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salantis [7]
3 years ago
13

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

ne from the center outward as the turntable spins counterclockwise at a constant 45 rpm. Its walking speed relative to the turntable is a steady 3.8 cm/s. Let its initial heading be in the positive x-direction, where the x- and y-directions are fixed relative to the laboratory. As the bug reaches the edge of the turntable (still traveling at 3.8 cm/s radially, relative to the turntable) what are the x and y components of the velocity of the bug?
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0

Answer:

v = \left[\begin{array}{c}0.66&0\end{array}\right]m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}

The velocity vector v is the first derivative of the position vector r with respect to time:

\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}

The given values are:

t=\frac{x}{v}=\frac{14}{3.8}=3.7 s

\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}

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Distance = average speed ×time

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ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

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The total weight increase by 42%

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Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

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200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

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F(h) = (0.81μ) (1.42 m g)

       = (0.81) (1.42) (μ m g)

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Why doesn't electric current flow through rubber?
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svetlana [45]

Answer:

(A) FM Radio had a somewhat shorter ranger than AM radio, but better sound quality.

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In FM Radio, the sound is transmitted through changes in frequency. Both FM and AM radio signals experience frequent change in amplitude, they are far less noticeable on FM.

When switching between stations, FM antenna is alternating between different frequencies, and not amplitudes and this produces a much clearer sound and allows for smoother transitions with little to no audible static.

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stiks02 [169]

Answer: An equation is missing in your question below is the missing equation

a) ≈ 8396

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attached below is the detailed solution

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