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LuckyWell [14K]
3 years ago
6

И СРСct to

Physics
1 answer:
oee [108]3 years ago
8 0

Crust

Explanation:

The crust is the layer of the earth with the least dense materials. The crust is made up of silicate materials.

  • During the early stage in the formation of earth.
  • The earth differentiated minerals based on their densities.
  • Lighter minerals stayed afloat while the denser ones sank to the core of the earth.
  • The core is made up of the most dense metallic minerals.
  • As we go deep into the earth density increases.
  • Density of crust ranges from 2.-3.3g/cm³
  • The core is about 13.8g/cm³ dense.

Learn more:

Layers of the earth brainly.com/question/13059175

#learnwithBrainly

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A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
8090 [49]

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

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3 0
3 years ago
The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.
Elena L [17]

Answer:

The magnitude of the acceleration is a_r = 1.50 \ m/s^2

The direction is  \theta =  32.5 6^o north of  east

Explanation:

From the question we are told that

   The force exerted by the wind is  F_{sail} =  (330 ) \ N \ north

   The force exerted by water is  F_{keel} =  (210  ) \ N \ east

      The mass of the boat(+ crew) is  m_b  =  260  \ kg

Now Force is mathematically represented as

      F =  ma

Now the acceleration towards the north is mathematically represented as

      a_n  =  \frac{F_{sail}}{m_b}

substituting values

       a_n  =  \frac{330 }{260}

      a_n  =  1.269 \ m/s^2

Now the acceleration towards the east is mathematically represented as

       a_e = \frac{F_{keel}}{m_b }

substituting values

      a_e = \frac{210}{260}

      a_e =0.808 \ m/s^2

The resultant acceleration is  

      a_r =  \sqrt{a_e^2 + a_n^2}

substituting values

     a_r =  \sqrt{(0.808)^2 + (1.269)^2}

      a_r = 1.50 \ m/s^2

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

        tan \theta =  \frac{a_e}{a_n}

       \theta = tan ^{-1} [\frac{a_e}{a_n } ]

substituting values

     \theta = tan ^{-1} [\frac{0.808}{1.269 } ]

    \theta = tan ^{-1} [0.636 ]

   \theta =  32.5 6^o

     

   

       

5 0
4 years ago
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