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vovangra [49]
1 year ago
11

As a train accelerates away from a station, it reaches a speed of 4.9 m/s in 5.4 s. if the train's acceleration remains constant

, what is its speed after an additional 5.0 s has elapsed?
Physics
1 answer:
romanna [79]1 year ago
3 0

The speed of the train after an additional 5 s has elapsed is 9.45 m/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the acceleration in 5.4 s</h3>
  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 4.9 m/s
  • Time (t) = 5.4 s
  • Acceleration (a) =?

a = (v – u) / t

a = (4.9 – 0) / 5.4

a = 4.9 / 5.4

a = 0.91 m/s²

<h3>How to determine the final velocity after 5 s</h3>
  • Initial velocity (u) = 4.9 m/s
  • Acceleration (a) = 0.91 m/s²
  • Time (t) = 5 s
  • Final velocity (v) = ?

a = (v – u) / t

0.91 = (v – 4.9) / 5

Cross multiply

v – 4.9 = 0.91 × 5

v – 4.9 = 4.55

Collect like terms

v = 4.55 + 4.9

v = 9.45 m/s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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Oxana [17]

Total hours is 3.35 hours

<u>Explanation:</u>

Given:

convert 3 hours 21 minutes to decimal hours

We know:

1 hour = 60 minutes

and

1 minute = 60 seconds

1 minute = 1 / 60 hours

So,

21 minutes = \frac{21}{60} hours

                  = 0.35 hours

Total hours would be = 3 hours + 0.35 hours

                                    = 3.35 hours

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RoseWind [281]
<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

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<h3> =1 m/s</h3>

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= (4 - 4)/(8-4)

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<h3> = 0</h3>

c) Average velocity for last 6 seconds

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= (2 - 4)/(14 - 8)

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<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

<h3> = 400 m</h3>

6 0
2 years ago
A satellite is always being pulled by gravity.<br> a. True<br> b. False
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3 0
3 years ago
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baherus [9]
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It doesn't matter whether you flick a marble horizontally from the roof,
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off the roof, or drop a bowling ball from the roof with zero horizontal speed. 
Their vertical speed is completely determined by gravity, (and it happens to
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Handy dandy formula for the distance covered by anything that starts out
with zero speed and accelerates to the end:

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

If the beginning of the journey is on Earth, then the acceleration is
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the 55-meter rooftop in the question is part of a building on Earth.

                       55 meters  =  (1/2) (9.8 m/s²) x (time)²           

Divide each side
by  4.9 m/s² :            55 m / 4.9 m/s²  =  (time)²

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Square-root
each side:                time  =  √(55/4.9 sec²)

                                           =      3.35 sec  .
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