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3 years ago

Why do moving back and forth, and moving in a circle look the same on a graph of position versus time?

Physics

1 answer:

Tomtit [17]3 years ago

3 0

Answer:

Because in both kind of motions, the position is periodically repeated. Both are simple harmonic motions.

Explanation:

Moving back and forth means that the moving object goes through the same position every time a certain period of time elapses.

The same is true for the circular motion, every time the object completes a turn, the object is in the same position.

They are examples of simple harmonic motions. The time it takes the object to repeat a complete cycle is called period.

Therefore, in a grahp of position versus time, the curve that represents the position repeats itself every so often equal to one period, both for moving back and forth and for moving in a circle.

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Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg = V²μ

g = V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

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3 years ago

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the arrangement of the outer planets is
1. Mercury
2. Venus 
3. Earth 
4. Mars 
5. Jupiter 
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7. Uranus 
8. Neptune
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3 years ago

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BabaBlast [244]

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3 years ago

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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

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3 years ago