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photoshop1234 [79]

3 years ago

5

Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,0

00 km from you, and you want to know your position to 2m, what percentage uncertainity is required?

1 answer:

o-na [289]3 years ago

8 0

Answer:

0.00001 %

Explanation:

The distance from satellite = 20000 km

Position range = 2 m

The percentage uncertainty is given by dividing the distance from satellite by the position range of desired accuracy.

Percentage uncertainty is given by

$\dfrac{2}{20000\times 10^3}\times 100=0.00001\ \%$

The percentage uncertainty that is required is 0.00001 %

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Shalnov [3]

A distance of d is covered with 53 mile/hr initially. Time taken to cover this distance t1 = d/53 hour Next distance of d is covered with x mile hours. Time taken to cover this distance t2 = d/x hours. We have average speed = 26.5 mile / hour

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= $\frac{2d}{\frac{d}{53}+\frac{d}{x}} = \frac{2}{\frac{1}{53}+\frac{1}{x} } = \frac{106x}{x+53}$

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3 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

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$\rho =$ Density

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Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

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For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

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$a_c = \frac{\Theta^2}{R}$

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Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

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