The tension in the cable is 23.2 N
<h3>What is the tension in the string?</h3>
The tension in the cable can be resolved into horizontal and vertical forces Tcosθ and Tsinθ respectively.
Tcosθ, is acting perpendicularly, Tcosθ = 0
Taking moments about the pivot:
Tsinθ * 2.2 = 4 * 9.8 * 0.7
Solving for θ;
θ = tan⁻¹(1.4/2.2) = 32.5°
T = 27.44/(sin 32.5 * 2.2)
T = 23.2 N
In conclusion, the tension in the cable is determined by taking moments about the pivot.
Learn more about moments of forces at: brainly.com/question/23826701
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Explanation:
Given that,
Mass of a freight car, 
Speed of a freight car, 
Mass of a scrap metal, 
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
![\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J](https://tex.z-dn.net/?f=%5CDelta%20K%3D%5Cdfrac%7B1%7D%7B2%7D%5B%28m_1%2Bm_2%29V%5E2-m_1u_1%5E2%29%5D%5C%5C%5C%5C%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20%5B%2830%2C000%2B110%2C000%20%290.182%5E2-30000%280.85%29%5E2%5D%5C%5C%5C%5C%3D-8518.82%5C%20J)
Lost in kinetic energy is 8518.82. Negative sign shows loss.
Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.
Answer:
Apply Newton's second law in the moving direction.
Explanation:

Friction force applies in the opposite direction of motion; as a restriction.