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Cerrena [4.2K]
4 years ago
10

A train moving with a velocity of 42.9 km/hour North, increases its speed with a uniform acceleration of 0.250 m/s^2 North until

it reaches a velocity of 160.0 km/hour North. What distance did the train travel while it was increasing its velocity, in units of meters? Give the answer as a positive number
Physics
1 answer:
ankoles [38]4 years ago
5 0

Answer:

3658.24m

Explanation:

Hello!

the first thing that we must be clear about is that the train moves with constant acceleration

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

\frac {Vf^{2}-Vo^2}{2.a} =X

Vf = final speed =160km/h=44.4m/s

Vo = Initial speed =42.9km/h=11.92m/s

A = acceleration =0.25m/s^2

X = displacement

solving

\frac {44.4^{2}-11.92^2}{2.(0.25)} =X\\X=3658.24m

the distance traveled by the train is 3658.24m

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LekaFEV [45]

Average speed = (total distance covered) / (time to cover the distance)

= (50 miles) / (1.5 hours)

= (50/1.5) miles/hours

= (33 and 1/3) mph .

We don't care about all that other "data" given earlier in the question.
We only need to know the total distance covered and the time it took
to cover the distance.
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3 years ago
Two 100kg bumper cars are moving towards eachother in oppisite directions. Car A is moving at 8 m/s and Car B at -10 m/s when th
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Answer:

-10 m/s

Explanation:

When two cars collide then the momentum of two cars will remains conserved

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  • After collision the speed of car B = +8 m/s

By momentum conservation equation

               m1v1i+m2v2i=m1v1f + m2v2f

               (100)(8)+(100)(-10)=(100v)+(100)(8)\\                            v=-10 m/s

7 0
3 years ago
A child pushes a 100 kg refrigerator with a force of 50 N, but the refrigerator does not move. Suppose the coefficient of static
faust18 [17]

Answer:

50 N

Explanation:

Since the refrigerator doesn’t move, that means the force of friction equals the amount of force the child exerts on the fridge. If the friction force were greater than the force by the child, the fridge would start accelerating towards the child. If it were less than the force the child exerted, the fridge would start accelerating away from the child. Therefore, the net force must be 0, in this case, the friction force is equal to the force the child exerted, for it to stay at rest (as Newton’s First Law stated).

I hope this helps! :)

8 0
3 years ago
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AlekseyPX
I believe it’s tackling but I’m not quite sure
4 0
3 years ago
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Suppose high tide is at midnight, the water level at midnight is 3 m, and the water level at low tide is 0.5 m. Assuming the nex
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We have that the time, to the nearest minute, when the water level is at 1.125 m for the second time after midnight is

t=10.0hours

From the Question we are told that

Maximum height h_{max}=3m

Minimum height  H_{min}=0.5m

Time for  next high tide will occurT=12 hours =>720 min

Generally Average Height

h_{avg}=\frac{3+0.5}{2}\\\\h_{avg}=1.75

Therefore determine Amplitude to be

A=h_{max}=j_{avg}\\\\A=3-1.75\\\\A=1.25

Generally, the equation for Time is mathematically given by

At t=0

h(x)=Acos(Bx)+h_{avg}

Where

B=\frac{2\pi}{P}\\\\B=\frac{2\pi}{720}\\\\B=8.73*10^{-3}

Therefore

h(t)=Acos8.73*10^{-3}(t)+h_{avg}

Hence the Time at T=1.125 is

1.125(t)=1.25cos(8.73*10^{-3})(t)+1.75

-0.1249t=1.75

t=10.0hours

For more information on this visit

brainly.com/question/22361343

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3 years ago
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