Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
I don't think you mean 'criteria'. I think you mean three occurrences or
observations that indicate the presence of acceleration.
They are:
-- an object is speeding up
-- an object is slowing down
-- the direction of an object's motion is changing .
Any one of these changes is acceleration.
There's a single term that covers them all. It is "change in velocity".
The square root of 80 is: 8.944
Answer:
The density of the metal is 5200 kg/m³.
Explanation:
Given that,
Weight in air= 0.10400 N
Weight in water = 0.08400 N
We need to calculate the density of metal
Let 
be the density of metal and 
be the density of water is 1000kg/m³.
V is volume of solid.
The weight of metal in air is
.....(I)
The weight of metal in water is
Using buoyancy force
We know that,
....(I)
Put the value of 
in equation (I)
Put the value of Vg in equation (II)
Hence, The density of the metal is 5200 kg/m³.
The car's speed is 240km/4hr= 60km/hr.
There's not enough information given in the question to determine its velocity.
