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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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Answer:

The answer is below

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3 years ago
An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi
Arturiano [62]

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  V_x'=0.8c\\V=0.6c\\m=4*10^5kg

  • Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}

  • So, to V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
  • We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V

  • Substituting values, we get,

          V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c

  • Thus, the KE will be,

              KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

3 0
2 years ago
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