Thank you for visiting How much more kinetic energy does a 6 kilogram bowling ball have when it is rolling at 16 mph 7 1 meters per second than. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
We start with the kinetic energy formula:
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6\text{ kg}$ rolling at $v = 7.1\text{ m/s}$, the kinetic energy is calculated as:
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the expression:
$$
(7.1)^2 \approx 50.41 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23\text{ J}.
$$
2. For the ball rolling at $v = 6.2\text{ m/s}$, the kinetic energy is:
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating further:
$$
(6.2)^2 \approx 38.44 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_2 \approx 3 \times 38.44 = 115.32\text{ J}.
$$
3. The increase in kinetic energy when the bowling ball goes from $6.2\text{ m/s}$ to $7.1\text{ m/s}$ is found by the difference:
$$
\Delta KE = KE_1 - KE_2 \approx 151.23\text{ J} - 115.32\text{ J} = 35.91\text{ J}.
$$
Thus, the bowling ball has approximately $35.9\text{ J}$ more kinetic energy when rolling at $7.1\text{ m/s}$ compared to $6.2\text{ m/s}$.
The correct answer is $\boxed{35.9\text{ J}}$.
$$
KE = \frac{1}{2} m v^2,
$$
where $m$ is the mass and $v$ is the speed.
1. For the bowling ball with mass $m = 6\text{ kg}$ rolling at $v = 7.1\text{ m/s}$, the kinetic energy is calculated as:
$$
KE_1 = \frac{1}{2} \times 6 \times (7.1)^2.
$$
Evaluating the expression:
$$
(7.1)^2 \approx 50.41 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_1 \approx 3 \times 50.41 = 151.23\text{ J}.
$$
2. For the ball rolling at $v = 6.2\text{ m/s}$, the kinetic energy is:
$$
KE_2 = \frac{1}{2} \times 6 \times (6.2)^2.
$$
Calculating further:
$$
(6.2)^2 \approx 38.44 \quad \text{and} \quad \frac{1}{2} \times 6 = 3,
$$
so
$$
KE_2 \approx 3 \times 38.44 = 115.32\text{ J}.
$$
3. The increase in kinetic energy when the bowling ball goes from $6.2\text{ m/s}$ to $7.1\text{ m/s}$ is found by the difference:
$$
\Delta KE = KE_1 - KE_2 \approx 151.23\text{ J} - 115.32\text{ J} = 35.91\text{ J}.
$$
Thus, the bowling ball has approximately $35.9\text{ J}$ more kinetic energy when rolling at $7.1\text{ m/s}$ compared to $6.2\text{ m/s}$.
The correct answer is $\boxed{35.9\text{ J}}$.
Thank you for reading the article How much more kinetic energy does a 6 kilogram bowling ball have when it is rolling at 16 mph 7 1 meters per second than. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
- You are operating a recreational vessel less than 39 4 feet long on federally controlled waters Which of the following is a legal sound device
- Which step should a food worker complete to prevent cross contact when preparing and serving an allergen free meal A Clean and sanitize all surfaces
- For one month Siera calculated her hometown s average high temperature in degrees Fahrenheit She wants to convert that temperature from degrees Fahrenheit to degrees
Rewritten by : Jeany
