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
A positive charge and a negative charge held a certain distance apart are released. as they move, the force on each particle increases
The most common charge carriers are the positively charged proton and the negatively charged electron. The movement of any of these charged particles constitutes an electric current
<h3>What is a Charge ?</h3>
When there are more or fewer protons in an atom than electrons, the substance has an electric charge. Protons have a positive charge, while electrons have a negative charge. If a substance has more protons than electrons, it is positively charged; if it has more electrons, it is negatively charged.
- The SI units for charge are ampere-second or coulomb. When one ampere of electric current goes through the conductor for one second, one coulomb of charge passes through it. Charge is denoted by the formula Q = I t.
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32.5 kg of air
Explanation:
To calculate the mass of the air, we use the density formula:
density = mass / volume
mass = density × volume
density of air = 1.3 kg/m³
volume = 5 × 3 × 2 = 25 m³
mass of the air = 1.3 kg/m³ × 25 m³
mass of the air = 32.5 kg
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density
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