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Free_Kalibri [48]
3 years ago
12

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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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

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Answer:

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From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

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Centripetal acceleration enables Rotation therefore?

     \omega ^2 r =Angular\ acc

Considering the outer ring,

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Generally the equation for solving Period T is mathematically given as

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Answer:

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