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kondor19780726 [428]
2 years ago
12

Light-rail passenger trains that provide transportation within and between cities are capable of modest accelerations. The magni

tude of the maximum acceleration is typically 1.3 m/s2 , but the driver will usually maintain a constant acceleration that is less than the maximum. A train travels through a congested part of town at 4.0m/s . Once free of this area, it speeds up to 13m/s in 8.0 s . At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed.
What is the final speed?
Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Mkey [24]2 years ago
3 0
During the first phase of acceleration we have:
v o = 4 m/s;  t = 8 s; v = 13 m/s, a = ?
v = v o + a * t
13 m/s = 4 m / s + a * 8 s
a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
The final speed:
v = ?;  v o = 13 m/s; a = 1.125 m/s² ;  t = 16 s
v = v o + a * t
v = 13 m/s + 1.125 m/s² * 16 s
v = 13 m/s + 18 m/s = 31 m/s 
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a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?
Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it.  Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s  ==>  (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out.  If you prefer to use the formula,
the first problem you run into is:  You need to remember the formula !

The formula is        D = 1/2 a T²

                   Distance = (1/2 acceleration) x (time in seconds)²

             Acceleration = (change in speed) / (time for the change)
                                  =        (18 m/s)            /        (12 sec)
                                  =                      1.5 m/s² .

                  Distance  =  (1/2 x 1.5 m/s²) x (12 sec)²
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5 0
3 years ago
Increase the slit width to 1050 nm. What happens to the width of the central bright fringe? (Always remember to wait a few secon
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Answer:

b.

Explanation:

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Then In Single slit Diffraction, width of central fringe is

x_c= 2D\lambda/a


where D= distance b/w screen and slit

a= slit width

\lambda = wavelength

Thus if Screen width increases keeping other factors same then width of central fringe becomes narrower as

x_c\propto 1/a

On increasing the slit width the central bright fringe width The width of the central bright fringe becomes narrower.

3 0
3 years ago
3. A 900N mountain climber scales a
Umnica [9.8K]

Answer:

<h2>135,000 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

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<h3>135,000 J</h3>

Hope this helps you

4 0
2 years ago
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