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Harman [31]
3 years ago
10

A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to the other, taking 2.90 s. Then security

agents appear and the man runs as fast as he can back along the sidewalk to his starting point, taking 10.7 s. What is the ratio of the man's running speed to the sidewalk's speed (running speed / sidewalk speed)?
Physics
1 answer:
Sindrei [870]3 years ago
4 0

Answer:

68:39

Explanation:

We have to find the ratio of the man's running speed to the sidewalk's speed.

Let running speed of man=x

Sidewalk's speed=y

When man running  in the same direction as side walk is  moving.

Then, total speed=x+y

Time=2.9 s

When a man running in opposite direction as the side walk is moving.

Then, total speed =x-y

Time =10.7 s

Distance traveled in both cases remain same.

Suppose , d is the distance from one end to another end.

Distance=speed\times time

(x+y)\times 2.9=(x-y)\times 10.7

2.9x+2.9y=10.7x-10.7y

10.7y+2.9y=10.7x-2.9x

13.6y=7.8 x

\frac{x}{y}=\frac{13.6}{7.8}

\frac{x}{y}=\frac{68}{39}

x:y=68:39

Hence, the ratio of the man's running speed to the sidewalk's speed =68:39

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How do I do these? My teacher didn’t show us how.
melisa1 [442]

Explanation:

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Δx =  -4 m − 0 m

Δx = -4 m

Between 2 s and 4 s, the position stays at -4 m.  The displacement is:

Δx = -4 m − (-4 m)

Δx = 0 m

Finally, between 4 s and 6 s, the position goes from -4 m to 6 m.  The displacement is:

Δx = 6 m − (-4 m)

Δx = 10 m

The net displacement is the change in position from 0 s to 6 s:

Δx = 6 m − 0 m

Δx = 6 m

In the second part of problem 1, we have a velocity vs time graph.

Car 1 starts with 0 velocity and ends with a velocity of 6 m/s, so it is accelerating and constantly moving to the right.

Car 2 starts with a velocity of -6 m/s and ends with a velocity of 6 m/s.  It is also accelerating, but first it is moving to the left, comes to a stop at t = 3 s, then moves to the right.

Car 3 starts with a velocity of 2 m/s and ends with a velocity of 2 m/s.  So it is moving constantly to the right, but never speeds up or slows down.

We want to know when two of the cars meet.  Unfortunately, this isn't as easy as looking for where the lines cross on the graph.  We need to calculate their displacements.  We can do this by finding the area under the graph (assuming all the cars start from the same point).

Let's start with Car 2.  Half of the area is below the x-axis, and half is above.  Without doing calculations, we can say the total displacement for this car is 0.  This means it ends back up where it started, and that it never meets either of the other cars, both of which have positive displacements.

So we know Car 1 and Car 3 meet, we just have to find where and when.  For Car 1, the area under the curve is a triangle.  So its displacement is:

Δx = ½ t v(t)

where t is the time and v(t) is the velocity of Car 1 at that time.  Since the line has a slope of 1 and y intercept of 0, we know v(t) = t.  So:

Δx = ½ t²

Now look at Car 3.  The area under the curve is a rectangle.  So its displacement is:

Δx = 2t

When the two cars have the same displacement:

½ t² = 2t

t² = 4t

t² − 4t = 0

t (t − 4) = 0

t = 0, 4

t = 0 refers to the time when both cars are at the starting point, so t = 4 is the answer we're looking for.  Where are the cars at this time?  Simply plug in t = 4 into either of the equations we found:

Δx = 8

So Cars 1 and 3 meet at 4 s and 8 m.

7 0
3 years ago
A ranger in a national park is driving at 52 km/h when a deer jumps onto the road 87 m ahead of the vehicle. After a reaction ti
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Answer:

Time, t = 0.23 seconds

Explanation:

It is given that,

Initial speed of the ranger, u = 52 km/h = 14.44 m/s

Final speed of the ranger, v = 0 (as brakes are applied)

Acceleration of the ranger, a=-4\ m/s^2

Distance between deer and the vehicle, d = 87 m

Let d' is the distance covered by the deer so that it comes top rest. So,

d'=\dfrac{v^2-u^2}{2a}

d'=\dfrac{-(14.44)^2}{2\times -4}

d' = 26.06 m

Distance between the point where the deer stops and the vehicle is :

D=d-d'

D=87 - 26.06 = 60.94 m

Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :

t=\dfrac{v}{D}

t=\dfrac{14.44}{60.94}

t = 0.23 seconds

Hence, this is the required solution.

4 0
3 years ago
Scientists have found that the most destructive and deadly tornadoes occur from rotating thunderstorms called
Julli [10]

Answer:

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AlexFokin [52]

The processes under chemical change are: cooking an egg, fireworks exploding, and rusting of a metal tool.

The processes under physical change are: melting ice, evaporating puddle of water, and dissolving of salt in water.

Given:

The following process:

  1. cooking an egg
  2. melting ice
  3. fireworks exploding
  4. evaporating puddle of water
  5. rusting of a metal tool
  6. Dissolving of salt in water

To find:

The classify the processes in whether it does or does not produce a new substance.

Solution:

  • Chemical change is the change when a new substance is formed due to a change in the chemical composition of the substance.
  • Physical change is the change in which no change in chemical composition takes place and the identity of the substance remains the same.
  • Cooking an egg

Chemical change, due to change in atomic arrangement egg get solidified on cooking, a new substance is formed.

  • Melting ice

Physical change, change in the state of matter is taking from solid to liquid, no new substance is formed.

  • Firework exploding

Chemical change. formation of carbon dioxide gas and sound energy, anew substance formed

  • Evaporating puddle of water

Physical change, change in the state of matter is taking from liquid to gas, no new substance is formed.

  • Rusting of metal tool

Chemical change, formation of oxides on the surface of the metal tool due to reaction between air and metal, a new substance is formed.

  • Dissolving of salt in water

Physical change, ions of salt in the solid state are changing into aqueous ions with no formation of new substance, no new substance is formed.

The processes under chemical change are: cooking an egg, fireworks exploding, and rusting of a metal tool.

The processes under physical change are: melting ice, evaporating puddle of water, and dissolving of salt in water.

Learn more about physical change and chemical change here:

brainly.com/question/864557?referrer=searchResults

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3 years ago
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son4ous [18]

Answer:

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

Explanation:

Given;

mass of the object, m = 2 kg

weigh of the object, W = 20 N

tension on the rope, T = 12 N

The acceleration of the object is calculated by applying Newton's second law of motion as follows;

T = F + W

T = ma + W

ma = T - W

a = \frac{T-W}{m} \\\\a = \frac{12 - 20}{2} \\\\a = -4 \ m/s^2 (the negative sign indicates deceleration of the object)

The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.

7 0
3 years ago
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