Answer:
Vf₂ = 2 Vf₁
It shows that final speed of Joe is twice the final speed of Jim.
Explanation:
First, we analyze the final speed of Jim by using first equation of motion:
Vf₁ = Vi + at
where,
Vf₁ = final speed of Jim
Vi = initial speed of Jim = 0 m/s
a = acceleration of Jim
t = time of acceleration for Jim
Therefore,
Vf₁ = at ---------------- equation (1)
Now, we see the final speed of Joe. For Joe the parameters will become:
Vf = Vf₂
Vi = 0 m/s
a = a
t = 2t
Therefore,
Vf₂ = 2at
using equation (1):
<u>Vf₂ = 2 Vf₁</u>
<u>It shows that final speed of Joe is twice the final speed of Jim.</u>
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!!!
I had this question before I think was it A or B.
Carbon-13 will have 6 protons (because it is carbon), 6 electrons, and 7 neutrons. Isotopes have a different number of neutrons.
