Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
Answer:
v₂ = v/1.5= 0.667 v
Explanation:
For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.
Initial moment before pushing
p₀ = 0
Final moment after they have been pushed
= m₁ v₁ + m₂ v₂
p₀ = 
0 = m₁ v₁ + m₂ v₂
m₁ v₁ = - m₂ v₂
Let's replace
M (-v) = -1.5M v₂
v₂ = v / 1.5
v₂ = 0.667 v
Answer: a) The Answer to the question is option a) None of it.
Explanation:
The reason is because according to the law of conservation of energy Energy can neither be created nor destroyed but can be transformed from one form to the other. Therefore none of the kinetic energy was dissipated, rather it was transformed to another form of energy.
Explanation:
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<em>F</em><em>=</em><em> </em><em>m.a</em>
<em>and</em><em> </em><em>momentum</em><em> </em><em>(</em><em>p</em><em>)</em><em>=</em><em> </em><em>m.v</em>
<em>where</em><em> </em><em>m</em><em>=</em><em> </em><em>mass</em>
<em>v</em><em>=</em><em> </em><em>velocity</em><em>.</em>
<em>a</em><em>=</em><em> </em><em>acceleration</em><em> </em><em>.</em>
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<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>
B The change in momentum stays the same because the ball still comes to a stop.