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Sagot :
Sure, let's determine the momentum of a 15 kg tire rolling down a hill at a speed of 3 m/s.
Momentum ([tex]\(p\)[/tex]) is defined as the product of mass ([tex]\(m\)[/tex]) and velocity ([tex]\(v\)[/tex]). The formula to calculate momentum is:
[tex]\[ p = m \times v \][/tex]
In this problem, we are given:
- The mass [tex]\(m\)[/tex] of the tire is [tex]\(15 \ \text{kg}\)[/tex]
- The velocity [tex]\(v\)[/tex] of the tire is [tex]\(3 \ \text{m/s}\)[/tex]
Now, substitute the given values into the formula:
[tex]\[ p = 15 \ \text{kg} \times 3 \ \text{m/s} \][/tex]
[tex]\[ p = 45 \ \text{kg} \cdot \text{m/s} \][/tex]
So, the momentum of the tire is:
[tex]\[ \boxed{45 \ \text{kg} \cdot \text{m/s}} \][/tex]
The correct answer from the given options is [tex]\(45 \ \text{kg} \cdot \text{m/s}\)[/tex].
Momentum ([tex]\(p\)[/tex]) is defined as the product of mass ([tex]\(m\)[/tex]) and velocity ([tex]\(v\)[/tex]). The formula to calculate momentum is:
[tex]\[ p = m \times v \][/tex]
In this problem, we are given:
- The mass [tex]\(m\)[/tex] of the tire is [tex]\(15 \ \text{kg}\)[/tex]
- The velocity [tex]\(v\)[/tex] of the tire is [tex]\(3 \ \text{m/s}\)[/tex]
Now, substitute the given values into the formula:
[tex]\[ p = 15 \ \text{kg} \times 3 \ \text{m/s} \][/tex]
[tex]\[ p = 45 \ \text{kg} \cdot \text{m/s} \][/tex]
So, the momentum of the tire is:
[tex]\[ \boxed{45 \ \text{kg} \cdot \text{m/s}} \][/tex]
The correct answer from the given options is [tex]\(45 \ \text{kg} \cdot \text{m/s}\)[/tex].
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