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Nezavi [6.7K]
3 years ago
6

When the dog shakes his body, his fur and the droplets of water clinging to it move in circular arcs. If the droplets leave the

fur, what will be the subsequent motion? (For the short times in the video, motion due to gravity is negligible) A) The droplets will move in parabolic arcs. B) The droplets will move in straight-line paths C) The droplets will travel in circular arcs.
Physics
1 answer:
AysviL [449]3 years ago
4 0

Answer:

A) True. The droplets will move in parabolic arcs.

Explanation:

When the dog shakes the drops that leave the fur they leave with a speed and an angle; These are now subject to the out of gravity that accelerates them in the vertical axis and has no outside the horizontal axis, so they must describe a parabolic movement.

Pair to describe a central force that changes its speed, but this force does not exist once they leave the dog

Let's check the answers

A) True. You agree with the analysis

B) False. There is an outside on the axis and therefore the line is deviating

C) False. There is no radial force that creates this movement

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Which image illustrates why dark clothing helps to keep you warm on a cool, sunny day?
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Answer:

Explanation:

I suppose it has to do with the way the diagram is drawn. The heat does not reflect which makes both A and B incorrect.

C would have nothing to do with either reflection or refraction.

That only leaves D which is the answer.

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3 years ago
The component of the ball's velocity whose magnitude is most affected by the collisions is
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3 years ago
A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top
Advocard [28]

Answer:

a)  Em₀ = 42.96 104 J , b)   W_{fr} = -2.49 105 J , c)  vf = 3.75 m / s

Explanation:

The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

        Em = K + U

a) Let's look for the initial mechanical energy

      Em₀ = K + U

      Em₀ = ½ m v2 + mg and

      Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

      Em₀ = 36 104 + 6.96 104

      Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

    W_{fr} = Em₂ -Em₀

     Em₂ = K + U

     Em₂ = ½ m v₂² + m g y₂

     Em₂ = ½ 50 85 2 + 50 9.8 427

     Em₂ = 180.625 + 2.09 105

     Em₂ = 1,806 105 J

     W_{fr} = Em₂ -Em₀

     W_{fr} = 1,806 105 - 4,296 105

     W_{fr} = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

       W_{fr} = ΔEm

       W_{fr} = Emf - Emo

       -1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

       ½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

       ½ 50.0 vf² = 0.561

       vf = √ 0.561 25

      vf = 3.75 m / s

6 0
3 years ago
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