22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Blue and yellow make green
Average speed of the runner is the rate at which the runner covers the total distance. Average speed of the runner in the race is given by,
Average speed = 
Where
Total distance = Distance covered by the runner from initial to final position
Total time = time taken by the runner to cover entire distance
Instantaneous speed is the speed of the runner at the particular moment in the given time. Instantaneous speed is given by,
Instantaneous speed = 
x = position of the runner at time t
t = time taken to cover distance x
Hence, Average speed and instantaneous speed are different for a runner running in the race.
Answer:
#_electrons = 2 10¹⁰ electrons
Explanation:
For this exercise we can use a direct rule of three proportions rule. If an electron has a charge of 1.6 10⁻¹⁹ C how many electrons have a charge of 3.2 10⁻⁹ C
#_electrons = 3.2 10⁻⁹ (
)
#_electrons = 2 10¹⁰ electrons