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2 years ago

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A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

d/s2. The child is standing 4.65 m from the center of the merry-go-round. What is the magnitude of the linear acceleration of the child?

1 answer:

skelet666 [1.2K]2 years ago

7 0

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

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1 year ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

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I hope it helps you!

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1 year ago

A superman cyclist rode a bike uphill at 20 miles/hour for two hours. To sustain this constant speed the cyclist was exerting 50

NeX [460]

Answer:

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Distance traveled by cyclist is given as

d = v t

d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

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W = work done by cyclist

Work done by cyclist is given as

W = F d

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W = 1.056 x 10⁷ lb-ft

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2 years ago

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valentina_108 [34]

Answer:

3. none of these

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$K=\frac{1}{2}I \omega^2$

where

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In this problem, we have two objects rotating, so the total rotational kinetic energy will be the sum of the rotational energies of each object.

For disk 1:

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For disk 2:

$K_2 = \frac{1}{2}I(-\omega)^2 = \frac{1}{2}I\omega^2$

so the total energy is

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So, none of the options is correct.

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2 years ago

Read 2 more answers