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Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
the angle of incidence θ is 45.56 º
Explanation:
Given data
strikes the mirror before wall x = 30.7 cm
reflected ray strikes the wall y = 30.1 cm
to find out
the angle of incidence θ
solution
let us consider ray is strike at angle θ so after strike on surface ray strike to wall at angle 90 - θ
we will apply here right angle triangle rule that is
tan( 90 - θ) = y /x
tan( 90 - θ) = 30.1 / 30.7
90 - θ = tan^-1 (30.1/30.7)
90 - θ = 44.4345
θ = 45.56 º
the angle of incidence θ is 45.56 º