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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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asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

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2 years ago

If a 580 N net force acts on a 40 kg what will the acceleration of the car be

myrzilka [38]

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2 years ago

Read 2 more answers

Suppose the glass paperweight has index of refraction n=1.38. a) find the value of θ for which the reflection on the vertical su

zimovet [89]

Answer:

a)θ=71.89°

b)NO

Explanation:

Given that

For glass n= 1.38

We know that for air n'=1

The angle for total internal reflection θc given as

sin θc=n'/n

By putting the values

sin θc=n'/n

sin θc=1/1.38

θc=46.43°

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sinθref = cosθc

n'sinθ = n cosθc

1 x sinθ =1.38 x cos 46.43°

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3 years ago

What is the strength of the electric field 3.0 cm from a small glass bead that has been charged to + 8.0 nc ?

Debora [2.8K]

Thinking the small glass bead as a single point charge, the electric field generated by it is given by
$E(r) = k_e \frac{Q}{r^2}$
where
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$Q=8.0 nC=8.0 \cdot 10^{-9} C$ is the charge of the bead
$r=3.0 cm=0.03 m$ is the distance at which we calculate the field.

Using these data, we find:
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