Charge will decreases.
A parallel plate capacitor when it is fully charged to voltage V is given as:
C = Q/V
The capacitance of parallel plate capacitor with two plates of Area A separated by distance d and no dielectric material between plates is
C = ε₀ A /d
since from above equation it shows C is proportional to Q and also C is inversely proportional to distance d.
So, ATQ when d increases C will decrease which in result decreases charge on the capacitor.
Thus, Charge will decrease.
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Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
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Answer:
Explanation:
We know that the pressure can be calculated in the following way:
p = d·g·h
with d being the density of the water, g the gravitational acceleration and h the depth.
Also d of the water = 1000 kg/m^3 circa and g = 9.8 m/s^2 circa
117,500 Pa = 1000kg/m³ · 9.8m/s² · h
Therefore h = 11,9 m