Answer:
option B
Explanation:
given,
diameter of the rotating space = 2 Km
Force exerted at the edge of the space = 1 g
force experienced at the half way = ?
As the object is rotating in the circular part
Force is equal to centripetal acceleration.
at the edge
g = ω² r
ω is the angular velocity of the particle
r is the radius.
now, acceleration at the half way
g' = ω² r'
People at the halfway experience g/2
hence, the correct answer is option B
since centripetal acceleration is always towards the center of the circle
so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle
also we know that
so the coordinates of the center will be
so the coordinate is (3.20 m, 4.04 m)
Answer:
375 m.
Explanation:
From the question,
Work done by the frictional force = Kinetic energy of the object
F×d = 1/2m(v²-u²)..................... Equation 1
Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.
Make d the subject of the equation.
d = 1/2m(v²-u²)/F.................. Equation 2
Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.
Note: If is negative because it tends to oppose the motion of the object.
Substitute into equation 2
d = 1/2(60)(0²-25²)/-50
d = 30(-625)/-50
d = -18750/-50
d = 375 m.
Hence the it will slide before coming to rest = 375 m
