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Sunny_sXe [5.5K]
3 years ago
14

Why does a coastal area have less variation in temperature than a noncoastal area? Temperatures feel cool all year round due to

the duration of daylight. Temperatures change from warm to cool depending on the season. The temperatures on land change more rapidly than temperatures on water. The air is warm because warm water moves from the equator to the poles.
Physics
2 answers:
lara31 [8.8K]3 years ago
4 0

Answer:

Yes coastal areas have less variation in temperature than non coastal area because of land breeze and sea breeze

Explanation:

In the coastal area temperature is maintained by land breeze and sea breeze in all the weathers. In day time land heats up faster than sea, so temperature is high and pressure is low and breeze from sea flows towards land. In the night time land cools faster than sea so low temperature and high pressure condition developed and land breeze flows towards sea.

iragen [17]3 years ago
3 0

<em>HERE'S YOUR ANSWER...</em>

• In coastal areas have less variation in temperature that. non-coastal area due to sea-breeze and land-breeze.

<em>EXPLANATION:</em>

• In day time the land heats up faster than the sea, so air above sand gets heated and rises up and cool air above ocean moves towards land to take its place.

• At night the land cools down faster than sea which causes hot air above sea to rise up and cool air above land moves in to take its place.

<em>HOPE</em><em> </em><em>IT</em><em> </em><em>HELPS</em><em>.</em><em>.</em><em>.</em>

care bear
2 years ago
what is the <em> for?</em>
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Answer:

Moment about SHOULDER  ∑ τ = 3.17 N / m,

Moment respect to ELBOW   Στ= 2.80 N m

Explanation:

For this exercise we can use Newton's second law relationships for rotational motion

         ∑ τ = I α

   

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They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

The mass of the forearm with the included weight is approximately 2.3 kg, with a length of about 50cm

Moment about SHOULDER

          ∑ τ = I α

           I = I_forearm + I_sphere

the forearm can be approximated as a fixed bar at one end

            I_forearm = ⅓ m L²

the moment of inertia of the mass in the hand, let's approach as punctual

            I_mass = m L²

we substitute

           ∑ τ = (⅓ m L² + M L²) α

let's calculate

          ∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

           ∑ τ = 3.17 N / m

Moment with respect to ELBOW

In this case, the arm exerts an upward force (muscle) that is about 3 cm from the elbow

         Στ = I α

         I = I_ forearm + I_mass

         I = ⅓ m (L-0.03)² + M (L-0.03)²

         

let's calculate

        i = ⅓ 2.3 0.47² + 0.5 0.47²

        I = 0.2798 Kg m²

        Στ = 0.2798 10

        Στ= 2.80 N m

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Given :

A 120 kg box is on the verge of slipping down an inclined plane with an angle of inclination of 47º.

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