The frictional force required is 9000 N
Explanation:
In order to keep the car in the turn in circular motion without sliding, the frictional force must provide the centripetal force necessary for the circular motion.
Therefore, we can write:
where the term on the left is the frictional force while the term on the right is the centripetal force, and where:
m is the mass of the car
v is its speed
r is the radius of the curve
For the car in this turn, we have
m = 1000 kg
v = 30 m/s
(since the diameter is 0.20 km, the radius is half that value)
And substituting, we find
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Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
Answer:
Saturated zone is area below the water table in which the soil is completely saturated with groundwater.
Explanation:
The saturated zone lies below the ground. It is mainly the lower zone of rock along with the water table where pore spaces are completely filled with water. Even the saturated zone is sometimes separated into 2 subzones: the phreatic zone and the capillary fringe.
The area where pores spaces are not saturated with water is also unsaturated zone. Localized saturated zones can occur within the unsaturated zone. The unsaturated zone lies above the groundwater table.
its period should be the amount it takes to cycle from cycle to cycle so it would be 10 and your frequency would have to be calculated by taking 10 and dividing by 2 since that is how many cycles you have done so your frequency is 5
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