Answer:
The gravitational field strength on the surface of Mars = 3.72 m/s²
Explanation:
Gravitational Field Strength: This can be defined as the force per unit mass which is exerted at that point. its direction is the force exerted on a mass in a gravitational field. The S.I unit of gravitational field strength is m/s²
Mathematically, Gravitational field is represented as,
g = F/m ..................... Equation 1.
m = F/g ..................... Equation 2.
Where g = gravitational Field Strength, F = force on the mass, m = mass of the body.
From the question,
Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.
On the Surface of the earth,
Given: F = 785 N, g = 9.8 m/s²
Substituting this values into equation 2,
m = 785/9.8
m = 80.10 kg.
On the surface of Mars.
Given: m = 80.10 kg, F = 298.
Substituting into equation 2
g = 298/80.1
g = 3.72 m/s²
Thus the gravitational field strength on the surface of Mars = 3.72 m/s²
True
The electromagnet will become stronger if we add more coils because there are more field lines in a loop then there is in a straight piece of wire. In a solenoid there are a lot of loops and they are concentrated in the middle, as more loops are added the field lines get larger, therefore making the electromagnet stronger.
The grains in igneous rocks are interlocking but the Metamorphic rocks are not.
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