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4 years ago

15

Global positioning satellites (GPS) can be used to determine positions w/great accuracy . The system works by determining the di

stance between the observer and each of several satellites orbiting earth . If one of the satellite is at a distance of 20000 km from you . What percent accuracy in the distance is required if we desire a 2 meter uncertainty ? How many significant figures do we need to have in the distance ?

1 answer:

prisoha [69]4 years ago

3 0

Answer:

uncertainity = 0.0000001%

the no of significant figures need to have is eight

Explanation:

given data:

satellite distance from earth = 20000 k = 20,000,000 m

we know that percntage of uncertainity of 2m can be calculated as $uncertainity =\frac{2m}{20,000,000} = 1*10^{-7}%$

uncertainity = 0.0000001%

the range of satellite from your location from km to m is

(20,000,000m -2m) to (20,000,000m + 2m) = 19999998 m +20,000,002 m)

the no of significant figures need to have is eight

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3 years ago

What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s 2 ) towards the Eart

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Answer:

The centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ is $1.020x10^{-3}m/s^{2}$

Explanation:

The centripetal acceleration is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude $71.9^{\circ}$ . Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

$\cos \theta = \frac{adjacent}{hypotenuse}$

$\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}}$ (3)

Where $r_{71.9^{\circ}}$ is the radius at the latitude of $71.9^{\circ}$ and $r_{e}$ is the radius at the equator ( $6.37x10^{6}m$ ).

$r_{71.9^{\circ}}$ can be isolated from equation 3:

$r_{71.9^{\circ}} = r_{e} \cos \theta$ (4)

$r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})$

$r_{71.9^{\circ}} = 1.97x10^{6} m$

Then, equation 2 can be used

$v = \frac{2 \pi (1.97x10^{6} m)}{24h}$

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

$T = 24h . \frac{3600s}{1h}$ ⇒ $84600s$

$v = \frac{2 \pi (1.97x10^{6} m)}{84600s}$

$v = 146.31m/s$

Finally, equation 1 can be used:

$a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}$

$a = 0.010m/s^{2}$

Hence, the centripetal acceleration is $0.010m/s^{2}$

To given the centripetal acceleration as a multiple of $g=9.8 m/s^{2}$ it is gotten:

$\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}$

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3 years ago

The electric potential difference between the ground and a cloud in a particular thunderstorm is 2.5 × 109 V. What is the magnit

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Answer:

U = - 4 x $10^{-10}$

Explanation:

ΔV = potential difference = $2.5 * 10^9$ Volts

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U = q ΔV = ( $-1.6 * 10^{-19}$ ) ( $2.5 * 10^9$ )

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3 years ago