Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover precise answers to your questions from a wide range of experts on our user-friendly Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Select the best answer for the question.

What is the velocity of a skateboarder whose momentum is [tex][tex]$100 \, \text{kg} \cdot \text{m/s}$[/tex][/tex] and mass is [tex][tex]$15 \, \text{kg}$[/tex][/tex]? Round to the nearest hundredth.

A. [tex][tex]$13.25 \, \text{m/s}$[/tex][/tex]
B. [tex][tex]$6.67 \, \text{m/s}$[/tex][/tex]
C. [tex][tex]$6.67 \, \text{kg}$[/tex][/tex]
D. [tex][tex]$3.33 \, \text{m/s}$[/tex][/tex]

Sagot :

To determine the velocity of a skateboarder whose momentum is [tex]\(100 \, \text{kg} \cdot \text{m} / \text{s}\)[/tex] and mass is [tex]\(15 \, \text{kg}\)[/tex], we use the formula for momentum, which is given by:

[tex]\[ p = mv \][/tex]

where:
- [tex]\( p \)[/tex] is momentum
- [tex]\( m \)[/tex] is mass
- [tex]\( v \)[/tex] is velocity

We aim to find the velocity [tex]\( v \)[/tex]. Rearranging the formula to solve for [tex]\( v \)[/tex], we get:

[tex]\[ v = \frac{p}{m} \][/tex]

Substitute the given values into the equation:

[tex]\[ v = \frac{100 \, \text{kg} \cdot \text{m} / \text{s}}{15 \, \text{kg}} \][/tex]

Simplifying this:

[tex]\[ v = 6.666666666666667 \, \text{m} / \text{s} \][/tex]

To give the answer rounded to the nearest hundredth, we round [tex]\( 6.666666666666667 \)[/tex] to [tex]\( 6.67 \)[/tex].

Therefore, the velocity of the skateboarder is [tex]\( 6.67 \, \text{m} / \text{s} \)[/tex].

The correct answer is:
B. [tex]\( 6.67 \, \text{m} / \text{s} \)[/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.