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Olenka [21]
3 years ago
11

A caterpillar climbs up a one-meter wall. For every 2 cm it climbs up, it slides down 1 cm. It takes 10 minutes for the caterpil

lar to climb to the top. Calculate the speed at which the caterpillar travels in cm/s.
Physics
2 answers:
GaryK [48]3 years ago
8 0
After 1 move, it reaches 2cm but ends up at 1 cm.
After 2 moves, it reaches 3cm but ends up at 2 cm.
After 3 moves, it reaches 4cm but ends up at 3 cm.
After 4 moves, it reaches 5cm but ends up at 4 cm.
After 5 moves, it reaches 6cm but ends up at 5 cm.
.
.
.
After 96 moves, it reaches 97cm but ends up at 96 cm.
After 97 moves, it reaches 98 cm but ends up at 97 cm.
After 98 moves, it reaches 99 cm but ends up at 98 cm.
Then, on the 99th move, it climbs up 2 cm, to 100 cm.

The critter has just momentarily reached the 100-cm point
after climbing 99 times.

If the sliding takes no time, then

   Speed = (total distance climbed) / (total time)

               = (99 x 2 cm)  /  (10 minutes)

               =  (198 cm) / (600 sec)

               =    0.33 cm/sec





Eddi Din [679]3 years ago
6 0
If the caterpillar goes up 2cm then slips down 1cm, it only got 1 cm
The poor caterpillar had to crawl 3m to get up 1m .

<span>average speed = distance travelled / time taken
</span>average speed = 3m / 10 minutes
average speed = 0.33 m / minutes

or
average speed = (9 x 2 cm)  /  (10 minutes)
average speed =  (198 cm) / (600 sec)
average speed =    0.33 cm/sec
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Natalija [7]

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I think it is the answer A

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8 0
3 years ago
A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo
Serggg [28]

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

5 0
3 years ago
An island reports that in the year 2000 there were 240 babies born. That same year, 100 individuals died.
lyudmila [28]

Answer:

+ 140

Explanation:

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Here it would be:

240 - 100 = + 140

// if you want to convert it to percentage, you need to know the size of the population

it would be

140 / (population size) * 100 %

4 0
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