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Answer:
A- mass and type of material
B- type of material
C- Temperature
Explanation:
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Here we must use the equation of motion
v^2 = u^2 + 2as; where v is final velocity, u is initial velocity, a is the acceleratoin and is the distance travelled.
We select this one because the time of collision is unknown to us.
We know the truck stopped so its final velocity is 0; thus v = 0.
Converting the initial velocity to SI units, we get 3.89 m/s.
The distance traveled, s, is 0.062 meters.
Inserting all of these values into the equation,
0 = (3.89)^2 + 2(a)(0.062)
and solving for a, we get a to be
-122.0 ms^(-2)
The negative sign indicates the acceleration is in the opposite direction to the initial motion, which means the truck decelerated. This is consistent with the given condition.
Explanation:
Let us assume that the maximum allowable horizontal distance be represented by "d".
Therefore, torque equation about A will be as follows.
d = ![\frac{[2 \times 75 \times (0.7+0.15+0.15) - 60 \times 0.15 - 252 \times 0.15 \times 2]}{252}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2%20%5Ctimes%2075%20%5Ctimes%20%280.7%2B0.15%2B0.15%29%20-%2060%20%5Ctimes%200.15%20-%20252%20%5Ctimes%200.15%20%5Ctimes%202%5D%7D%7B252%7D)
d = 0.409 m
Thus, we can conclude that the maximum allowable horizontal distance from the axle A of the wheelbarrow to the center of gravity of the second bag if she can hold only 75 N with each arm is 0.409 m.
Answer:
from the formula of work done =Force *Distance 500*100=50000
