Answer:
Explanation:
1) The moment of inertia of the grindstone is:
Which of the following quantities is a scalar? I need help plz someone help me!!!!!
2 answers:
You might be interested in
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
A)
The strength of the magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
In this situation,
I = 5.20 A
r = 4.60 cm = 0.046 m is the distance of the electron from the wire
Therefore the magnetic field strength at the electron's location is
The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by
where
q is the magnitude of the charge of the particle
v is its velocity
B is the magnetic flux density
For the electron of this problem,
is the charge
is the speed
is the magnetic field
Substituting,
B) Same direction as the current in the wire
First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:
- The thumb gives the direction of the current -> upward
- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)
Now the direction of the force can be found by using the right-hand rule. We have:
- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)
- middle finger = direction of the magnetic field (to the north)
- Thumb = direction of the force --> downward
However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.
Answer: 428.57 Mpc
Explanation:
Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance", and enunciated the now called Hubble–Lemaître Law.
This is mathematically expressed as:
(1)
Where:
is the recession velocity of the galaxy
is the Hubble constant
is the distance
Isolating from (1):
(2)
(3)
Finally: