<span>Since youc oncetrate all your force directly towards the moment arm it means that you push it at an angle of your force is directed to the left or the right and I bet that it must be 90</span> degrees to the bar. Obviuosly, if you are about to push it you will do it straight up but not in a zig zag way. In other words, it should be perpendicular to the arm because the<span> torque can be produced only if force is applied at a constant index (90).
Hope that helps! Regards.</span>
Answer:
m g sin theta = force of object along incline due to gravity
N μ = frictional of incline on object where N is the normal force
N = m g cos theta force perpendicular to incline
m g sin theta = N μ = μ m g cos theta
μ = tan theta = tan 38 = .78
Answer:
The distance traveled during its acceleration, d = 214.38 m
Explanation:
Given,
The object's acceleration, a = -6.8 m/s²
The initial speed of the object, u = 54 m/s
The final speed of the object, v = 0
The acceleration of the object is given by the formula,
a = (v - u) / t m/s²
∴ t = (v - u) / a
= (0 - 54) / (-6.8)
= 7.94 s
The average velocity of the object,
V = (54 + 0)/2
= 27 m/s
The displacement of the object,
d = V x t meter
= 27 x 7.94
= 214.38 m
Hence, the distance the object traveled during that acceleration is, a = 214.38 m
There are many people in the city and so much traffic that it pollutes the air and makes it warmer where the country dosent have much traffic so there for its cooler
20.9m = 1s
286.33m = 13.7s
To answer this you would multiply both sides by the amount of seconds she ran. The answer however is that she ran as far as 286.33m.
