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3 years ago

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

Physics

2 answers:

maria [59]3 years ago

8 0

Answer: A

Explanation:

Kruka [31]3 years ago

3 0

Answer: A is the spot that is prone to hurricane mostly.

Explanation:

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, an ocean basin closer to the Atlantic ocean, surrounded with the North American continent, the Hawaiian island is also within that region.

The SPOT B, is the southern Pacific ocean with countries like BRAZIL, PERU, ECUADOR, CHILE, in that same region. This is also visited by the hurricane.

The SPOT C, is within the regions of AFRICA, countries at the edge are also affected by hurricane.

The SPOT D, is Asian continent with Russia at the edge. They are also prone to hurricane.

The SPOT E, is the North Pacific ocean region located at the top left hand side of the world map, with countries like Canada, Alaska and co. Close to the artic circle.

With the observation of all this spot, THE SPOT A, is most prone to hurricane, and its effect are very high there....

Country thee are greatly affected by it,

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Read 2 more answers

An object is moving along a straight line, and the uncertainty in its position is 1.90 m.

just olya [345]

Answer:

$2.78\times 10^{-35}\ \text{kg m/s}$

$6.178\times 10^{-34}\ \text{m/s}$

$0.31\times 10^{-4}\ \text{m/s}$

Explanation:

$\Delta x$ = Uncertainty in position = 1.9 m

$\Delta p$ = Uncertainty in momentum

h = Planck's constant = $6.626\times 10^{-34}\ \text{Js}$

m = Mass of object

From Heisenberg's uncertainty principle we know

$\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}$

The minimum uncertainty in the momentum of the object is $2.78\times 10^{-35}\ \text{kg m/s}$

Golf ball minimum uncertainty in the momentum of the object

$m=0.045\ \text{kg}$

Uncertainty in velocity is given by

$\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}$

The minimum uncertainty in the object's velocity is $6.178\times 10^{-34}\ \text{m/s}$

Electron

$m=9.11\times 10^{-31}\ \text{kg}$

$\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}$