Answer:
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
Explanation:
The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
A = Cross sectional Area of the capacitor
d = separation between the capacitor
So,
U = CV²/2
Substituting for C
U = ϵAV²/2d
Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵAV²/2d₁
U₁ = ϵAV²/(2(3d))
U₁ = (ϵAV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡ HI ≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡
∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴∵∴
Due to Rectilinear propagation of light
- A shadow is formed
- Formation of Day and Night
- An Image in the pinhole camera is formed
∞║║ HOPE THIS HELPS PLEASE MARK AS BRAINLIEST║║∞
<h3><u>Mass of an object affect the outcome of unbalanced forces:</u></h3>
Newton’s second law of motion deals with the result of motion of an object when unbalanced forces are applied. The second law of motion establish the relationship between mass, acceleration and unbalanced forces of the object. The below points explains the relation between unbalanced forces acting on mass of the object. The acceleration of object would be higher when the unbalanced force is higher.
The equation for the unbalanced force is,
For Example: Two masses of 500 kg and 1 kg is applied with same unbalanced forces. The change in motion of 500 kg would be very much less than the change in motion of 1 kg.
Given:
V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
V2 = final velocity = ?
For linear-motion problems with those given terms, the following formula is used:
V2 = V1 + as
Substituting the given values:
V2 = 24 + 2(8)
V2 = 24 + 16
V2 = 40 m/s
Therefore the car will have a speed of 40 m/s as it leaves the tunnel.
