Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹ 
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction
By Boyle's law:
P₁V₁ = P₂V₂
70*8 = P<span>₂*4
</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>
Answer:
Yes the body will receive a dangerous shock in both cases.
Explanation:
Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.
Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.
Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:
I = V/R= 120/ 500*10^3
I= 0.24 mA
Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.
Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:
I = 120 / 1000
I = 12 mA
The current is higher than safe zone so the body will receive a dangerous shock.
