Answer:
Time of race = 10.18 s
Explanation:
She keeps this acceleration for 17 m and then maintains the velocity for the remainder of the 100-m dash
Time to travel 17 m can be calculated
s = ut + 0.5at²
17 = 0 x t + 0.5 x 3.89 x t²
t = 2.96 s
Velocity after 2.96 seconds
v = 3.89 x 2.96 = 11.50 m/s
Remaining distance = 100 - 17 = 83 m
Time required to cover 83 m with a speed of 11.50 m/s
![t=\frac{83}{11.50}=7.22s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B83%7D%7B11.50%7D%3D7.22s)
Time of race = 2.96 + 7.22 = 10.18 s
Answer:
e = k q / r² then,
e = 9×10^9 * 1×10^-6 / 50² = 3.6 N/C
hope this helps ❤.
Answer:
Explanation:
Given
Motorcyclist speed=12 m/s
maximum acceleration![=-6 m/s^2](https://tex.z-dn.net/?f=%3D-6%20m%2Fs%5E2)
distance=39 m
Let x be the distance traveled by motorist in his reaction time
therefore remaining 39-x will be traveled with
acceleration
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
s=39-x
v=0
u=12 m/s
![0-12^2=2\left ( -6\right )\left ( 39-x\right )](https://tex.z-dn.net/?f=0-12%5E2%3D2%5Cleft%20%28%20-6%5Cright%20%29%5Cleft%20%28%2039-x%5Cright%20%29)
x=27 m
Therefore he traveled 27 m in his reaction time
![27=12\times t](https://tex.z-dn.net/?f=27%3D12%5Ctimes%20t)
t=2.25 s
(b)If his reaction time is 2.56 sec
then distance traveled in his reaction time
![x_0=12\times 2.56=30.72 m](https://tex.z-dn.net/?f=x_0%3D12%5Ctimes%202.56%3D30.72%20m)
Remaining distance 39-30.72=8.28 m
therefore its velocity when it reaches the deer
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
![v^2=12^2+2\times \left ( -6\right )\times 8.28=44.64](https://tex.z-dn.net/?f=v%5E2%3D12%5E2%2B2%5Ctimes%20%5Cleft%20%28%20-6%5Cright%20%29%5Ctimes%208.28%3D44.64)
v=6.681 m/s