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:
F=ma is the relationship where, F is force, m is mass and a is acceleration.
Newton's second law states that the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.
If we apply force to a toy car then It will accelerate.
This is how Newton's second law of motion is verified.
A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>
Answer:
The answer is "
".
Explanation:
Formula for calculating the mass in He:


Formula for calculating the mass in
:


by using the temperature balancing the equation:


Sum the vector components:
Dx = 225* Cos(180)+ 78*Cos(225)= -280.154 km
Dy = 225* Sin(180)+ 78*Sin(225) = -55.154 km
displacement:
Sqrt(Dx^2+Dy^2) = 285.532 km
Arctan(Dy/Dx) = 191.137degrees CCW
OR:
11.137 degrees South of West