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Aloiza [94]
3 years ago
12

How can being near a large body of water affect the climate of the region?

Physics
2 answers:
shepuryov [24]3 years ago
7 0

Answer:

yes

Explanation:

If you've ever heard of lake effect snow that is when cold air mass moves across long expanses of warmer lake water. The lower layer of air, heated up by the lake water, picks up water vapor from the lake and rises up through the colder air above; the vapor then freezes and is deposited on the leeward (downwind) shore. this is most common with lake Michigan when cold air travels down from Canada making the neighboring states colder, so YES, being near any body of water can effect the climate.

I hope this helps

bija089 [108]3 years ago
7 0

Answer:

Explanation:

Hurricane winds whip up large

waves in the ocean. These waves cause

a bulge of water in the ocean known

as a storm surge (stawrm surj). As

a storm moves over a coast, the

storm surge can cause water levels to

suddenly rise, or surge, several meters.

The main dangers in a hurricane

are flying objects, strong winds, and

flooding. If possible, leave areas

threatened by a hurricane. Take

prescription medicines with you.

If you stay in a building, board up

windows and stay away from windows

and doors. Store food, bottled water,

flashlights, and battery operated

radios. Stay away from beaches

and areas that may become flooded

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Q1 is located at the origin, Q2 is located at x = 2.50 cm and Q3 is located at x = 3.50 cm. Q1 has a charge of +4.92μC and Q3 ha
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Answer:

+1.11\mu C

Explanation:

A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.

Q_1 is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to Q_2 and Q_3 at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of Q_1.

Let the electric field intensity due to Q_2 be +E_2 and that due to Q_3 be -E_3 since the charge is negative. Hence at the origin;

+E_2-E_3=0..................(1)

From equation (1) above, we obtain the following;

E_2=E_3.................(2)

From Coulomb's law the following relationship holds;

+E_2=\frac{kQ_2}{r_2^2}\\  

-E_3=\frac{kQ_3}{r_3^2}

where r_2 is the distance of Q_2 from the origin, r_3 is the distance of Q_3 from the origin and k is the electrostatic constant.

It therefore means that from equation (2) we can write the following;

\frac{kQ_2}{r_2^2}=\frac{kQ_3}{r_3^2}.................(3)

k can cancel out from both side of equation (3), so that we finally obtain the following;

\frac{Q_2}{r_2^2}=\frac{Q_3}{r_3^2}................(4)

Given;

Q_2=?\\r_2=2.5cm=0.025m\\Q_3=-2.18\mu C=-2.18* 10^{-6}C\\r_3=3.5cm=0.035m

Substituting these values into equation (4); we obtain the following;

\frac{Q_2}{0.025^2}=\frac{2.18*10^{-6}}{0.035^2}\\\\hence;\\\\Q_2=\frac{0.025^2*2.18*10^{-6}}{0.035^2}\\

Q_2=\frac{0.00136*10^{-6}}{0.00123}=1.11*10^{-6}C\\\\Q_3=+1.11\mu C

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