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belka [17]
4 years ago
5

Two passenger trains are passing each other on adjacent tracks. train a is moving east with a speed of 13 m/s, and train b is tr

aveling west with a speed of 28 m/s.(a) what is the velocity (magnitude and direction) of train a as seen by the passengers in train b? (b) what is the velocity (magnitude anddirection) of train b as seen by the passengers in train a?
Physics
2 answers:
harkovskaia [24]4 years ago
8 0

Answer:

B and C

Explanation:

alukav5142 [94]4 years ago
7 0

Since the two trains are passing in opposite directions, so this means that their relative velocities will be the sum of the two trains that is: 

<span>
relative velocity = (13 + 28) = 41m/s</span>

<span>
a. The passengers aboard on train B will see that train A is moving at 41m/sec due east</span>

 

<span>b. The passengers aboard on train A will see that train B is moving at 41m/sec due west</span>

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a car takes off from rest and covers a distance of 80m on a straight road in 10s.calculate the magnitude of its acceleration
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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Here's the solution :

Let's find the final velocity :

  • \dfrac{displacement}{time}

  • \dfrac{80}{10}

  • 8 \:  \: ms {}^{ - 1}

Initial velocity (u) = 0 (cuz it started from rest)

Final velocity (v) = 8 m/s

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now, we know that :

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3 years ago
What is escape velocity?
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6 0
3 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

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7 0
3 years ago
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