Answer:
120 is the correct answer
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.
Ummm what does this even mean I need a picture to help u
Answer:
1626.4 N
Explanation:
Given that a 82 kg man, at rest, drops from a diving board 3.0 m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?
The parameters to be considered are:
Distance S = 3m
Time t = 0.55s
Since the man started from rest, initial velocity u = 0
Using second equation of motion
S = Ut + 1/2at^2
3 = 1/2 × a × 0.55^2
3 = 1/2 × a × 0.3025
a = 3/ 0.15125
a = 19.83 m/s^2
Force = mass × acceleration
Force = 82 × 19.83
Force = 1626.4 N
Therefore, the force that water exerted on him is 1626.4 N
