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.
a. The direction of the stone's velocity changes as it moves around the circle.
b. The magnitude of the stone's velocity does not change.
d. The change in direction of the stone's motion is due to the centripetal force acting on the stone.
Above given are true for the given situation.
<u>Answer:</u> Option A, B and D
<u>Explanation:</u>
Circular motion may be characterized as the moving of an objects along the diameter of the circle or any circular direction. It may be standardized and non-uniform based on whether or not the rate of rotation is unchanged.
The velocity, a vector quantity is constant in a uniform circle motion speed is constant as its direction continues to change. Centripetal force works inward toward the core to counterbalance the centrifugal force from the center moving outward.
Answer:
more than 500 n i think the answer will
Explanation:
Q1) What is the speed of the tip of the minute hand of a clock where the hand is of length 7cm?
Ans1) speed, v=st=2πrT=2×227×7×10-260×60=119×10-4=1.22×10-4m/s
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<em><u>Hope it helps</u></em></h2>
