8 0

Answer:

MKS stands for Meter, Kilogram and second. In this system of unit mass is given in Kilogram, length in meter and time in second. ... CGS system stands for Centimeter- Gram- Second system. In CGS system, length is measured in centimeters mass is measured in grams and time is in seconds.

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How much work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s?

soldi70 [24.7K]

Answer:

The correct option is (b).

Explanation:

We need to find the work done to increase the speed of a 1 kg toy car by 5 m/s.

We know that, the work done is equal to the kinetic energy of an object i.e.

$W=\Delta K\\\\W=\dfrac{1}{2}mv^2\\\\W=\dfrac{1}{2}\times 1\times 5^2\\W=12.5\ J$

So, 12.5 J of work is done to increase the speed of a 1.0 kg toy car by 5.0 m/s.

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Jesse wants to know how well a particular brand of car wax protects his car from dirt. What is the independent variable ?

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2 years ago

how would the velocity of the book change if the applied force were equal to the sliding friction force

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3 years ago

A block of concrete has a mass of 48kg a crane lifts the block to a height of 12m above the ground calculate the gravitational p

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Answer:

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From the question given above, the following data were obtained:

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3 years ago

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

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3 years ago

Difference between MKS system and CGS system​

Difference between MKS system and CGS system