Answer :
To solve for the walker's momentum, we will use the fundamental formula for momentum in physics. The formula for momentum ([tex]\( p \)[/tex]) is given by:
[tex]\[ p = m \cdot v \][/tex]
where:
- [tex]\( p \)[/tex] is the momentum,
- [tex]\( m \)[/tex] is the mass,
- [tex]\( v \)[/tex] is the velocity.
Given in the problem:
- The mass ([tex]\( m \)[/tex]) of the walker is [tex]\( 15 \, \text{kg} \)[/tex],
- The velocity ([tex]\( v \)[/tex]) of the walker is [tex]\( 10 \, \text{m/s} \)[/tex].
Substitute the given values into the formula:
[tex]\[ p = 15 \, \text{kg} \cdot 10 \, \text{m/s} \][/tex]
By performing the multiplication, we get:
[tex]\[ p = 150 \, \text{kg} \cdot \text{m/s} \][/tex]
So, the walker's momentum is [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex].
Looking at the given choices:
- A. [tex]\( 75 \, \text{kg} \cdot \text{m/s} \)[/tex]
- B. [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex]
- C. [tex]\( 15 \, \text{kg} \cdot \text{m/s} \)[/tex]
- D. [tex]\( 5 \, \text{kg} \cdot \text{m/s} \)[/tex]
The correct answer is B. [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex].
[tex]\[ p = m \cdot v \][/tex]
where:
- [tex]\( p \)[/tex] is the momentum,
- [tex]\( m \)[/tex] is the mass,
- [tex]\( v \)[/tex] is the velocity.
Given in the problem:
- The mass ([tex]\( m \)[/tex]) of the walker is [tex]\( 15 \, \text{kg} \)[/tex],
- The velocity ([tex]\( v \)[/tex]) of the walker is [tex]\( 10 \, \text{m/s} \)[/tex].
Substitute the given values into the formula:
[tex]\[ p = 15 \, \text{kg} \cdot 10 \, \text{m/s} \][/tex]
By performing the multiplication, we get:
[tex]\[ p = 150 \, \text{kg} \cdot \text{m/s} \][/tex]
So, the walker's momentum is [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex].
Looking at the given choices:
- A. [tex]\( 75 \, \text{kg} \cdot \text{m/s} \)[/tex]
- B. [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex]
- C. [tex]\( 15 \, \text{kg} \cdot \text{m/s} \)[/tex]
- D. [tex]\( 5 \, \text{kg} \cdot \text{m/s} \)[/tex]
The correct answer is B. [tex]\( 150 \, \text{kg} \cdot \text{m/s} \)[/tex].