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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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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A
ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is m  =  1612.5  \  kg

Explanation:

From the question we are told that

   The radius of the grinding wheel is R =  0.330 \ m

   The  tangential force is F_t =  250 \ N

    The angular acceleration is  \alpha  =  0.940 \ rad/s^2

The torque experienced by the wheel is mathematically represented as

     \tau  =  I  *  \alpha

Where  I  is the moment of inertia

The torque experienced by the wheel can also be  mathematically represented as

       \tau  =  F_t  * r

substituting values

       \tau  =  250 * 0.330

      \tau  = 82.5  \ N\cdot m

So

   82.5  =  I *  \alpha

    82.5  =  I *  0.940

So

   I  =  87.8 \ kg \cdot m^2

This moment of inertia can be mathematically evaluated as

     I  =  \frac{1}{2} * m* r^2

substituting values

  87.8  =  \frac{1}{2} * m* (0.330)^2

=>   m  =  1612.5  \  kg

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3 years ago
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