What is the magnitude of the force of a +25 microCoulomb charge exerts on a +2.5 miliCoulomb charge 28 cm away?

7.175 N

Explanation:

From coulomb's law of electrostatics,

F = kQq/r² Equation 1

Where F = magnitude of the force, M = first Charge, q = second Charge, r = distance between the charges and k = coulomb's constant

Given: Q = +25 micro-Coulomb = + 25×10⁻⁶ C, q = +2.5 micro-coulomb = +2.5×10⁻⁶ C, r = 28 cm = 0.28 m.

Constant: k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = 9×10⁹(25×10⁻⁶)(2.5×10⁻⁶)/ 0.28²

F = 562.5×10⁻³/0.0784

F = 7.175 N

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.