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marusya05 [52]
2 years ago
15

Five locations are marked on the world map. which location is most prone to hurricanes? five locations are marked on the world m

ap. which location is most prone to hurricanes?
a.

b.

c.

d.

e.
Physics
1 answer:
Alex17521 [72]2 years ago
5 0

Five locations are marked on the world map. The spot that is prone to hurricane mostly A.

<h3>What is hurricane?</h3>

A hurricane is a cyclone with winds of 74 miles (119 kilometers) each hour or more prominent that is typically joined by downpour, thunder, and lightning, and that occasionally moves into calm scopes.

The locations on the world map spot A, B, C, D, E are regions with country prone to hurricane.

The SPOT A, is GULF of Mexico, a sea bowl nearer to the Atlantic sea, encompassed with the North American landmass, the Hawaiian island is likewise inside that locale.

The SPOT B, is the southern Pacific sea with nations like BRAZIL, PERU, ECUADOR, CHILE, in that equivalent locale. This is likewise visited by the tropical storm.

The SPOT C, is inside the locales of AFRICA, nations at the edge are likewise impacted by typhoon.

The SPOT D, is Asian landmass with Russia at the edge. They are likewise inclined to typhoon.

The SPOT E, is the North Pacific sea locale situated at the upper left hand side of the world guide, with nations like Canada, Alaska and co. Near the artic circle.

Thus, location A is the most prone to hurricanes.

Learn more about hurricane.

brainly.com/question/18950883

#SPJ1

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A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

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