Answer:
The speed after being pulled is 2.4123m/s
Explanation:
The work realize by the tension and the friction is equal to the change in the kinetic energy, so:
(1)
Where:

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.
Additionally, the kinetic energy is equal to
, so if the initial velocity
is equal to zero, the initial kinetic energy
is equal to zero.
Then, replacing the values on the equation and solving for
, we get:


So, the speed after being pulled 3.2m is 2.4123 m/s
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Answer:
Explanation:
When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.
This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ = 
Hence for a given w , θ depends upon g or weight .
I think it is False because as the Gad relajases fuel it doesn’t move as much anymore