Answer:
t=L/
Explanation:
<u>solution:</u>
Let E be an observer, and B a second observer traveling with velocity 
as measured by E. If E measures the velocity of an object A as 
then B will measure A velocity as
= -
Applied here,
the walkway (W) and the man (M) are moving relative to Earth (E}, the velocity of the man relative to the moving walkway is
= 
- 
,
The time required for the woman, traveling at constant speed relative to the ground, to travel distance L relative to the ground is
:
t=L/
The answer is 2. Magnetite. Hope I could help you :)
Answer:
A
Explanation:
ewan ang hirap naman nyan
Refer to the diagram shown below.
Let ω = the angular velocity (rad/s) of the wheel.
At the topmost point, the passenger, with mass m, will feel weightless if the passenger's weight matches the centripetal force.
The passenger's weight is mg.
The centripetal force is mω²r.
Therefore
mω²r = mg
ω = √(g/r)
Because g = 9.8 m/s² and r = 25/2 = 12.5 m, therefore
ω = √(9.8/12.5) = 0.8854 rad/s
Because 1 revolution per second is 2π rad/s, therefore
Answer: 8.455 rev/min
It’s to push an object in it’s direction of where it’s leading to.
