Why are mineral bands in metamorphic rocks light and dark colored?
1 answer:
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Answer:
Explanation:
Given
Force
one at an angle of East of North and another at
West of North
Net Force is in North Direction
Forces in horizontal direction will cancel out each other
thus Work done will be by north direction forces
here
Given data
The angle of inclination of the plane is theta = 30 degree
The applied force in the inclined plane is F = 94 N
The distance moved in the inclined plane is d = 2.30 m
The coefficient of kinetic friction is u_k = 0.280
The free-body diagram of the above configuration is shown below:
Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.
(a)
The expression for the work done by the pushing force is given as:
Substitute the value in the above equation.
Thus, the work done by the pushing force is 216.2 J.
(b)
The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.
The expression for the normal reaction force is given as:
The expression for the mass of the box is given as:
Substitute the value in the above equation.
Thus, the mass of the box is 12.9 kg.
Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system