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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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What is one characteristic of a sample of matter that indicates it is a pure substance? A) It can be separated using by a physic
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D) If the composition of a sample is fixed, the sample is a pure substance.

Explanation:

A) It can be separated by using physical means, such as filtering.

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8 0
3 years ago
Read 2 more answers
A cyclist traveling at constant speed of 12m/s when he passes a stationary bus.The bus starts moving just as the cyclist passes
Bogdan [553]

Answer:

A.) 8 seconds

B.) 16 seconds

C.) 48 m

Explanation:

Given that a cyclist traveling at constant speed of 12 m/s

and the bus accelerates uniformly at 1.5ms²

A.) The bus has the following parameters

Acceleration a = 1.5 m/s^2

Initial velocity U = 0. Since the bus is starting from rest.

Final velocity V = 12 m/s

Use equation one of linear motion.

V = U + at

Substitute V, U and a into the formula

12 = 0 + 1.5t

1.5t = 12

t = 12/1.5

t = 8 seconds

Therefore, the bus reach the same speed as the cyclist at 8 seconds.

B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion

h = Ut + 1/2at^2

Since a = 0, the equation is reduced to:

h = Ut.

Also, for the bus,

h = Ut + 1/2at^2

Equate the two equations since the h is the same

Ut = Ut + 1/2at^2

Substitute all the parameters into the formula

12t = 0 + 1/2 × 1.5t^2

12t = 0.75t^2

0.75t = 12

t = 12/0.75

t = 16 seconds

Therefore, the bus takes 16 seconds to catch the cyclist

C.) Use third equation of linear motion.

V^2 = U^2 + 2as

Where s = distance

Substitute V, U and a into the formula

12^2 = 0 + 2 × 1.5 S

144 = 3S

S = 144/3

S = 48 m

8 0
3 years ago
a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how
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Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = 6\times 10^6\ J

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s

The time taken to lift the load is 141.26626 seconds

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