The dimension of force, F is ML/T².
<h3>What is force?</h3>
Force is a push o push agent which causes a change in the state of rest or motion of an object.
From a fundamental law of motion states that the acceleration of an object is directly proportional to the resultant force exerted on the object and inversely proportional to its mass.
Mathematically; acceleration ∝ F/m
F = ma
dimension of Mass = M
dimension of acceleration = L/T²
dimension of force, F = ML/T²
In conclusion, the dimension of force is obtained from the dimensions of mass and acceleration.
Learn more about dimension of force at: brainly.com/question/28243574
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Your answer is friction. Friction between to object slows the object down.
Answer:
sorry mate i dont know
Explanation:
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Heat is added to the mass as follows:
Q_-14-0 = mC_iΔT = 39*2.06*(0--14) = 39*2.06*14 = 1124.76 J
Q_0 = mC_f = 39*334 = 13026 J
Q_0-100 = mC_wΔT = 39*4.18*100 = 16302 J
Q_100 = mC_v = 39*2230 = 86970 J
Q_100-108 = mC_sΔT = 39*2.03*(108-100) = 39*2.03*8 = 633.36J
Q = Summation of all the heats added = 1124.76+13026+16302+86970+633.36 = 118056.12 J ≈ 118.06 kJ
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