The Virtual Laboratory is an interactive environment for creating and conducting simulated experiments: a playground for experimentation. It consists of domain-dependent simulation programs, experimental units called objects that encompass data files, tools that operate on these objects
The longer you spend reading and thinking about this question,
the more defective it appears.
-- In each case, the amount of work done is determined by the strength
of
the force AND by the distance the skateboard rolls <em><u>while you're still
</u></em>
<em><u>applying the force</u>. </em>Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.<em>
</em>
-- We know that the forces are equal, but we don't know anything about
how far each one rolled <em>while the force continued</em>. All we know is that
one force must have been removed.
-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.
-- How far did that one roll while you were still pushing it ?
-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?
-- Did each skateboard both roll the same distance while you continued pushing it ?
I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.
Answer:
E = 2k 
Explanation:
Gauss's law states that the electric flux equals the wax charge between the dielectric permeability.
We must define a Gaussian surface that takes advantage of the symmetry of the problem, let's use a cylinder with the faces perpendicular to the line of charge. Therefore the angle between the cylinder side area has the same direction of the electric field which is radial.
Ф = ∫ E . dA = E ∫ dA = q_{int} /ε₀
tells us that the linear charge density is
λ = q_ {int} /l
q_ {int} = l λ
we substitute
E A = l λ /ε₀
is area of cylinder is
A = 2π r l
we substitute
E = 
E = 
the amount
k = 1 / 4πε₀
E = 2k
You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.
-- While it's drifting in space, it's a <em>meteoroid</em>.
-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.
-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.
We need to be careful here.
The calculation of the gravitational force between two objects
refers to the distance between their centers.
The minimum possible distance between the Earth's and moon's
centers is the sum of their radii (radiuses).
Earth's radius . . . . . 6,360 km = 6.36 x 10⁶ meters
Moon's radius . . . . . 1,738 km = 1.738 x 10⁶ meters
Sum of their radii = 8.098 x 10⁶ meters
Also:
Earth's mass . . . . . 5.972 x 10²⁴ kg
Moon's mass . . . . . 7.348 x 10²² kg
<span>
and now we're ready to go !
Gravitational force =
G M₁ M₂ / R²
= (6.67 x 10⁻¹¹ N-m²/kg²)(</span><span>5.972 x 10²⁴ kg)(7.348 x 10²² kg)/</span>(8.098 x 10⁶ m)²
= (6.67 · 5.972 · 7.348 / 8.098²) · (10²³) Newtons
= (I get ...) 4.463 x 10²³ Newtons
That's almost exactly 10²³ pounds
= 50,153,000,000,000,000,000 tons.
Those are big numbers.
All I can say is: I wouldn't exactly call that "resting" on the surface".
