A 1900kg airplane is flying at an altitude of 510 above the ground. What is the gravitational potential energy in Joules?
1 answer:
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Answer:
D. 0.9
Explanation:
Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.
Resolving into x component:
∑
------(1)
Resolving into y component:
∑
------(2)
Static frictional force, μ
------(3)
substituting from equation (1) and
from equation (2) into equation (3)
μ
μ
μ
μ
The angle the vehicles make with the horizontal α = 42°
μ ≥ tan 42°
μ ≥ 0.9
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