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NemiM [27]
3 years ago
10

Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the ot

her travels at 60 miles an hour. A bird starts flight at the same location as the faster train, flying at a speed of 90 miles per hour. When it reaches the slower train, it turns around, flying the other direction at the same speed. When it reaches the fastest train again, it turns around -- and so on. When the trains collide, how many miles the bird has flown.
Physics
2 answers:
Paladinen [302]3 years ago
7 0

D = distance between th two trains at the start of the motion = 100 miles

V = speed of the faster train towards slower train = 60 mph

v = speed of the slower train towards faster train = 40 mph

t = time taken by the two trains to collide = ?

time taken by the two trains to collide is given as

t = D/(V + v)

t = 100/(60 + 40) = 1 h

v' = speed of the bird = 90 mph

d = distance traveled by the bird

distance traveled by the bird is given as

d = v' t

d = 90 x 1

d = 90 miles

xz_007 [3.2K]3 years ago
3 0

-- One train travels at 60 mph.  The other one travels at 40 mph.  So the space between them shrinks at the rate of 100 mph.

-- They start 100 miles apart, so they meet in 1 hour.

-- The bird and the trains all start moving at the same time, and the bird keeps flying in the space between them until they meet.  That's 1 hour.

-- The bird's speed is 90 mph.  So the bird covers <em>90 miles</em> in that hour.

-- After flying at 90 mph for a whole hour without resting, we can be sure that the bird is grateful to be put out of its misery.

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a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

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friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

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b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

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