Answer:
S = V0 t + 1/2 a t^2
S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)
S = 1500 m + .6 * 90000 m = 55,500 m
Check: V0 = 5 m/s
V2 = V0 + a t = 5 + 1.2 * 300 = 365 m/s
Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s (note uniform motion)
S = 185 * 300 = 55,500 m
We calculated V2 above at 365 m/s the speed after 300 sec
Allied is the right answer, ig.
Answer:
A
Explanation:
the answer would be basicly A
Answer: 4.29 m/s
Explanation:
Given
Depth of the well, s = 8.23 m
Time taken to reach the well, t = 0.93 s
Speed of sound = 343 m/s
To solve this, we would be using one of l the laws of motion.
S = ut + 1/2gt², where
S = depth of the well
u = initial speed of toss
g = acceleration due to gravity
t = time taken to reach the well
We would then have
8.23 = 0.93 u + 1/2 * 9.8 * 0.93²
8.23 = 0.93 u + 4.9 * 0.8649
8.23 = 0.93 u + 4.23801
0.93 u = 8.23 - 4.23801
0.93 u = 3.99199
u = 3.99199 / 0.93
u = 4.29 m/s
Therefore, the initial speed of the coin is 4.29 m/s
