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Andre45 [30]
3 years ago
13

PLZ HELP ON #22-26!!!! Please explain why and how you got your answer.

Physics
1 answer:
AleksAgata [21]3 years ago
5 0
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!! 

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Answer:

245.45km in a direction 21.45° west of north from city A

Explanation:

Let's place the origin of a coordinate system at city A.

The final position of the airplane is given by:

rf = ra + rb + rc    where ra, rb and rc are the vectors of the relative displacements the airplane has made. If we separate this equation into its x and y coordinates:

rfX = raX+ rbX + rcX = 175*cos(30)-150*sin(20)-190 = -89.75km

rfY = raY + rbY + rcT = 175*sin(30)+150*cos(20) = 228.45km

The module of this position is:

rf = \sqrt{rfX^2+rfY^2} = 245.45km

And the angle measure from the y-axis is:

\alpha =atan(rfX/rfY) = 21.45\°

So the answer is 245.45km in a direction 21.45° west of north from city A

6 0
3 years ago
An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand
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Answer:

(A) Reading will be 65 N

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We have given the balance force = 65 N

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So net acceleration = 9.8-2.4=7.4m/sec^2

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Answer:

The answer

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