Answer:
(B) Dark energy does not exist and there is much more matter than current evidence suggests.
Explanation:
The repulsive force which is accelerating expansion of the universe is called as dark energy. Most of matter present in the universe is the dark matter of about eighty five percent.
So, a collapsing universe would not have the dark energy and there is more matter which is not the dark matter. This theory is rejected because expansion of the universe is observable.
Answer:
it have Potential energy
Explanation:
given data
Drag the pendulum to an angle 30∘
to find out
what form of energy does it have
solution
we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.
we know that the total energy is always conserve
so it have potential energy
Electrons would likely pass through the plasma channel when the formation of a lightning takes place. Lightning is primarily caused by the formation of an ionised cloud in the atmosphere. This cloud was due to the attractions between molecules of rain drops wherein produce a negative charge.
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA =
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
So this is the info that's given to us:
m1 = 4.87 x 10^24 kg
m2 = 4.87 x 10^24kg
F = 2.58 x 10^3 N
Now we can use this equation to solve:

You may be asking what g is. Well, it is equal to 6.67x10^-11 m3 /kg s2. G is gravitational constant. So :
As you can see we only have one variable, so we solve for r!

I hope this helps you out. I remember when I was first learning Newton's law<span> of universal </span><span>gravitation and I had some trouble. But with some practice, I got better. =) </span>