It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
The correct answer is
a) 1, 2, 3
Explanation:
In rolling down an inclined plane, the potential energy is Transferred to both linear and rotational kinetic energy thus
PE = KE or mgh = 1/2×m×v² + 1/2×I×ω²
The transformation equation fom potential to kinetic energy is =
m×g×h = 
= 
= 
=
Therefore the order is with increasing rotational kinetic energy hence
the first is the sphere 1 followed by the disc 2 then the hoop 3
the correct order is a, 1, 2, 3
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
A) lighting an electric lamp as it becomes darker
Wrap around a metal with wire instead of using wire alone.
