Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Answer:
Figure E is the correct representation of the first part of the motion. When in a hanging position from the chin-up bar, the bicep muscles are stretched beyond their normal length already. So at this point they are at the peak of their capacity and you are at rest (this corresponds to the velocity v = 0 at t = 0). On contracting the bicep muscles and pulling your whole body up, you begin to gain speed and v increases. This increase in velocity is exponential. Soon the bicep muscles contract up to 80% their normal length reducing the force they can produce to keep you rising up to zero. The velocity change happens because the body is accelerating and the muscles can still supply a net force to lift you up. The acceleration is present because of this net force. The moment this force reduces to zero, the acceleration too reduces to zero. (From Newton's second law of motion). This reduction in acceleration is responsible for the reduction of the curvature of the v curve in figure E above. The point where the velocity becomes horizontal corresponds to the point where the muscles reach their maximum contraction unit and can supply no more net force and as a result no acceleration. This further results inba constant velocity which is the flat nature of the curve seen in diagram E.
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Explanation:
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