Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m
The complete sentence is:
In a third class lever, the distance from the effort to the fulcrum is SMALLER the distance from the load/resistance to the fulcrum.
In fact, in a third class lever, the fulcrum is on one side of the effort and the load/resistance is on the other side, so the effort is located somewhere between the two of them. This means that the distance effort-fulcrum is smaller than the distance load-fulcrum.
