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iris [78.8K]
3 years ago
14

What is the most common state of matter in the universe

Physics
1 answer:
ch4aika [34]3 years ago
7 0

Answer:

<u><em>Plasma</em></u>

Explanation:

<u><em>Plasma</em></u> is the most common because plasma is a gas that has been energized to the point that some of the electrons break

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Which of the following is the visible part of the spectrum of light?
mestny [16]
The visible spectrum<span> is portion of the </span>electromagnetic spectrum<span> that is </span>visible <span>to the </span><span>human eye.
</span>gamma rays, <span>sound waves, </span><span>microwaves are not the part of spectrum of light, whereas color is the portion of spectrum of light.</span>
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3 years ago
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An object is represented by the dot on a motion map. What is the best description for the motion of this object? The object is m
ankoles [38]

The object is standing still.

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The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizon
kirza4 [7]

Answer:

A) 3.79 m/s  B) 1.33 m

Explanation:

A)

  • Horizontal movement:
  • Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
  • This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):

        v_{ox} = v_{o} *cos (58.0 deg)

  • The horizontal displacement can be simply calculated as follows:

        x = v_{ox} *t

  • Vertical movement:
  • As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
  • This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):

        v_{oy} = v_{o} *sin (58.0 deg) (1)

  • Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
  • In that moment, we can apply the following kinematic equation:

        v_{fy} ^{2} -v_{oy} ^{2} = 2*g*h_{max}

  • where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
  • Replacing by the givens, we can solve for voy:

        v_{oy} =\sqrt{2*g*h_{max}} = \sqrt{2*9.8m/s2*0.527m} =3.21 m/s

  • From the equation (1), we can solve for the magnitude of the initial velocity, v₀:

        v_{o} = \frac{v_{oy}}{sin 58.0} =\frac{3.21m/s}{0.848} = 3.79 m/s

B)

  • With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:

        v_{ox} = v_{o} *cos (58.0 deg) =\\  \\ 3.79 m/s * cos (58.0deg) = 2.01 m/s

  • In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
  • We can use the equation for the vertical displacement, replacing this value by 0, as follows:

       y = 0 = v_{oy} *t -\frac{1}{2} * g *t^{2}

  • Replacing by  the givens, and rearranging terms, we can solve for t:

        t_{air} =\frac{2*v_{oy} }{g} = \frac{2*3.21 m/s}{9.8 m/s} = 0.66 s

  • Finally, we find the horizontal displacement, as follows:

       x_{max}  = v_{ox} *t_{air} = 2.01 m/s * 0.66 s \\ \\ x_{max} = 1.33 m

  • The horizontal distance covered by  the froghopper was 1.33 m.
4 0
3 years ago
5g of sugar is dissolved in "50g" of water what is the mass of the sugar water
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55g because you would still have the weight of the sugar but it combines with the liquid and spreads out into the water.

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slega [8]

The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ?           I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed.   (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = <em>1.449 seconds</em> (rounded)

At that point, since it has no more upward speed, it can't go any higher.  Right ?

(crickets . . .)

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3 years ago
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