Calculate the rate, in watts, at which this heat transfer through radiation occurs (almost entirely in the infrared) from 1.0 m2 to the atmosphere at night. assume the emissivity is 0.90, the temperature of the surface of the earth is 15°c, and that of outer space is 2.7 k. q/δt = 376.4|

Explanation:

As we know that heat transfer is given as

so we will have

now we will have

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft