Why is it difficult to define psychology disorder
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Answer:
a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing
b) E_total = 1.89 10⁶ N / C, field is incoming radial
c) E_total = 0
Explanetion:
For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is
Ф = E .dA = q_{int} /ε₀
inside the spherical shell there are no charges
The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.
We have two sphere shells with radii 0.050m and 0.15m respectively
a) point where you want to know the electric field d = 0.20 m
shell 1
the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center
E₁ = k q₁ / d²
shell 2
the point is on the outside,d>ro
E₂ = k q₂ / d²
the total camp is
E_total = -E₁ + E₂
E_total = k ( )
E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²
E_total = 6,525 10⁵ N /C
The field direction is radial and outgoing ti the shells
b) the calculation point is d = 0.10m
shell 1
point outside the shell d> ro
E₁ = k q₁ / d²
shell 2
the point is inside the shell d <ro
Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero
E₂ = 0
E_total = E₁
E_total = k q₁ / d²
E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²
E_total = 1.89 10⁶ N / A
As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1
c) the point of interest d = 0.025 m
shell 1
point is inside the shell d< ro
as there are no charges inside
E₁ = 0
shell 2
point is inside the radius of the shell d <ro
E₂ = 0
the total field is
E_total = 0
Answer:
The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is
Explanation:
The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,
Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.
The expression for static friction over a horizontal surface is,
Here, is the coefficient of static friction, mm is mass of the object, and
is the acceleration due to gravity.
Use the expression of static friction and solve for maximum static friction for box of mass
Substitute for in the expression of maximum static friction
Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.
Substitute for , [/tex]{m_1}[/tex] for in the equation .
Substitute for
in the equation
Rearrange for a.
The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.
The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.
Use the Newton’s second law for the system of box and the board.
Substitute for for in the equation .
Substitute for in the above equation .
The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is
There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.
Complete question:
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.
a) 60 ⁰
b) 90 ⁰
c) 120 ⁰
Answer:
(a) When the angle, θ = 60 ⁰, force = 4.07 N
(b) When the angle, θ = 90 ⁰, force = 4.7 N
(c) When the angle, θ = 120 ⁰, force = 4.07 N
Explanation:
Given;
length of the wire, L = 2.8 m
current carried by the wire, I = 5.6 A
magnitude of the magnetic force, F = 0.3 T
The magnitude of the magnetic force is calculated as follows;
(a) When the angle, θ = 60 ⁰
(b) When the angle, θ = 90 ⁰
(c) When the angle, θ = 120 ⁰