Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
The forces are applied in opposite direction.
So the net force will be the difference of both forces.
net force =5-3=2N
This force will be in direction of 5N(bigger force) means in direction which Andy is pushing.
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I think it’s B I could be wrong but I tried lol
Answer:
74.52s
Explanation:
The solution is shown in the picture below