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.
Moving an object up an inclined plane<span> requires </span>less<span>force </span>than<span> lifting it straight up, at a cost of an increase in the distance moved. The </span>mechanical advantage<span>of an </span>inclined plane<span>, the factor by which the force is reduced, is equal to the ratio of the length of the sloped surface to the height it spans.</span>
Answer: option d. 
Explanation:1) The
direction of the
field lines inform about the
sign of the charges.
The field lines <span>
extend from the positive charges to the negative charges, so you can conclude that the charge C is positve and both charge A and charge B are negative:
</span><span>
</span><span>
</span><span>Charge C: positive
</span><span>
</span><span>Charge A: negative
</span><span>
</span><span>Charge B: netative
</span>
2) The
density of the lines (number of lines in a region) inform about the
magnitude of the electric field.
Since the charges are at the same distance, the magnitude of the electric field informs directly about the magnitude of the force and that about the magnitude of the charges.
Since, there are the
double of lines between C and B than between C and A, the magnitude of
charge B is the double than the magnitud of charge A.
From the five options given (a throug e) the only that is consistent with that charges A and B have the same sign, that charge C has different sign, and that charge B is the double of charge A is:
Here, K.E. = 1/2 * mv²
So, K.E. = 1/2 * (1200) * (24)²
K.E. = 1/2 * 1200 * 576
K.E. = 600 * 576
K.E. = 345,600 J
Hope this helps!
