So the equation used in this problem is ΔX=V0*T+1/2AT^2 the X is the distance, v0 is initial velocity, T is time, and a is acceleration. So when we plug these values it we get: 108= 0•T+1/2•3•T^2,the 0•t disappears, and the 1/2•3 gets us 1.5, so we have 108=1.5T^2, then we divide 108 by 1.5 which gets us 72=t^2, and we then take the square root and get 8.49=T so the answer is 8.49 seconds.
Answer:
a. you speak clearly and well
Explanation:
When you rub a balloon on a sweater, for example, some electrons come off and end up on the balloon. The fibers have lost electrons giving them a positive charge. The rubber gained electrons giving it a negative charge. ... The positively charged fibers are now attracted to the negatively charged balloon.
The answer is A. subduction
Answer:
(a). The value of angular speed of the merry-go-round 
= - 5.82 × 
(b). The linear speed of the girl after the rock is thrown V = -1.89 × 
Explanation:
Given data
Mass of the girl 
= 50.6 kg
Mass of merry-go-round 
= 827 kg
Radius r = 3.72 m
The speed of the rock relative to the ground 
= 7.82
(a). The angular speed of the merry-go-round is given by
![\omega = - [\frac{m_{r}v_{r} }{r} ] \frac{2}{m_{m} + 2m_{g} }](https://tex.z-dn.net/?f=%5Comega%20%3D%20-%20%5B%5Cfrac%7Bm_%7Br%7Dv_%7Br%7D%20%20%7D%7Br%7D%20%5D%20%5Cfrac%7B2%7D%7Bm_%7Bm%7D%20%2B%202m_%7Bg%7D%20%7D)
Put all the values in above formula
= - 5.82 ×
This is the value of angular speed of the merry-go-round.
(b). The liner speed of the girl is given by
⇒ V = r ×
⇒ V = - 3.72 × 5.82 ×
⇒ V = -1.89 ×
This is the linear speed of the girl after the rock is thrown.
