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Mama L [17]
2 years ago
9

A 12,000 kg railroad car is traveling at 2 m/s when it strikes another 10,000 kg railroad car that is at rest. If the cars lock

together, what is the final speed of the two
Physics
1 answer:
Degger [83]2 years ago
4 0

Answer:

The final speed of the railroad car

V= 1.14 \frac{m}{s}

Explanation:

v_{1}=2.1\frac{m}{s} \\m_{1}=12000kg\\v_{2}=0\frac{m}{s} \\m_{2}=10000kg \\v_{t}=?

m_{1}*v_{1}+m_{2}*v_{2}= (m_{1}+m_{2})*v_{t}\\v_{t}=\frac{m_{1}*v_{1}}{(m_{1}+m_{2})} \\v_{t}=\frac{12000kg*2.1\frac{m}{s} }{(12000+10000)kg} \\v_{t}=1.14 \frac{m}{s}

That's the final speed of the both railroad car

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You have found a treasure map that directs you to start at a hollow tree, walk 300 meters directly north, turn and walk 500 mete
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Answer:633 m

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let say it as point a and its vector is 300\hat{j}

after that we have moved 500 m northeast

let say it as point b

therefore position of b with respect to a is

r_{ba}=500cos(45)\hat{i}+500sin(45)\hat{j}

Therefore position of b w.r.t to origin is

r_b=r_a+r_{ba}

r_b=300\hat{j}+500cos(45)\hat{i}+500sin(45)\hat{j}

r_b=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}

after this we moved 400 m 60^{\circ} south of east i.e. 60^{\circ} below from positive x axis

let say it as c

r_{cb}=400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=r_{b}+r_{cb}

r_c=500cos(45)\hat{i}+\left [ 250\sqrt{2}+300\right ]\hat{j}+400cos(60)\hat{i}-400sin(60)\hat{j}

r_c=\left [ 250\sqrt{2}+200\right ]\hat{i}+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]\hat{j}

magnitude is \sqrt{\left [ 250\sqrt{2}+200\right ]^2+\left [ 250\sqrt{2}+300-200\sqrt{3}\right ]^2}

=633.052

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tan\theta =\frac{307.139}{553.553}

\theta =29.021^{\circ} with x -axis

7 0
1 year ago
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