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Kipish [7]
3 years ago
15

Two trains are traveling between city A and city B a distance of 4000 kilometers (km). The regular train travels at a speed of 8

0 kilometers per hour (km/h). The express train leaves at the same time on a parallel track and travels at 100 km/h. The formula for speed (S) is S=DT where S is speed, D is distance and T is time. How much earlier will the express train arrive at city B? * 10 points
Physics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

10 hours earlier than regular train

Explanation:

In this case you are already giving the expression to be used which is:

S = D/t   (1)

The problem is giving us the data of the speed of both trains, and we also know the distance between City A and B, which is 4000 km, therefore, we just need to solve for t in the above expression for both trains, and then, do the difference between their times and see how much earlier the express train arrives.

Solving for t, we have:

t = D/S   (2)

For Train 1 (The regular):

t₁ = 4000 / 80

t₁ = 50 h

For Train 2 (Express):

t₂ = 4000 / 100

t₂ = 40 h

Now, as expected express train arrives earlier, now let's see how much:

T = t₁ - t₂

T = 50 - 40

<h2>T = 10 h</h2><h2></h2>

Therefore, Express train arrives 10 hours earlier than regular train.

Hope this helps

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A ball is tossed vertically upwards with a speed of 5.0 m s–1. After how many seconds will the ball return to its initial positi
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Read 2 more answers
2 kg<br> Use 10 m/s2 for g
Iteru [2.4K]

Answer:

See below

Explanation:

At point A    the PE = mgh = 2 * 10 * 1 = 20 J

  at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2     v = sqrt 20 = 4.47 m/s

for the friction part

 vf = vo t  + 1/2 a t^2      vf = final velocity = 0 (stopped)

                                       vo = original velocity = 4.47 m/s  

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  0 = 4.47 t + 1/2 (-1) t^2

            - .5t^2  + 4.47 t = 0

                 t ( -.5t+ 4.47) = 0    shows t =  4.47/.5 = 8.9 seconds

6 0
2 years ago
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