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Karolina [17]
2 years ago
12

You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal

culate the Rotational Kinetic Energy of the bowling ball as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. A bowling ball can be modeled as a solid sphere rotating about its center. This bowling ball has a mass of 6.40 kg and a radius of 0.130 m. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with an angular velocity of 16.0 radians / second in the counter-clockwise (or positive) direction. You can use this to determine the linear velocity of the bowling ball (since it is rolling without slipping). What is the Total Kinetic Energy of the bowling ball
Physics
1 answer:
RideAnS [48]2 years ago
6 0

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

         K_{total} = K_{traslation}  + K_{rotation}

let's look for each energy

linear

        K_{traslation} = ½ m v²

rotation

        K_{rotation} = ½ I w²

the moment of inertia of a solid sphere is

       I = 2/5 m r²

we substitute

       K_{total} = ½ mv² + ½ I w²

           

angular and linear velocity are related

           v = w r

we substitute

           K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

           K_{total} = m w² r² (½ + 1/5)

           K_{total} = \frac{7}{10} m w² r²

let's calculate

           K_{total} = \frac{7}{10}   6.40 16.0² 0.130²

           K_{total} = 19.4 J

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xenn [34]

Answer:

Explanation:

1. The amount of heat needed to melt ice at 0°C is equal to the mass of the ice times the latent heat of fusion.

q = mL

q = (450 g) (334 J/g)

q = 150,300 J

q = 150 kJ

2. The amount of heat released by the condensation of steam at 100°C is equal to the mass of the steam times the latent heat of vaporization.

q = mL

q = (325 g) (2260 J/g)

q = 734,500 J

q = 735 kJ

3. q = mL

q = (85 g) (2260 J/g)

q = 192,100 J

q = 190 kJ

4. q = mL

q = (225 g) (334 J/g)

q = 75,150 J

q = 75.2 kJ

5. Above 100°C, water is steam.  The amount of heat needed to increase the temperature of steam is equal to its mass times its specific heat times the change in temperature.

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q = (20.0 g) (2.03 J/g/°C) (303.0°C − 283.0°C)

q = 812 J

6. q = mCΔT

q = (15.0 g) (2.03 J/g/°C) (250.0°C − 275.0°C)

q = -761 J

7. q = mCΔT

q = (10.0 g) (0.90 J/g/°C) (55°C − 22°C)

q = 297 J

8. q = mCΔT

198 J = (55.0 g) C (15°C)

C = 0.24 J/g/°C

9. q = mCΔT

41,840 J = m (4.184 J/g/°C) (28.5°C − 22.0°C)

m = 1540 g

10. q = mCΔT

q = (193 g) (2.46 J/g/°C) (35°C − 19°C)

q = 7600 J

11. First, the temperature of the ice must be raised to 0°C.

q = mCΔT

q = m (2.09 J/g/°C) (0°C − (-23.0°C))

q/m = 48.1 J/g

Next, the ice must be melted.

q = mL

q/m = 334 J/g

Then, the water must be heated to 100°C.

q = mCΔT

q = m (4.184 J/g/°C) (100°C − 0°C)

q/m = 418.4 J/g

The water is then vaporized.

q = mL

q/m = 2260 J/g

Finally, the steam is heated to its final temperature.

q = mCΔT

q = m (2.03 J/g/°C) (118°C − 100°C)

q/m = 36.5 J/g

So the total amount of energy needed is:

q/m = 48.1 J/g + 334 J/g + 418.4 J/g + 2260 J/g + 36.5 J/g

q/m = 3100 J/g

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