Question

$\huge \boxed{hola} \blue{users}$


Please do solve que no 8, 9 & 10 i solve it.but not sure tho.
so please i need a neat & best quality of answer :)
Do fast.....

pls don't spam & don't be silly ans for points.i need best & i'll mark best one's ans as brainliest ans☆☆...

Answer 1

9.The equation for a falling object is h(t)=gt²/2 +vt+c. In this case, h(t)=0, v=0, and c=1. So:

0=-4.9t²+1

4.9t²=1

t²=1/4.9

t=0.4518s

It’s velocity at this time is gt, or -9.8 m/s² * 0.4518s=4.43 m/s
10.We know that the radius of moon s orbit is 3.8 * 108 m

Period of moon s revolution = 27.3 days = 2.4 * 106 seconds

The velocity = ωR = 2 * 22/7 * R / T = 2 * 22/7 * 3.8 * 108 / 2.4 * 106 = 103 m/s

The acceleration is centripetal acceleration = v2/R = (103)2 / 3.8 * 108 m/s2 = 2.7 * 10-3 m/s2

At a given time, the horizontal distance covered is given by vt and the vertical distance covered is ½ at2

X= vt = 103 * 2 * 24 *60 *60 = 1728 * 105 m  (9 digit number in m)

Y = ½ at2 = ½ * 2.7 * 10-3 * (2 * 24 * 60 *60)2 =   4,03,107.84 m (6 digit number)

Thus, the horizontal distance covered is more than the vertical distance covered.

Answer 2

Step-by-step explanation:

Google has been a bit more people searching and how he was a great way of doing it