Answer:
44.4m/s^2
Explanation:
Utilize the equation...S = ut + 1/2at^2
where...S = 32m...u = 0m/s....t = 1.20s
32 = (0)(1.20) + 0.5(1.20^2)a
; The acceleration due to gravity is 44.4m/s^2
<span>The partial pressure of A = 1.06 atm and the partial pressure of B = 0.53 atm</span>
Given: Speed of the sports car, v = 85 mph = 37.99 m/s. Radius of curvature, r = 525 m. Let normal weight be denoted as n and apparent weight as a. The apparent weight can be described by:... or... Consequently, this provides the necessary solution.
To counteract a 58 mph crosswind, the western component of the trajectory must be accounted for. Consequently, directing towards the northwest creates a 45-degree angle, aligning with the destination. This triangle's third vertex is located at the destination, with the right angle positioned there. The western aspect of their flight represents the triangle's base, while the vertical side reflects the resultant path, and the hypotenuse indicates the actual distance traveled. Since the 58 mph crosswind was countered by flying in a northwest direction, the distance from the starting point to the destination should equal the westward segment of their journey. The hypotenuse can be determined via the square root of twice the dimension of the identical sides.
c = sqrt (58^2 + 58^2) = sqrt (6728) = 82.02
An alternative method:
c = sqrt (2) * 58 = 1.414 * 58 = 82.02
Thus, they must fly at 82.02 mph.
Response:
83.1946504051 m
Rationale:
u = Starting velocity = 
s = Distance traveled = 
= Incline = 
Friction coefficient
The calculated stopping distance is 83.1946504051 m
