Explanation:
Particle moving in a circular path with a constant speed.
Answer:
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Explanation:
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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:
Explanation:
Magnitude of frictional force = μ mg
μ is either static or kinetic friction.
To start the crate moving , static friction is calculated .
a ) To start crate moving , force required = μ mg where μ is coefficient of static friction .
force required =.517 x 56.6 x 9.8 = 286.76 N .
b ) to slide the crate across the dock at a constant speed , force required
= μ mg where μ is coefficient of kinetic friction , where μ is kinetic friction
= .26 x 56.6 x 9.8 = 144.21 N .