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allsm [11]
3 years ago
5

Two trains are headed towards each other on the same track unbeknownst to the engineers. One departs San Francisco. Its average

speed is 55 miles per hour. The 2nd departs Seattle at the same time that the first one departs San Francisco. Its average speed is 65 miles per hour. The length of track separating the two locations is 860 miles. How many minutes (from departure time) before the two trains collide?
Physics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

7,166 hrs =430  minutes

Explanation:

Since both train are on the same track, going one towards the other, the relative speed is the addition of both, then the time they need to meet, and consistently crash, is the time that (65mph + 55 mph)=120mph need to travel the total distance of 860 miles, of course in this case one part is traveled by the first train and the rest by the other. Then to find the time we use a three rule

1 h --->120mi

X ---->860mi, then X=(860 mi* 1h)/120 mi = 43/6 hrs= 7,16666 hrs, turning this into minutes need that we notice 1h=60min, then 43/6 hrs *60 min/hrs = 430 minutes.

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Pretend you (80 kg) are making repairs on the outside of the International Space Station. You are floating 20 meters away from t
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Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part
Fantom [35]

a) Angular acceleration: 17.0 rad/s^2

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

a)

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

W=mg

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

W_p =mg sin \theta

where \theta is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta

where L is the length of the pencil.

The relationship between torque and angular acceleration \alpha is

\tau = I \alpha (1)

where

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Substituting into (1) and solving for \alpha, we find:

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b)

There are only two forces acting on the pencil here:

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The torque exerted by each force is given by

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Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0
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