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Free_Kalibri [48]
3 years ago
6

A 20-car train standing on the siding is started in motion by the train’s engine. 2 cm slack is between each of the cars, which

are 11 m long. The engine is tightly connected to the first car and moves at a constant speed of 21 cm/s. How much time is required for the pulse to travel the length of the train? Answer in units of s.
Physics
1 answer:
My name is Ann [436]3 years ago
6 0

Answer:

1.81 s

Explanation:

From the given information:

The links between the cars are:

1-2, 2-3, 3-4, 4-5, 5-6, 6-7, 7-8, 8-9, 9-10, 10-11, 11-12, 12-13, 13-14, 14-15, 15-16, 16-17, 17-18, 18-19, 19-20.

where;

1 denotes the first car that joins strongly attached to the engine.

There are 19 links from the above connections and each one of them has a 2 cm slack.

Thus; the total slack = 2 × 19 = 38 cm

However, the speed of the train = 21 cm/s

Then, the time taken by the pulse to travel the length of the train is expressed by using the formula:

= length / speed

= 38 / 21

= 1.81 seconds

Thus, the required time = 1.81 s

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A motorcycle is capable of accelerating at 1.5 m/s2. Starting from rest, how far can it travel in 0.5 seconds?
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Answer:
Here,
Initial velocity(u)=0 m/s
acceleration(a)=1.5m/s
time(t)=0.5s
Now,
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6 0
4 years ago
A block is given a very brief push up a 20.0 degree frictionless incline to give it an initial speed of 12.0 m/s.(a) How far alo
Orlov [11]

Explanation:

(a)   Net force acting on the block is as follows.

           F_{net} = -mg Sin (\theta)

or,           ma = -mg Sin (\theta)[/tex]

                 a = -g Sin (\theta)

                    = -9.8 \times Sin (20^{o})

                    = -3.35 m/s^{2}

According to the kinematic equation of motion,

             v^{2} - v^{2}_{o} = 2as

Distance traveled by the block before stopping is as follows.

     s = \frac{v^{2} - v^{2}_{o}}{2a}

        = \frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}

        = 21.5 m

According to the kinematic equation of motion,

               v = v_{o} + at

      0 = 12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t

   t_{1} = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

(b)    When the block is moving down the inline then net force acting on the block is as follows.

                 F_{net} = -mg Sin (\theta)

                ma = mg Sin (\theta)

                    a = g Sin (\theta)

                       = 9.8 m/s^{2} \times Sin (20^{o})

                       = 3.35 m/s^{2}

Kinematics equation of the motion is as follows.

                   s = v_{o}t + \frac{1}{2}at^{2}

      21.5 m = 0 + \frac{1}{2} \times 3.35 m/s^{2} \times t^{2}

     t_{2} = \sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}

             = 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

               t = t_{1} + t_{2}

                 = 7.16 sec + 3.58 sec

                 = 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

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<em>t</em> ≈ 0.452 s

In this time, the horse reaches the tree, so its distance from it is

(20 m/s) * (0.452 s) ≈ 9.04 m

7 0
4 years ago
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