(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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Explanation:
It is given that, the position of a particle as as function of time t is given by :

Let v is the velocity of the particle. Velocity of an object is given by :

![v=\dfrac{d[(8t+9)i+(2t^2-8)j+6tk]}{dt}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bd%5B%288t%2B9%29i%2B%282t%5E2-8%29j%2B6tk%5D%7D%7Bdt%7D)

So, the above equation is the velocity vector.
Let a is the acceleration of the particle. Acceleration of an object is given by :

![a=\dfrac{d[8i+4tj+6k]}{dt}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bd%5B8i%2B4tj%2B6k%5D%7D%7Bdt%7D)

At t = 0, 

Hence, this is the required solution.
Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.
In a series circuit, the sum of the voltages consumed by each individual resistance is equal to the source voltage. ... In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component.
Answer:
a) Diffusion coefficient, D = 1.5 in/hr
b) Mean jump frequency, f = 0.0833 Hz
Explanation:
a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:
..........(1)
Where <r> = mean displacement
D = Diffusion coefficient
t = time = 12 hrs
sum of the squares of the distance divided by 100 is 36 in2.
<r>²= 36 in²
Substituting these values into equation (1) above

b) Mean jumping distance, <r> = 0.1 inches
Applying equation (1) again
Where D = 1.5 in/hr


The mean jump frequency, f = 1/t
f = 1/12
f = 0.0833 Hz