Answer:
The mother has to sit 2.17 ft from the center on the other side of the seesaw.
Explanation:
We are trying to find the sum of torques given by the weights of mother and daughter to be zero.
If the torque of the daughter on one side of the pivoting point is given by:
5.5 ft x 63.5 lb x g = 349.25 g ft lb
we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:
d x 160.9 lb x g = 349.25 g ft lb
d = 2.17 ft
Answer:
Explanation:
For the cat to stay in place on the merry go round without sliding the magnitude of maximum static friction must be equal to magnitude of centripetal force
Where the r is the radius of merry-go-round and v is the tangential speed
but
So we have
Substitute the given values
So
-- She went up for 0.4 sec and down for 0.4 sec.
-- The vertical distance traveled in gravity during ' t ' seconds is
D = (1/2) x (g) x (t)²
= (1/2) (9.8 m/s²) (0.4 sec)²
= (4.9 m/s²) x (0.16 s²)
= 0.784 meter ( B )
