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Grace [21]
3 years ago
6

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

light years in radius. (A light year is the distance traveled by light in 1 y.)
Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?
Physics
1 answer:
Mamont248 [21]3 years ago
8 0

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

a_c = \frac{v^2}{R}

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

v = R\omega

Where,

R = Radius

\omega = Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

\omega = \frac{2\pi}{T}

T = Period

So replacing we have to

\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s

Since 1 Light year = 9.48*10^{15}m

Then the radius in meters would be

R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})

R = 2.844*10^{20}m

Then the centripetal acceleration would be

a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2

From the result obtained, considering that it is an unimaginably low value of an order of less than 10^{-10} it is possible to conclude that it supports the assertion on the inertial reference frame.

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Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging,
katrin2010 [14]

Answer:

<em>380 kHz</em>

<em></em>

Explanation:

The speed of sound is taken as 1500 m/s

The length of the fetus is 1.6 cm long

The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.

For this 1.6 cm baby, the wavelength must not exceed

λ = \frac{1}{4} of 1.6 cm = \frac{1}{4} x 1.6 cm = 0.4 cm =

0.4 cm = 0.004 m   this is the wavelength of the required ultrasonic sound.

we know that

v = λf

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λ is the wavelength of the wave

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f = v/λ

substituting values, we have

f = 1500/0.004 = 375000 Hz

==> 375000/1000 = 375 kHz ≅ <em>380 kHz</em>

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Answer:

a.) The main scale reading is 10.2cm

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d.)  10.31 cm

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Explanation:

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