The image of the thin square metal plate showing the force and the lines are missing, so i have attached it.
Answer:
Line B represents the lever arm
Explanation:
From the attached image, we can see 4 different dashed lines between the force F and the moment centroid O.
Now, we want to know which of the dashed lines represents lever arm.
Now, in physics/engineering, the lever arm which is also known as moment arm is defined as the perpendicular distance between the line of action of the acting force and the centre of the moments.
Now, from the image attached, the only dashed line that is perpendicular to the acting Force F and the moment centre O is Line B.
Thus,line B represents the lever arm
C should be it. Correct me if im wrong xd
Answer: 50J
Explanation:
Mechanical energy follows the same principles of kinetic energy and potential energy, it is conserved. So Ei = Ef.
Mechanical energy is the sum of ALL energy's. There is no friction, so its just kinetic plus potential.
37.5 + 12.5 = 50J
Since the particle has not touched the ground, it has not transferred any energy to the ground yet, therefore the mechanical energy must still be 50J; mostly in kinetic energy with a very small amount of potential because of the low height relative to the ground.
A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at rest (at x = 900 m). Through the first 1/4 of that distance, its acceleration is +6.25 m/s2. Through the next 3/4 of that distance, its acceleration is -2.08 m/s2. What are (a) its travel time through the 900 m and (b) its maximum speed?
<span>Solve for the time at the 1/4 mark. That's 225 m. How? d = (1/2)at^2 ( initial velocity zero). Thus 225 = (1/2) 6.25 t^2. t^2 = ( 225 * 2 ) / 6.25. t = 8.5 sec. </span>
<span>At the other end t^2 = (675 * 2) / 2.08 -- we reversed the sign and ran time backwards. t = 25.5 sec. </span>
<span>So total time is 8.5 + 25.5 or 34 sec. </span>
<span>Since zero initial velocity: v^2 = 2 a d. Here, v^2 = 2 * 6.25 * 225. v = 53 m/s. That's the fastest speed since braking then occurs.</span>
