Answer:
zero
Explanation:
q = 6.4 nC = 6.4 x 1 0^-9 C
d = 16 cm = 0.16 m
r = 16 / 2 = 8 cm = 0.08 m
Electric field at P due to the charge placed at A
Ea = k q / r^2
Ea = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards right
Ea = 9000 Towards right
Electric field at P due to the charge placed at B
Eb = k q / r^2
Eb = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards left
Eb = 9000 Towards left
The magnitude of electric field is same but teh direction is opposite, so the resultant electric field at P is zero.
I recently received the following question regarding EMP effects: “What will happen to vehicles with electronic ignitions, a Chevy with an ignition module, but they are not hooked to a battery, with no path for electricity to follow, can it do damage? There may not be power on the grid, but what about a generator? Can a drill that was not plugged in still be able to run?”
The short answer is, maybe.
the answer is a!! its pretty simple I just read the graph.
Answer:
8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2
Explanation:
Exact number is 8.9875517923(14)×109 kg⋅m3⋅s−2⋅C−2
Answer: have "cis C=C double bonds" and "liquid" at room temperature.
Explanation:
The unsaturated fatty acids have one or more C=C double bonds in the cis formation. Thus, this results in the molecules not been as stable as the saturated fats. They have weaker intermolecular bonds thus resulting in lower melting point . The consequently results in it being liquid at room temperature.
