<h3>

</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a = 
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
52m/s is the answer because before it was gonna accelerate it was 52 m/s
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
D is your answer hope this helps
Answer:
the higher the ramp the less distance it will travel