E = <u>kQ</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>
(r + h)²
where,
k = 9 × 10^9Nm²C^-2
Q = total charge, 300uC = 300 × 10^ -6C
r = 8 × 10^ -2m
h = 16 × 10^ -2m
then,
E = <u>9</u><u>e</u><u>9</u><u> </u><u>*</u><u> </u><u>3</u><u>0</u><u>0</u><u>e</u><u>^</u><u>-</u><u>6</u><u> </u><u> </u><u> </u><u> </u>
(8e^-2 + 16e^-2)²
E = 4687500N/C
For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
The total displacement of the toy car at the given positions is 0.
The given parameters;
- <em>First displacement of the car, = 5 cm left</em>
- <em>Second displacement of the car, = 8 cm right</em>
- <em>Third displacement of the car, = 3 cm to the left</em>
The total displacement of the car is calculated as follows;
- <em>Let the </em><em>left </em><em>direction be "</em><em>negative </em><em>direction"</em>
- <em>Let the </em><em>right </em><em>direction be "</em><em>positive </em><em>direction"</em>

Thus, the total displacement of the toy car at the given positions is 0.
Learn more about displacement here: brainly.com/question/18158577
Your answer is A
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