The magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
<h3>
Work done by the force experienced  by the object</h3>
The magnitude of the work done by force experience by the object is calculated as follows;
W = f.d
where;
- F is the applied force (2xyi + 3yj), where x and y are in meters
- d is the displacement of the object = (a, b)
The work done by the force is determined from the dot product of the force and the displacement of the object.
F = (2xyi + 3yj).(a + b)
W = (2abi + 3bj).(ai + bj)
W = (2a²b + 3b²)J
Thus, the magnitude of the work done by force experience by the object is (2a²b + 3b²)J.
The complete question is below:
The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis.
How much work does the force do?
Learn more about work done here: brainly.com/question/8119756
 
        
             
        
        
        
Answer:
True 
Explanation:
Electronegativity difference of less than 0.4 characterized covalent bonds. For two atoms with an electronegativity difference of between 0.4 and 2.0, a polar covalent bond is formed-one that is neither ionic nor totally covalent. 
 
        
                    
             
        
        
        
The correct answer is the third, It reflects the green light waves and absorbs most of the rest.
        
                    
             
        
        
        
Answer:
a)  y₂ = 49.1 m
,    t = 1.02 s
, b)   y = 49.1 m
, t= 1.02 s
Explanation:
a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero
              ² =
² =  ² - 2 g (y –yo)
² - 2 g (y –yo)
The origin of the coordinate system is on the floor and the ball is thrown from a height
            y-yo = ![v_{oy}² /2 g
            y- 0 = 10.0²/2 9.8
            y - 0 = 5.10 m
            
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
             y₂ = 5.1 + 44
             y₂ = 49.1 m
Let's use the other equation to find the time
              [tex]v_{y}](https://tex.z-dn.net/?f=v_%7Boy%7D%C2%B2%20%2F2%20g%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y-%200%20%3D%2010.0%C2%B2%2F2%209.8%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20y%20-%200%20%3D%205.10%20m%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%0A%3C%2Fp%3E%3Cp%3EThe%20height%20from%20the%20ground%20is%20the%20height%20that%20rises%20from%20the%20reference%20system%20plus%20the%20depth%20of%20the%20ground%20from%20the%20reference%20system%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%205.1%20%2B%2044%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20y%E2%82%82%20%3D%2049.1%20m%0A%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ELet%27s%20use%20the%20other%20equation%20to%20find%20the%20time%0A%3C%2Fp%3E%3Cp%3E%20%20%20%20%20%20%20%20%20%20%20%20%20%20%5Btex%5Dv_%7By%7D) =
 =  - g t
 - g t
               t =  / g
 / g
               t = 10 / 9.8
               t = 1.02 s
b) the maximum height
             y- 44.0 =  ² / 2 g
² / 2 g
             y - 44.0 = 5.1
             y = 5.1 +44.0
             y = 49.1 m
The time is the same because it does not depend on the initial height
               t = 1.02 s