The object would have a centripetal acceleration <em>a</em> of
<em>a</em> = (12 m/s)² / (0.7 m) ≈ 205.714 m/s²
so that the required tension in the string would be
<em>T</em> = (0.5 kg) <em>a</em> ≈ 102.857 N ≈ 100 N
(rounding to 1 significant digit)
(C). Remember gravity provides an acceleration of 9.81m/s^2, so the y component of velocity initial is zero because it isn’t already falling, and we have the height, so basically we use the kinematic equation vf^2=vi^2+2ad, substitute given values and you get vf^2=2(9.81)(65) which is 1275, when you take the square root you get 35.7m/s for final velocity
(B). Then you use vf=vi+at to get the equation 35.7=(9.81)t, when you divide out you get 3.64s for time t
(A). Finally, since we assume that there is no acceleration or deceleration horizonatally, we just multiply the time taken for it to hit the ground and the initial speed ((3.64)(35.7)) to get 129.96, with significant figures I would round that to 130 metres.
**this is in the order that I felt was easiest to answer**
Answer:
fit of the continents, paleoclimate indicators, truncated geologic features, and fossils.
Explanation:
