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Dmitry_Shevchenko [17]
4 years ago
10

Stars and planets form when dust and gas from space is pulled together by.. what?

Physics
1 answer:
NemiM [27]4 years ago
5 0

Answer:

gravity?

Explanation:

i dont have a lot of context but this is what i thought of so...

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Answer: C) Conduct the experiment in a vacuum.
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
Which of the following BEST describes what occurs in a fusion reaction?
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Answer:

Two mass nuclei are joined to form one nucleus

Explanation:

As the name implies "fusion", it is the joining of lighter nucleus to form a heavier nucleus to produce energy

4 0
3 years ago
The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If
ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0
3 years ago
Imagine that you’re observing a collision. Which action would allow you to determine whether the collision is inelastic?
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The answer is B;) in plato

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