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!!
337493603.8m/s²
Explanation:
Radius of the earth = 6.38 x 10¹⁶m
time = 24hr (86400s)
Unknown:
Centripetal acceleration = ?
Solution:
The centripetal acceleration is directed inward to keep the body from falling off the surface of the earth.
centripetal acceleration = 
where v is the velocity and r is the radius
also;
v = wr
where w is the angular velocity
substituting in the equation for centripetal acceleration gives;
a = w²r
also w = 
therefore;
a = 
a = 
a = 337493603.8m/s²
learn more:
Acceleration brainly.com/question/3820012
#learnwithBrainly
Answer:
Utilization, effects
Explanation:
The conductors that carry the current to electrical devices and utilization equipment are the heart of all electrical systems. There are associated effects whenever current flows through a conductor.
I’m pretty sure u have it right
