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
Circularity system........…….......
Answer:
The angular acceleration is 
Explanation:
From the question we are told that
The moment of inertia is 
The net torque is 
Generally the net torque is mathematically represented as
Where 
is the angular acceleration so
substituting values
Answer:
Explanation:
AB = 110 miles
Let the distance of the western station from fire is d.
As according to the diagram, use Sine law
d = 110 x 0.2588 / 0.73
d = 39 miles
La velocidad vertical del tanque después de caer 10 m es 14 m/seg .
La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :
Vfy=?
h = 10 m
Fórmula de Velocidad vertical Vfy:
Vfy² = 2*g*h
Vfy= √(2*9.8m/seg2* 10m )
Vfy= 14 m/seg
