Eight and I don’t know what else to say but for sure 8
Using the equation v(avg)=distance/time
and the equation v=v(original)+a(t)
solve for acceleration
2600=0+a(12)
a=216.66666 m/s^2
Then, you use the equation
v^2=v(original)+2a*(change in x)
2600^2=2(216.666666)*change in x
6760000/2/216.666666 = 15600 meters which is the length of the race
Then using v(avg)=x/t
15600/12= 1300 m/s
Answer:
7 meters per second or 7 m/s
Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²
Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m
(a) Angular frequency w= 2πf
w= 2π × 760 = 4775.2Hz
(b) Maximum speed v is given as the product of angular frequency and maximum displacement
V=wx
V= 4775.2 × 0.00085
V= 4.06m/s
(c) The maximum acceleration a
= w²x
= (4775.2)² × (0.00085)
a= 19382.15m/s².
