Answer:
3.6ft
Explanation:
Using= 2*π*sqrt(L/32)
To solve for L, first move 2*n over:
T/(2*π) = sqrt(L/32)
Next,eliminate the square root by squaring both sides
(T/(2*π))2 = L/32
or
T2/(4π2) = L/32
Lastly, multiply both sides by 32 to yield:
32T2/(4π2) = L
and simplify:
8T²/π²= L
Hence, L(T) = 8T²/π²
But T = 2.1
Pi= 3.14
8(2.1)²/3.14²
35.28/9.85
= 3.6feet
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The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
Answer: 1608.39 J
Explanation: Given that the
mass M = 42kg
U = 11.5m/s
V = 3.33m/s
how much work did friction do
Work done = Force × distance
Work done = Ma × distance
But acceleration a = V/t
Work done = M × V/t × d
Work done = M × V × d/t
Where d/t = velocity
Therefore,
Work done = M × U × V
Work done = 42 × 11.5 × 3.33
Work done = 1608.39 J
To solve this problem we will apply the linear motion kinematic equations. On these equations we will define the speed as the distance traveled in a space of time, and that speed will be in charge of indicating the reaction rate of the individual. In turn, using the ratio of speed, position and acceleration, we will clear the position and determine the distance necessary for braking.
The relation to express the velocity in terms of position for constant acceleration is as follows

Here,
u = Initial velocity
v= Final velocity
a = Acceleration
= Initial position
s = Final position
PART 1) Calculate the displacement within the reaction time



In this case we can calculate the shortest stopping distance


PART 2)
PART 1) Calculate the displacement within the reaction time



In this case we can calculate the shortest stopping distance


While a person without alcohol would cost 517ft to slow down, under alcoholic substances that distance would be 616ft