Answer:
1: an object will stay still until acted on by another force
2: acceleration of an object depends on the mass of the object and the force applied
3: for every action there is an equal and opposite reaction
Explanation:
physical quantity is any physical property that can be qualified that,is, be measured using numbers e.g mass, amount of substance,time and length
Answer:
0.001 s
Explanation:
The force applied on an object is equal to the rate of change of momentum of the object:

where
F is the force applied
is the change in momentum
is the time interval
The change in momentum can be written as

where
m is the mass
v is the final velocity
u is the initial velocity
So the original equation can be written as

In this problem:
m = 5 kg is the mass of the fist
u = 9 m/s is the initial velocity
v = 0 is the final velocity
F = -45,000 N is the force applied (negative because its direction is opposite to the motion)
Therefore, we can re-arrange the equation to solve for the time:

When an item is raised, the work is done in opposition to gravity. When an item is worked on, energy is transmitted to it, and it develops gravitational potential energy. If the same thing falls from that height, gravity must do the same amount of effort to bring it back to the Earth's surface.
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