Answer:
The answer would be 0.04ohms.
Explanation:
Hopefully this helps
Answer:
The rock will reach 9 m from the ground at eaxactly 5.06 s after it was initially thrown upwards.
Explanation:
We will use the equations of motion for this.
u = initial velocity of the rock = 22 m/s
g = acceleration due to gravity = -9.8 m/s²
y = vertical position of the rock at a time t = 9 m
y₀ = initial height of the rock = 25 m
t = time it takes for the rock to reach height of 9 m.
(y-y₀) = ut + 0.5gt²
(9 - 25) = 22t + 0.5(-9.8)t²
- 14 = 22t - 4.9t²
4.9t² - 22t - 14 = 0
solving this quadratic equation,
t = 5.055 s or - 0.565 s
Since time cannot be negative,
t = 5.055 s = 5.06 s
Hope this Helps!!!
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
I believe it is noise pollution
2.27 mps repeating.
This is the last question ill ever answer here. Thanks for being the last.
