Answer:
No, the farmer is not able to move the mule.
Explanation:
Mass =100 kg
Force=F=800 N
The coefficient between the mule and the ground=

Static friction force,f=
Normal force=N=mg
Static friction force,f=
Using 
F<f
Static friction force is greater than applied force.
Therefore , the farmer is not able to move the mule.
I think you would use F = ma
F = 65*10
F = 650N
(The 10m/s is from acceleration due to gravity)
Answer:
E1 = 2996.667N/C E2 = 11237.5N/C
Explanation:
E1 = kQ1/r^2
=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2
= 2996.667N/C
E2 = kQ2/r^2
= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2
= 11237.5N/C
The direction are towards the point a
The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s
<h3>Conservation of Linear Momentum</h3>
Given Data
- Mass of cart one M1 = 150kg
- Initial Velocity U1 = 8m/s
Mass of cart two M2 = 150kg
Velocity U2 = 6m/s
Applying the principle of conservation of linear momentum we have
M1U1+M2U2 = M1V1+ M2V2
a. what is the speed of cart b after collision
substituting our given data we have
150*8+ 150*6 = 150*5+150*V2
1200 + 900 = 1200+ 150V2
2100 - 1200 = 150V2
900 = 150V2
Divide both sides by 150
V2 = 900/150
V2 = 6m/s
b. what is the total momentum of the system before and after collision
Total Momentum in the system is
Total momentum = Momentum before Impact+ Momentum after Impact
Total momentum = M1U1+M2U2 + M1V1+ M2V2
Total momentum = 1200 + 900 + 1200+ 900
Total momentum = 4200 kg m/s
Learn more about Conservation of Linear Momentum here:
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