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
<span>I would say greater than because as you do deeper, the pressure strengthens. If you were in a 10 ft deep pool and you dive all the way to the bottom, the ears usually pop. That's because of the pressure. Whereas if you were to go five feet, your ears wouldn't. It depends on the age of the person.
Hope this helps.</span>
Answer:
The Moon revolves or moves around the Earth in a path called its orbit and rotates, or spins, in space.
Explanation: The Moon's movements cause the phases of the Moon and the Earth's ocean tides.
I'll tell you how I look at this, although I may be missing something important.
Position = x(t) = 0.5 sin(pt + p/3)
Speed = position' = x'(t) = 0.5 p cos(pt + p/3)
Acceleration = speed' = position ' ' = x ' '(t) = -0.5 p² sin(pt + p/3)
At (t = 1.0),
x ' '(t) = -0.5 p² sin( 4/3 p )
In order to evaluate this, don't I still have to know what 'p' is ? ?
I don't think it can be evaluated with the information given in the question.
Answer:
C sedimentary rocks is the answer.
Explanation:
