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Sagot :
To find the impulse John applies to increase his speed, we can use the relationship between impulse, mass, and change in velocity. Impulse ([tex]\(J\)[/tex]) is defined as the change in momentum of an object. Momentum ([tex]\(p\)[/tex]) is the product of mass ([tex]\(m\)[/tex]) and velocity ([tex]\(v\)[/tex]).
The formula for impulse is:
[tex]\[ J = \Delta p = m \cdot \Delta v \][/tex]
where:
- [tex]\( \Delta p \)[/tex] is the change in momentum,
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( \Delta v \)[/tex] is the change in velocity.
Given:
- John's mass ([tex]\( m \)[/tex]) = 103 kg,
- Initial speed ([tex]\( v_{\text{initial}} \)[/tex]) = 4 m/s,
- Final speed ([tex]\( v_{\text{final}} \)[/tex]) = 5 m/s.
First, we calculate the change in velocity ([tex]\( \Delta v \)[/tex]):
[tex]\[ \Delta v = v_{\text{final}} - v_{\text{initial}} \][/tex]
[tex]\[ \Delta v = 5 \, \text{m/s} - 4 \, \text{m/s} = 1 \, \text{m/s} \][/tex]
Now, we use the mass and the change in velocity to find the impulse:
[tex]\[ J = m \cdot \Delta v \][/tex]
[tex]\[ J = 103 \, \text{kg} \cdot 1 \, \text{m/s} = 103 \, \text{kg} \cdot \text{m/s} \][/tex]
Therefore, the impulse that John applies is:
[tex]\[ J = 103 \, \text{kg} \cdot \text{m/s} \][/tex]
So, the correct answer is:
B. 103 kg-m/s
The formula for impulse is:
[tex]\[ J = \Delta p = m \cdot \Delta v \][/tex]
where:
- [tex]\( \Delta p \)[/tex] is the change in momentum,
- [tex]\( m \)[/tex] is the mass of the object,
- [tex]\( \Delta v \)[/tex] is the change in velocity.
Given:
- John's mass ([tex]\( m \)[/tex]) = 103 kg,
- Initial speed ([tex]\( v_{\text{initial}} \)[/tex]) = 4 m/s,
- Final speed ([tex]\( v_{\text{final}} \)[/tex]) = 5 m/s.
First, we calculate the change in velocity ([tex]\( \Delta v \)[/tex]):
[tex]\[ \Delta v = v_{\text{final}} - v_{\text{initial}} \][/tex]
[tex]\[ \Delta v = 5 \, \text{m/s} - 4 \, \text{m/s} = 1 \, \text{m/s} \][/tex]
Now, we use the mass and the change in velocity to find the impulse:
[tex]\[ J = m \cdot \Delta v \][/tex]
[tex]\[ J = 103 \, \text{kg} \cdot 1 \, \text{m/s} = 103 \, \text{kg} \cdot \text{m/s} \][/tex]
Therefore, the impulse that John applies is:
[tex]\[ J = 103 \, \text{kg} \cdot \text{m/s} \][/tex]
So, the correct answer is:
B. 103 kg-m/s
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