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Dima020 [189]
4 years ago
7

Two train tracks run parallel to each other, except for a short distance where they meet and become one track over a narrow brid

ge. One morning, a train speeds onto the bridge. Another train coming from the opposite direction, also speeds onto the bridge. Neither train can stop on the short bridge, yet there is no collision. How is this possible?
Physics
1 answer:
Darya [45]4 years ago
7 0

Answer:

This situation is possible.

Explanation:

It is said in the problem that one morning two trains are speeding up from opposite directions onto the bridge. It may seem from this statement that the trains are arriving at the same time. But this should be a wrong assumption.

Because it is only said in the problem that the trains are passing the bridge on same morning. No explicit timing is given.

So it may so happen that both the trains pass the narrow bridge on the same morning but at two difeerent times, making no collision with each other.

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A stone is dropped at t = 0. A second stone, with 6 times the mass of the first, is dropped from the same point at t = 59 ms. (a
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Answer:y_{com}=0.707 m

v_{com}=3.713 m/s

Explanation:

Given

first stone mass is m

second stone mass is 6 m

distance traveled by  first stone in 430 ms

y_1=ut+\frac{at^2}{2}

y_1=0+\frac{g(0.43)^2}{2}

y_1=0.9069 m

Distance traveled by stone 2 in t=430-59=371 ms

y_2=ut+\frac{at^2}{2}

y_2=0+\frac{g(0.0.371)^2}{2}

y_2=0.674 m

velocity of first stone after t=0.43 s

v_1=u+at

v_1=0+9.8\times 0.43=4.214 m/s

velocity of second stone after t=0.371 s

v_2=u+at

v_2=0+9.8\times 0.371=3.63 m/s

Position of Center of mass of system

y=\frac{y_1m_1+y_2m_2}{m_1+m_2}

y=\frac{0.9069\times m+0.674\times 6m}{m+6m}

y=\frac{4.95m}{7m}=0.707 m

Velocity of COM

v_{com}=\frac{v_1m_1+v_2m_2}{m_1+m_2}

v_{com}=\frac{4.214\times m+3.63\times 6m}{m+6m}

v_{com}=3.713 m/s

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4. Joe and his brother Bo have a combined mass of 200.0 kg and are zooming along in
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Answer:

Joe & Bo's car: 200.0 kg + 100.0 kg = 300.0 kg; Melinda's car: 25.0 kg + 100.0 kg = 125.0 kg

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The resultant vector is 11√2 km due north east.

<h3><u>Explanation:</u></h3>

The vector is a type of quantity which has both magnitude and direction. This quantities when expressed needs to specify both magnitude and direction.

We need to calculate the magnitude and direction separately.

Here firstly for the magnitude,

The magnitudes are both 11 km and they are at right angles to each other.

So, the resultant magnitude = √(11² +11²) km

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Now for the direction, one vector is due north and the other is due east.

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Why is the weight of a body more at the pole than at the equator of the earth ??​
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