A student drops the same 10 kilogram bowling ball three separate times. She drops the ball from heights of 1 meter, 2 meters, and 3 meters. Which graph correctly shows the ball’s kinetic energy when dropped from each height?(1 point)
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Sorry, but I'm unable to provide the graph as I'm a text-based AI assistant. However, the graph should show that as the height increases, the kinetic energy of the ball also increases. It should be a positive relationship, with the kinetic energy increasing linearly or exponentially with the height.
To determine the kinetic energy of the bowling ball when dropped from different heights, we can use the formula:
KE = 0.5 * m * v^2
where KE is the kinetic energy, m is the mass of the bowling ball, and v is the velocity.
Since the mass of the bowling ball is 10 kilograms, the kinetic energy will depend on the velocity at each height. The velocity can be found using the principle of conservation of energy:
mgh = 0.5 * m * v^2
where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), h is the height, and v is the velocity.
For each height, we can calculate the velocity and then use it to find the kinetic energy.
For 1 meter height:
Potential energy (mgh) = 10 kg * 9.8 m/s^2 * 1 m = 98 J
Velocity (v) = sqrt(2 * (potential energy/m)) = sqrt(2 * (98 J / 10 kg)) = 4.43 m/s
Kinetic energy (KE) = 0.5 * 10 kg * (4.43 m/s)^2 = 97.94 J (approximately)
For 2 meters height:
Potential energy (mgh) = 10 kg * 9.8 m/s^2 * 2 m = 196 J
Velocity (v) = sqrt(2 * (potential energy/m)) = sqrt(2 * (196 J / 10 kg)) = 6.26 m/s
Kinetic energy (KE) = 0.5 * 10 kg * (6.26 m/s)^2 = 196.5 J (approximately)
For 3 meters height:
Potential energy (mgh) = 10 kg * 9.8 m/s^2 * 3 m = 294 J
Velocity (v) = sqrt(2 * (potential energy/m)) = sqrt(2 * (294 J / 10 kg)) = 8.08 m/s
Kinetic energy (KE) = 0.5 * 10 kg * (8.08 m/s)^2 = 326.72 J (approximately)
Based on these calculations, the graph showing the ball's kinetic energy correctly would have the highest value (326.72 J) for the 3 meters height, followed by the 2 meters height (196.5 J), and the lowest value (97.94 J) for the 1-meter height.
To calculate the kinetic energy of an object, we use the formula KE = 1/2 * m * v^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity of the object.
In this case, the mass of the bowling ball is given as 10 kilograms, and we are dropping it from heights of 1 meter, 2 meters, and 3 meters.
We can calculate the velocity of the ball at each height using the formula v = sqrt(2 * g * h), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height.
For the first drop, the height is 1 meter. Therefore, the velocity of the ball just before it hits the ground can be calculated as v = sqrt(2 * 9.8 * 1) = sqrt(19.6) m/s.
Using this value of velocity, we can calculate the kinetic energy as KE = 1/2 * 10 * (sqrt(19.6))^2 = 1/2 * 10 * 19.6 = 98 Joules.
Similarly, for the second drop from a height of 2 meters, we can calculate the velocity as v = sqrt(2 * 9.8 * 2) = sqrt(39.2) m/s.
Using this value of velocity, the kinetic energy is KE = 1/2 * 10 * (sqrt(39.2))^2 = 1/2 * 10 * 39.2 = 196 Joules.
For the third drop from a height of 3 meters, the velocity can be calculated as v = sqrt(2 * 9.8 * 3) = sqrt(58.8) m/s.
Using this value of velocity, the kinetic energy is KE = 1/2 * 10 * (sqrt(58.8))^2 = 1/2 * 10 * 58.8 = 294 Joules.
The correct graph would show the kinetic energy on the y-axis (vertical axis) and the height of the drop on the x-axis (horizontal axis). The values for the kinetic energy for each height would be plotted as points on the graph and then connected to form a line. The graph should show an increasing trend since the kinetic energy increases with the height of the drop.
Please note that without the graph options provided, I cannot directly point out the correct graph. However, you can use the information I provided to analyze the given options and identify the graph that correctly shows the ball's kinetic energy when dropped from each height.