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
Endo I think but look it up jus in case
Answer:
Weight = 966 Newton.
Explanation:
Given the following data;
Length = 1.2 m
Width = 2.3 m
Pressure = 350 Pa
To find the weight of the tank;
We know that weight is the force of gravity acting on an object multiplied by its mass.
Weight = mg = force
Hence, we would determine the force using the parameters that were given.
But we would first determine the area of the rectangular tank.
Area of rectangle, A = length * width
A = 1.2 * 2.3
A = 2.76 m²
Mathematically, pressure is given by the formula;
Pressure = force/area
Force = pressure * area
Substituting into the formula, we have;
Force = 2.76 * 350
Force = 966 Newton
Therefore, the weight of the tank is 966 Newton.
Answer:
W = 30 N
Explanation:
Applying the summation of torques about the wedge for equilibrium, taking the clockwise direction as negative. Since the ruler is balanced horizontally about the wedge. Therefore, the summation of all torques acting about the wedge must be equal to zero.
<u>W = 30 N</u>
