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Ierofanga [76]
3 years ago
13

A jetliner leaves san francisco for new york, 4600 km away. with a strong tailwind, its speed is 1100 km/hr. at the same time, a

second jet leaves new york for san francisco. flying into the wind, it makes only 700 km/hr. when and where do the two planes pass?

Physics
1 answer:
Troyanec [42]3 years ago
8 0
First, make an illustration of the problem. In reality, the direction traveled by the planes is not linear. Strictly speaking, it should curve out following the shape of the sphere. But I suppose, we could do away with that because we are given with the quantity for distances. I'm just showing a guide to see where the planes would meet.

So, we know that the total distance is 4,600 km. It would consequently mean that x and y would equal to 4,600. The distances x and y could be determined by multiplying their respective speeds to the time. Since they meet at the same time, let's denote it as t.

1,100t + 700t = 4,600
Solving for t,
1,800t = 4600
t = 4600/1800
t = 2.556 hours

So, we know that the two planes would meet after 2.56 hours from the time they leave the airports. Next, we have to know where they meet exactly.

1,100(2.556) = 2,812 km
700(2.556) = 1,788 km

Therefore, they meet at approximately 2,812 km from San Francisco airport, or 1,788 km from New York airport after 2.56 hours since they left their respective ports.

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zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

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1 year ago
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