Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V
•wind
•snow
•high tide/low tide
•thunder/lightning storms
Answer:
the center of mass is 7.07 cm apart from the bend
Explanation:
the centre of mass of a wire of length L is L/2 ( assuming uniform density). Then initially the x coordinate of the centre of mass is
x₁ = L/2 = 20 cm /2 = 10 cm
when the wire is bent in a right angle the coordinates of the new centre of mass will be
x₂ = L₂/2
y₂= L₂/2
where L₂ is the length of the horizontal piece and vertical piece . Then L₂=L/2
x₂ = L₂/2 = L/4 = 20 cm/4 = 5 cm
y₂= L₂/2 = L/4 = 20 cm/4 = 5 cm
x₂=y₂=X
locating the bend in the origin (0,0) the distance to the centre of mass is
d = √(x₂²+y₂²) = √(2X²) = √2*X=√2*5cm = 7.07 cm
d = 7.07 cm
Answer:
The number of turns is 
Explanation:
From the question we are told that
The inner radius is 
The outer radius is 
The current it carries is 
The magnetic field is 
The distance from the center is 
Generally the number of turns is mathematically represented as
Generally 
is the permeability of free space with value
So
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
