Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3
= 5 3 2
x_{f} = 45 m
we substitute

v = 15 m / s
> Non-zero numbers (like 1,2,3,4...) are always significant
> A zero sandwiched between two non-zero numbers is always significant
> Trailing zeros in a decimal (not whole number like million) are always significant.
<span>0,020170 = 2.0170 × 10^-2
5 sig-figs
</span>
Answer:
-30°C
Explanation:
F-32/180 =C-0/100
or, -22-32/180=C/100
or, -54/180*100=C
or, -0.3*100=C
therefore, C= -30
-22°F = -30°C
PLEASE MARK ME AS BRAINLIEST!
v^2 = v0^2 +2ad
v^2 = 22^2 + 2*3.78*45 = 824.2
v= √824.2 = 28.7 m/s