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3 years ago

If a 5 kg bowling ball is lifted to a height of 2 meters, how much potential

Physics

1 answer:

natka813 [3]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 98.1 \ Joules}}$

Explanation:

The formula for potential energy is:

$E_p=mgh$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass is 5 kilograms and the height is 2 meters. Assuming this ball is on Earth, then the gravitational acceleration is 9.81 meters per square second.

$m= 5 \ kg \\g= 9.81 \ m/s^2 \\\h= 2 \ m$

Substitute the values into the formula.

$E_p= ( 5 \ kg )(9.81 \ m/s^2)( 2 \ m)$

Multiply the first two numbers.

$E_p=49.05 \ kg*m/s^2 (2 \ m )$

Multiply again.

$E_p= 98.1 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer is equal to 98.1 Joules

$E_p= 98.1 \ J$

The ball has 98.1 Joules of potential energy.

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A particle undergoes a constant acceleration of 3.90 m/s2. After a certain amount of time, its velocity is 12.2 m/s. (Where appl

Law Incorporation [45]

Answer:

$s_{f} = 14.312 m$

Explanation:

Since the particle is experimenting a constant acceleration, the displacement can be found by using this formula:

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Then, the displacement is now isolated:

$s_{f} = \frac{v_{f}^{2}-v_{o}^{2}}{2 \cdot a}$

Terms are replaced herein:

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

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Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

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3 years ago