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ziro4ka [17]
3 years ago
12

(a) What is the angular speed ω about the polar axis of a point on Earth's surface at a latitude of 55° N? (Earth rotates about

that axis.) (b) What is the linear speed v of the point? What are (c) ω and (d) v for a point at the equator? (Note: Earth radius equals 6370 km and let one day be 24 hours)
Physics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer

given,

radius of earth = 6370 km

earths latitude = 55° N

time  = 24 hour = 86400 s

angular speed = ω

a) \omega = \dfrac{2\pi}{T}

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

b) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 55^0

v = 266 m/s

c) ω at equator

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

\omega = \dfrac{2\pi}{86400}

\omega = 7.27 \times 10^{-5}rad/s

d) v = r\omega cos\theta

v = 6370\times 7.27 \times 10^{-5}\times cos 0^0

v = 463 m/s

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The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 \frac{N}{C}.

First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.

In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.

The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

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Replacing in the definition of electric field:

E=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{20x10^{-6} C}{(1 m)^{2} }

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