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tekilochka [14]
3 years ago
10

Describe the parts of a comet.

Physics
1 answer:
scZoUnD [109]3 years ago
5 0
Here are the parts of the comet:
1. NUCLEUS: This is the frozen part of the comet. It is also known as the core. It is made up of ice and dust which are completely covered by organic matter. The nucleus usually consist of frozen water but other materials that are in frozen forms can be found in it. Comet nuclei are usually less than 16 kilometer in diameter.
2. COMA: The atmosphere of dust and gases formed when the nucleus vaporize. The coma refers to the envelope of gases that surround the comet's nucleus. The coma plus the nucleus forms the head of the comet. The coma is about a million kilometer in diameter and is made up of gases and dust which sublime from the comet's nucleus.
3. ION TAIL: Tail made of ions that appear to point away from the comet's orbit. The charged solar particles convert the gases found in the comet to ions thus forming an ion tail.  The ion tail can measure over 100 million kilometer long and it accelerate much faster than the dust tail.
4. DUST TAIL: Tail made up of small solid dust particles. It is formed by radiation from the sun, which forces dust particles away from the coma. It usually point away from the sun because the tail are shaped by the solar wind. As the distance from the sun increases, the dust tail usually become faint and diminished.

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What is the energy source of the convection cells in the earth's mantle?
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A. Radioactive decay does not result in heat. Gravity also does not result in heat. In the Earth's mantle, it is highly unlikely that the radiation from the sun would reach that deep into the earth.
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What four things<br> are required for photosynthesis?
drek231 [11]

Explanation:

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A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent
anzhelika [568]

Answer:

The position is  P  =  47.4 \ m relative to the base of the ocean

Explanation:

From the question we are told that

    The angle made by the incline with the horizontal is \theta =  24.0 ^o

    The constant acceleration is a =  3.82 \ m/s^2

    The distance covered is d =  60.0 \ m

    The height of the cliff is h  =  50 .0 \ m

The velocity of the car is mathematically represented as

      v^2 = u^2 +  2ad

The initial velocity of the car is  u= 0

So

     v^2 =  2ad

substituting values

     v^2 = 2 *  3.82 * 60

    v =  21.4 \ m/s

The vertical component of this velocity is

    v_v  =  -v * sin(\theta )

substituting values

    v_v  = -21.4 * sin(24.0)

    v_v  = -8.7 \ m/s

The negative sign is because is moving in the negative direction of the y-axis

The horizontal  component of this velocity is

     v_h  =  v * cos (\theta)

    v_h  =  21.4 * cos (24.0)

    v_h  = 19.5 \ m/s

Now according to equation of motion we have

     h  =  v_v*t  - \frac{1}{2} *  g t^2

substituting  values

    50  =  -8.7 t  - \frac{1}{2} *  9.8 t^2

    4.9t^2  +8.7t -50 = 0

using quadratic equation we have that

  t_1 = 2.42\ s \ and\   t_2 =  -4.20\ s

given that time cannot be negative

      t = 2.42 \ s

The  car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

           P  =  v_h * t

substituting values

          P  =  19.5 *  2.43

         P  =  47.4 \ m

5 0
3 years ago
If two balls have the same volume,
Lena [83]

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

  • 1. Ball A will have the greater density
  • 2. Ball C and Ball D have the same density.
  • 3. Ball Q will have the greater density.
  • 4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

  • Density = Mass/Volume.

For Case 1:

  • Va = Vb and Ma = 2Mb
  • D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
  • Therefore, the density of ball A,
  • D(a) = 2D(b).
  • Therefore, ball A has the greater density.

For Case 2:

  • Vc = 3Vd,

  • Vd = (1/3)Vc

  • Md = (1/3)Mc

  • D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

  • D(c) = D(d).

  • Therefore, ball C and D have the same density

For Case 3:

  • Vp = 2Vq and Mp = Mq
  • D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
  • Therefore, the density of ball P is half the density of ball Q
  • Therefore, ball Q has the greater density.

For case 4:

  • Mx = (1/2)My
  • Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

8 0
3 years ago
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