The term <u>c</u><u>e</u><u>l</u><u>l</u> is incorrect because, the average voltage of a cell is 3.40 V and as already mentioned in question, the volt of source is 18 V, so cell couldn't be the appropriate word, actually the source is <u>b</u><u>a</u><u>t</u><u>t</u><u>e</u><u>r</u><u>y</u><u>.</u>
Now, the source is most likely related to cell because a <u>b</u><u>a</u><u>t</u><u>t</u><u>e</u><u>r</u><u>y</u> which is the source here, is made up of a "group of cell".
Answer: B)To the left of the charges.
Explanation: between the charges the electric field will not cancel but will be added since electric field lines from both charges point in the same direction. To the right of the charge the -4q will take over as it’s strength overcomes the strength of the +q charge. At this point the magnitude of +q will never reach a magnitude strong enough to cancel the -4q. To the left, it is further away from -4q and is closer to +q and electric field lines point in different direction
Answer:
<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>
Explanation:
Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.
v² = u² + 2gs.......................... Equation 1
Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball
<em>Given: s = 20 m, u = 0 m/s</em>
<em>Constant: g = 9.8 m/s²</em>
<em>Substituting these values into equation 1,</em>
<em>v² = 0 + 2×9.8×20</em>
<em>v² = 392</em>
<em>v = √392</em>
<em>v = 19.799 m/s.</em>
<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>