Respuesta :
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed
[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]
[tex]v_{1}=0.833\ m/s[/tex]
The force F₁is constant acceleration is also a constant.
[tex]F_{1}=ma_{1}[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a_{1}=\dfrac{v}{t}[/tex]
[tex]a_{1}=\dfrac{0.833}{10}[/tex]
[tex]a_{1}=0.083\ m/s^2[/tex]
Similarly,
[tex]F_{2}=ma_{2}[/tex]
For total force,
[tex]F_{3}=F_{2}+F_{1}[/tex]
[tex]ma_{3}=ma_{2}+ma_{1}[/tex]
The speed of second tugboat is
[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]
[tex]v=3.05\ m/s[/tex]
We need to calculate total acceleration
[tex]a_{3}=\dfrac{v}{t}[/tex]
[tex]a_{3}=\dfrac{3.05}{10}[/tex]
[tex]a_{3}=0.305\ m/s^2[/tex]
We need to calculate the acceleration a₂
[tex]0.305=a_{2}+0.083[/tex]
[tex]a_{2}=0.305-0.083[/tex]
[tex]a_{2}=0.222\ m/s^2[/tex]
We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂
[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]
[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]
[tex]F_{1}=3.7F_{2}[/tex]
Hence, The magnitude of F₁ is 3.7 times of F₂
