An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object

T_object = 380.35 K

Explanation:

From Stefan–Boltzmann law, the power output is given by the formula:

P = σAT⁴

where;

σ is Stefan-Boltzmann constant

A is area of the radiating surface.

T is temperature of the body

Now, we are told that the power the object emitted is 3 times the power absorbed from the room.

Thus, we have;

P_e = 3P_a

Where P_e is power emitted and P_a is power absorbed.

So, we have;

σA(T_object)⁴ = 3σA (T_room)⁴

σA will cancel out to give;

(T_object)⁴ = 3(T_room)⁴

We are given T_room = 289 K

Thus;

(T_object)⁴ = 3 × 289⁴

(T_object) = ∜(3 × 289⁴)

T_object = 380.35 K

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.