A transfer by work from the system, an elevation decrease of 50 m, and an increase in velocity from 15 m/s to 30 m/s. The specific internal energy, u, decreases by 5 kJ/kg and the acceleration of gravity is constant at 9.7 m/s². Determine the satellite’s heat transfer in the process. magnitude is about 50 kJ.

-50 kJ.

Explanation:

W = 0.147 * 10

= 1.47 kJ

ΔU = -5 * 10

= - 50 kJ

Potential energy, P = m*g*h

= 50*9.7*(-50)

= -4850 J

= -4.85 kJ

Kinetic energy, K = 1/2 * m * (vf^2 - vi^2)

= 1/2 * 10 * (30^2 - 15^2)

= 3375 J

= 3.375 kJ

ΔE = Ein - Eout

Q - W = ΔK + ΔP + ΔU

Q = 1.47 + 3.375 - 4.85 - 50

= -50.005 kJ

= -50 kJ.

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.

increasing transnational kinetic energy. because the transnational kinetic energy formula consists of only two variables, mass and velocity, increasing one of those properties is the only way to increase an object's transnational kinetic energy. increases to mass and velocity, however, do not have the same impact.