Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
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Answer:
Explanation: do y like me yes or no
Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
