Part A
F = 50 N is the force applied along the purple vector
r = 1.5 is the radius (half the diameter 3)
theta = 110 is the angle in which the force vector is applied
Use this formula to plug in the values to find the torque T
T = F*r*sin(theta)
T = 50*1.5*sin(110)
T = 70.4769
<h3>Answer: The torque applied is approximately 70.4769 Newton-meters</h3>
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Part B
Refer back to the formula in part A. If theta is the variable, then T maxes out when theta = 90 degrees, because sin(theta) is maxed out at 1 here. If theta = 90, then T = F*r. The torque is maxed out when the force vector is perpendicular to the original position vector, this way you get the most push leading to the highest twisting or turning force possible.
<h3>Answer: 90 degrees</h3>
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Part C
Use the values from part A, but make theta = 90 so that the torque T is maxed out. So we would get the following
T = F*r*sin(theta)
T = 50*1.5*sin(90)
T = 50*1.5*1
T = 75
<h3>Answer: The max torque possible is 75 Newton-meters</h3>
Answer:
A: No, because the bike order does not meet the restrictions of 4c + 6a ≤ 120 and 4c + 4a ≤ 100
Step-by-step explanation:
c = child bike
a = adult bike
Child bike requirement:
4 hours build time
4 hours test time
Adult bike requirement:
6 hours build time
4 hours test time
Available Building time per week ≤ 120 hours
Available Testing time per week ≤ 100 hours
Therefore, company's work schedule is;
Building:
4c + 6a ≤ 120
Testing:
4c + 4a ≤ 100
Combined:
4c + 6a ≤ 120 and 4c + 4a ≤ 100
Building 20 Child bike and 6 adult bike in a week:
4(20) + 6(6) ≤ 120 and 4(20) + 4(6) ≤ 100
80 + 36 ≤ 120 - - - - 80 + 24 > 100
Therefore Building 20 Child bike and 6 adult bike in a week is not feasible.
Graph the absolute value using the vertex and a few selected points.
x: -2, -1, 0, 1, 2
y: 4, 2, 0, 2, 4
Hope this helps! :)
Answer:
x8 y6
Step-by-step explanation: