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3 years ago

A proposed space station includes living quarters in a circular ring 45.0 m in diameter.

Physics

1 answer:

siniylev [52]3 years ago

6 0

Answer:

Angular speed, $\omega=0.65\ rad/s$

Explanation:

Given that,

Diameter of the circular ring, D = 45 m

Radius, r = 22.5 m

Let $\omega$ is the angular speed should the ring rotate so the occupants feel that they have the same weight as they do on Earth. It can be given by providing the centripetal force to the normal force as :

$mg=m\omega^2r$

$\omega=\sqrt{\dfrac{g}{r}}$

$\omega=\sqrt{\dfrac{9.8\ m/s^2}{22.5\ m}}$

$\omega=0.65\ rad/s$

So, the angular speed of the ring is 0.65 rad/s. Hence, this is the required solution.

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How do physicists distinguish one type of electromagnetic wave from another?

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3 years ago

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

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Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

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$d=3.4(10)^{-5}m$ is the width of the slit

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Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

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3 years ago

How is energy transferred through a generator that produces electric current? A.from armature to commutator to brushes to turbin

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2 years ago

Read 2 more answers

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

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

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

6 0

3 years ago