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lora16 [44]
3 years ago
11

A dog walks a distance of 55.5 meters in 120 seconds. What was its speed?

Physics
1 answer:
Nitella [24]3 years ago
8 0

Answer: ...

Explanation: 1: The dog is walking 0.4625MPS (Meters per second. M/S)

To calculate speed you divide the distance by time so 55.5 Meters/120 Seconds is 0.4625 MPS

2: The car traveled 15 KM

To calculate speed you need to multiply the speed by time. (Since it's multiplication it can be put in any order) So 600 Secondsx25 Meters is 15,000 M which is 15 KM.

3: It would take 8 Minutes and 20 Seconds to travel 1500 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 1500 Metersx3 MPS is 8 Minutes and 20 seconds.

4: It was going 2.7775 MPS.

To calculate the speed you divide the distance over time. So 9999 Meters/3600 seconds is 2.7775 MPS.

5: It was going 1.1222... (Infinite)

To calculate the speed you divide the distance over time

So 8080 Meters/ 7200 Seconds is 1.222...

6: It would cover 1.375 Meters.

To calculate the Distance you multiply Speed by time (Order doesn't count.)

So 10.25 Secondsx5.5 MPS is 1.375 Meters.

7: It would take 1 Minute and 31 Seconds to travel 10 Meters.

To calculate the time you multiply Speed by distance (Order doesn't count.)

So 10 Metersx0.11 MPS is 1 Minute and 31 seconds

Hopes this helps because this took over 10 minutes.

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The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

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As in part a, the angular speed of the CD is given by

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When scanning the outermost part of the track, we have

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Therefore, the angular speed is

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The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

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If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

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The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

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