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1 year ago

Which physical activity would a global positioning system resource help you with?

Physics

1 answer:

PolarNik [594]1 year ago

4 0

The physical activity that a global positioning system resource can help me with is : ( A ) Hiking

<h3>Function of Global positioning system ( GPS ) </h3>

GPS helps with the accurate measurement of physical activities and factors such as location, time, elevation and so on. When hiking, the time, distance covered and location of the hiker can be accurately measured with the use of the GPS.

Hence we can conclude that The physical activity that a global positioning system resource can help me with is Hiking.

Learn more about GPS : brainly.com/question/9795929

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On a distant planet, a rock falls in 48.4 s from the top of a 1.10e+02 m cliff to the planet surface below. What is the accelera

pav-90 [236]

this can be solve using the formala of free fall

t = sqrt( 2y/ g)

where t is the time of fall

y is the height

g is the acceleration due to gravity

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G = 0.0930 m/s2

The velocity at impact

V = sqrt(2gy)

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V = 4.523 m/s

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3 years ago

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IgorLugansk [536]

Answer:

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3 years ago

A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i

olasank [31]

Answer:

The direction of the wave propagation is the negative z -axis

The amplitude of electric and magnetic field are $A_E= 3.35*10^5 V/m$ ,

$A_M= 1.12 *10^{-3} T$ respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

$I = \frac{1}{2} c \epsilon _O E_{rms}^2$

Given that $I = 7.43 \frac{kW}{cm^2} = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}$

Making $E_{rms}$ the subject we have

$E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

$E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

$= 2.37*10^5 \ V/m$

The amplitude of the electric field is mathematically represented as

$A_E = \sqrt{2} * E_{rms}$

$= \sqrt{2} * 2.37*10^5$

$A_E= 3.35*10^5 V/m$

The amplitude of the magnetic field is mathematically represented as

$A_M = \frac{A_E}{c}$

Substituting value

$A_M = \frac{3.35 *10^5}{3.0*10^8}$

$A_M= 1.12 *10^{-3} T$

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3 years ago

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DIA [1.3K]

Answer:

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if C = k - 273

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substitute 22 for C

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