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icang [17]
3 years ago
15

A caterpillar tries to climb straight up a wall a meter high, but for every 2 cm up it climbs, it slides down 1 cm. Eventually,

it reaches the top. When it reaches the top, it does not pull itself over so it will slide down 1 cm. What is the total displacement traveled? (Include direction, whether up, down, or n/a.)
Physics
2 answers:
tiny-mole [99]3 years ago
5 0

<u>Answer:</u>

Total displacement traveled = 298

<u>Explanation:</u>

According to the given information, to actually climb for 1 cm, the caterpillar has to travel for 3 cm (2 cm upwards and 1 cm downwards).

So in order to climb straight up a one meter (100 cm) high wall, it needs to travel for 99 × 3 = 297 cm.

Then after a little it can travel up another cm to reach the top.

Therefore, the total displacement traveled = 297 + 1 = 298 cm

olga nikolaevna [1]3 years ago
3 0

"Displacement" means the distance and direction from start to finish, regardless of what happened in between.

The caterpillar's displacement is 99 cm straight up.

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4 0
3 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

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\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5
Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

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time = 3.5s

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time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

8 0
3 years ago
 An ultrasound pulse used in medical imaging has a frequency of 5 MHz and a pulse width of 0.5 us. Ap- proximately how many osci
podryga [215]

At the frequency of 5 MHz, the period of the oscillations is 1/5meg. That's a period of 1/5 microsecond.

There are 5 full cycles in one full microsecond, and there are 2.5 full cycles in a 0.5 us pulse.

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4 0
3 years ago
Mia plays soccer for her high school team. generally, they start practice by stretching. what kind of stretching should they do
Keith_Richards [23]

Mia plays soccer for her high school team. generally, they start practicing by stretching. The kind of stretching should they do before they are warmed up is Dynamic Stretching

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We require our muscles' flexibility to maintain the range of motion in our joints, so stretching keeps them strong, flexible, and healthy. The muscles shorten and tighten without it. The muscles are then weak and unable to fully expand when you want them to work.

Learn more about dynamic stretching here

brainly.com/question/14635799

#SPJ4

8 0
2 years ago
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