Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
mass of iron block given as
density of iron block is
now the volume of the iron piece is given as
Now when this iron block is complete submerged in oil inside the beaker the buoyancy force on the iron block will be given as
here we know that
= density of liquid = 916 kg/m^3
Now for the reading of spring balance we can say the spring force and buoyancy force on the block will counter balance the weight of the block at equilibrium
So reading of spring balance will be 16.45 N
Now for other scale which will read the normal force of the surface we can write that normal force on the container will balance weight of liquid + container and buoyancy force on block
So the other scale will read 36.47 N
Answer:
The lungs get rid of carbon dioxide and water vapor. The liver gets rid of bile, which, in addition to breaking down fats, is partially made up of the breakdown of red blood cells. The kidneys get rid of toxins from the blood. The large intestine gets rid of undigested food
Answer:
Electric field, E = 0.064 V/m
Explanation:
It is given that,
Resistivity of silver wire, 
Current density of the wire, 
We need to find the magnitude of the electric field inside the wire. The relationship between electric field and the current density is given by :
E = 0.0636 V/m
or
E = 0.064 V/m
So, the magnitude of electric field inside the wire is 0.064 V/m. Hence, this is the required solution.
