Since the electric field between the plates is constant, If the two plates are brought closer together, the potential difference between the two plates decreases
The relation between potential difference and the electric field is given by ΔV = E.d
Since the electric field is maintained constant, the potential difference is directly inversely proportional to the distance between the plates.
The potential difference between the plates will therefore likewise decrease if the distance between the plates is reduced, we will state in this case.
The energy required to move a unit charge, or one coulomb, from one point to the other in a circuit is measured as the potential difference between the two points. Potential difference is measured in volts or joules per coulomb.
Refer to more about the potential difference here
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Answer:
The final image relative to the converging lens is 34 cm.
Explanation:
Given that,
Focal length of diverging lens = -12.0 cm
Focal length of converging lens = 34.0 cm
Height of object = 2.0 cm
Distance of object = 12 cm
Because object at focal point
We need to calculate the image distance of diverging lens
Using formula of lens



The rays are parallel to the principle axis after passing from the diverging lens.
We need to calculate the image distance of converging lens
Now, object distance is ∞
Using formula of lens


The image distance is 34 cm right to the converging lens.
Hence, The final image relative to the converging lens is 34 cm.
Answer:
8.8 kN
Explanation:
V = 2 m³, W = 40 kN, SG = 1.59
Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN
So the weight becomes 40 - 31.2 = 8.8 kN
Answer:
W = 6642 J
Explanation:
Given that,
Mass of a crate, m = 67 kg
Force with which the crate is pulled, F = 738 N
It is moved 9 m across a frictionless floor
We need to find the work done in moving the crate. Let the work done is W. It is given by :
W = F d
W = 738 N × 9 m
= 6642 J
So, the work done is 6642 J.