Weight = (mass) x (gravity).
It always acts downward.
On Earth, the acceleration of gravity is 9.807 m/s².
On the Moon, the acceleration of gravity is 1.623 m/s².
On Earth, the rocket's weight is (0.8kg) x (9.8 m/s²) = 7.84 newtons
On the Moon, the rocket's weight is (0.8kg) x (1.62 m/s²) = 1.3 newtons
The force of the rocket engine acts upward.
Its magnitude is 12 newtons. (From the burning chemicals.
Doesn't depend on local gravity. Same force everywhere.)
Now we have all the data we need to mash together and calculate the
answers to the question. You might choose a different method, but the
machine that I have selected to do the mashing with is Newton's 2nd law
of motion:
Net Force = (mass) x (acceleration).
Since the question is asking for acceleration, let's first solve Newton's law
for it. Divide each side by (mass) and we have
Acceleration = (net force) / (mass) .
On Earth, the forces on the rocket are
(weight of 7.84 N down) + (blast of 12 N up) = 4.16 newtons UP (net)
Acceleration = (4.16 newtons UP) / (0.8 kg) = 5.2 m/s² UP .
On the moon, the forces on the rocket are
(weight of 1.3 N down) + (blast of 12 N up) = 10.7 newtons UP (net)
Acceleration = (10.7 newtons UP) / (0.8 kg) = 13.375 m/s² UP
6 degrees
4N
.....................
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
Answer:
1.6×10⁻⁶ N.
Explanation:
From the question,
F = (V/r)q......................... Equation 1
Where F = Electric force on the raindrop, V = Potential difference between the base of the cloud and the ground, r = distance between the base of the cloud and the ground, q = the charge on a rain drop.
Given: V = 200MV = 200×10⁶ V, r = 500 m, q = 4.0×10⁻¹² C.
Substitute these values into equation 1
F = [(200×10⁶ )/500]×4.0×10⁻¹²
F = 1.6×10⁻⁶ N.
Answer:
Mix
Explanation:
A battery has two electrodes, at one end it has the anode and the other end has the cathode. Electrons travel through the circuit from the anode (negative) to the cathode (positive), and this is the driving force that provides electricity to flow through circuits.
The anode needs to have a low electron affinity because it needs to readily release electrons, and the cathode needs to have a high electron affinity because it needs to readily accept electrons.
