Answer:
<u>We are given: </u>
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
<u>Solving for 'a'</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
A.the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system I hope this helps
Answer:
R = 2216m and The normal force of the seat on the pilot is 5008N
Explanation:
See attachment below please.
Answer:
The maximum safe speed of the car is 30.82 m/s.
Explanation:
It is given that,
The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :
.........(1)
A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet
Put the value of r in equation (1) as :

v = 30.82 m/s
So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:
d = vt = (22 m/s)(12 s) = 264 m
For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²
Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m
Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.