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dimaraw [331]
3 years ago
15

Turner's treadmill runs with a velocity of -1.3 m/s and speeds up at regular intervals during a half-hour workout. after 25 min,

the treadmill has a velocity of -6.5 m/s. what is the average acceleration of the treadmill during this period ?

Physics
2 answers:
sammy [17]3 years ago
8 0

Answer:

The average acceleration of the treadmill during this period is 0.00346 m/s²

Explanation:

It is given that,

Initial velocity of the treadmill, u = 1.3 m/s

After 25 minutes, the treadmill has a velocity of 6.5 m/s

Final velocity of the treadmill, v = 6.5 m/s

Time taken, t = 25 minutes = 1500 seconds

We need to find the average acceleration of the treadmill during this period. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{6.5\ m/s-1.3\ m/s}{1500\ s}

a=0.00346\ m/s^2

So, the average acceleration of the treadmill during this period is 0.00346 m/s². Hence, this is the required solution.

Travka [436]3 years ago
3 0

Given : Initial velocity = -1.3 m/s

            Final Velocity = -6.5 m/s.

            Time = 25 minutes.

To find : Average acceleration.

Solution: We are given units in meter/second (m/s).

So, we need to convert time 25 minutes in seconds.

1 minute = 60 seconds.

25 minutes = 60*25 = 1500 seconds.

Formula for average acceleration is given by,

\frac{Final \ velocity -Initial \ velocity}{Final \ time - Initial \ time }

We are not given intial time, so we can take initial time =0.

Plugging values in the above formula.

Average \ acceleration = \frac{-6.5 -(-1.3)}{1500-0}

= \frac{-5.2}{1500}

= -0.003467

or Average \ acceleration = -3.467 \times 10^{-3}\ m/s^2..


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A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
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Answer:

21.6\ \text{kg m}^2

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Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

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