A spring stretches by 0.014 m when a 3.7-kg object is suspended from its end. How much mass should be attached to this spring so
that its frequency of vibration is f
1 answer:
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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:
(a) Acceleration of the bag will be a=16.214m/sec^2
(B) Weight of the bag will be 137.2 N
Explanation:
We have given mass of the bag m = 14 kg
Force with which bag is lifted = 227 N
(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration
So
(b) Acceleration due to gravity
We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity
So weight
So weight of the bag will be 137.2 N