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SCORPION-xisa [38]
3 years ago
5

What is one limitation of using the jawbreaker as a model for Earth's layers? A. The layers are all made out of the same materia

l. B. The outside layer is the thinnest layer. C. The jawbreaker has four different layers. D. The jawbreaker can be split to show the inside.
Physics
2 answers:
schepotkina [342]3 years ago
7 0
The layers of a jawbreaker is all made of the same material. The layers of the Earth are not, it is made of different kinds of rock, magma, and metal is inside of the core
Margarita [4]3 years ago
4 0

The correct answe is A

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Why is the term minority group no longer effective
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The term minority group is no longer effective because these groups now make up significant percentages of the total population
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65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight
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Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

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A student attaches a block of mass M to a vertical spring so that the block-spring system will oscillate if the block-spring sys
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2 years ago
A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

                 I = mr^2

Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

      Now expressing the above statement mathematically

            potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}

               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

Substituting values

                v = \sqrt{9.81 * 1.7}

                  v=4.08m/s

4 0
2 years ago
Read 2 more answers
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