The net force on the charge at the origin is -1.2×10-8
<u>Explanation:</u>
Solving the problem,
- Draw the x-axis and the locations of the given three charges.
- The forces applied on the charge at the origin and there are two of them, and since all the changes are positive, all the forces are repulsive.
- we have the formula, F = kq1Q/r².
- F1 = kq1Q/r²1 = (9.0*109Nm²/C²)(2.2*10^-9C)(3.5*10^-9C)/(1.5m)² = 31*10-9N = 3.1*10-8N. F1 points to the right (+x direction).
- F2 = kq2Q/r²2 = (9.0*109Nm²/C²)(5.4*10^-9C)(3.5*10^-9C)/(2.0m)² = 43*10^-9N = 4.3*10^-8N.
- F2 points to the left (-x direction).
- To find the net force we have to subtract the force F1 and force F2 .
- The net force is F(origin) = F1 - F2 = -1.2×10-8N.
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Answer:
W
Explanation:
= Temperature of the room = 22.0 °C = 22 + 273 = 295 K
= Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
= Surface area = 1.50 m²
= emissivity = 0.97
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as
W
Inversely proportional
I hope this helps you!
Answer:
A. 70 m OS the correct one
