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
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency 
Speed of sound in air 
The definition of wavelength is,

Here,
v = Velocity
f = Frequency
Replacing,


Therefore the wavelength of that tone in air at standard conditions is 0.589m
Answer:
Charge of the alpha particle
Explanation:
Answer:
Speed at which it will reach the ground is given as

Total time for which it will remain in air is given as
t = 6.3 s
Explanation:
As we know that the object is projected upwards with speed


now when it will reach the ground then we have

so we have


so we have

Now speed of the object when it reaches the ground is given as



Answer:
-387883.3 m/s²
0.000286168546055 seconds
Explanation:
t = Time taken
u = Initial velocity = 370 m/s
v = Final velocity = 259 m/s
s = Displacement = 9 cm
a = Acceleration

The acceleration of the bullet is -387883.3 m/s²

The time taken is 0.000286168546055 seconds