The dependent variable was the time and the independent variable was thecars
Velocity is defined by rate of change in the position
which we can also write as

while acceleration is defined as rate of change in velocity

so acceleration and velocity both are rate of change in position and rate of change in velocity with respect to time respectively
out of all above statement the correct answer must be
<u>Acceleration equals change in velocity divided by time. </u>
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:

Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m

The view factor between two coaxial parallel disks would be


Then the view factor between base to top surface of the cylinder becomes
. From the summation rule


Then the net rate of radiation heat transfer from the disks to the environment is calculated as





Therefore the rate heat radiation is 780.76W
That's false. No medium = no sound.
Average speed = distance / time
Average speed = 90 km / 0.5 hrs
Average speed = 180 km per hour