Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with  decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
 
        
             
        
        
        
Answer:
if the object is not in motion
Explanation:
 
        
                    
             
        
        
        
Answer:
v₂ = 97.4 m / s
Explanation:
Let's write the Bernoulli equation
          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Index 1 is for tank and index 2 for exit
We can calculate the pressure in the tank with the equation
         P = F / A
Where the area of a circle is
        A = π r²
E radius is half the diameter
       r = d / 2
       A = π d² / 4
We replace
     P = F 4 / π d²2
     P₁ = 397 4 /π  0.058²
     P₁ = 1.50 10⁵ Pa
 
The water velocity in the tank is zero because it is at rest (v1 = 0)
The outlet pressure, being open to the atmosphere is P1 = 1.13 105 Pa
Since the pipe is horizontal y₁ = y₂
We replace on the first occasion
    P₁ = P₂ + ½ ρ v₂²
   v₂ = √ (P1-P2) 2 / ρ
   v₂ = √ [(1.50-1.013) 10⁵ 2/1000]
   v₂ = 97.4 m / s
 
        
             
        
        
        
Answer:
The rider experiences weight or does not experience weight depends upon the direction of the elevator in which it moves and its acceleration.   
Explanation:
Consider the mass of a rider as m. So its actual weight is mg and it acts vertically downward. The apparent weight of the rider is depended on the acceleration of the elevator and its direction of movement.
When elevator moves with a constant speed i.e. its acceleration is zero, the apparent weight of the rider is equal to the actual weight. Thus the rider's sensation is normal. 
If the elevator is moving upward with an acceleration a, then the apparent weight of the rider will be more and the rider will experience an increase in weight or the sensation is heavy.
Or when the elevator moves downward with an acceleration a, the apparent weight of the rider is less. And the rider sensation is lighter.
 
        
             
        
        
        
Answer:
2/3V
Explanation:
Given that a ball of mass M and speed V collides with another ball of mass 2M and velocity v/2 . After collision they stick together and we need to find their speed after collision . According to Law of Conservation of Momentum , <em>T</em><em>h</em><em>e</em><em> </em><em>t</em><em>o</em><em>t</em><em>a</em><em>l</em><em> </em><em>m</em><em>o</em><em>m</em><em>e</em><em>n</em><em>t</em><em>u</em><em>m</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>s</em><em>y</em><em>s</em><em>t</em><em>e</em><em>m</em><em> </em><em>b</em><em>e</em><em>f</em><em>o</em><em>r</em><em>e</em><em> </em><em>a</em><em>n</em><em>d</em><em> </em><em>a</em><em>f</em><em>t</em><em>e</em><em>r</em><em> </em><em>c</em><em>o</em><em>l</em><em>l</em><em>i</em><em>s</em><em>i</em><em>o</em><em>n</em><em> </em><em>i</em><em>s</em><em> </em><em>c</em><em>o</em><em>n</em><em>s</em><em>t</em><em>a</em><em>n</em><em>t</em><em> </em><em>.</em><em>T</em><em>h</em><em>a</em><em>t</em><em> </em><em>i</em><em>s</em><em> </em><em>;</em>
 
 
 
 
 
 
 
 
 
 
 
 
<u>H</u><u>e</u><u>n</u><u>c</u><u>e</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>v</u><u>e</u><u>l</u><u>o</u><u>c</u><u>i</u><u>t</u><u>y</u><u> </u><u>a</u><u>f</u><u>t</u><u>e</u><u>r</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>c</u><u>o</u><u>l</u><u>l</u><u>i</u><u>s</u><u>i</u><u>o</u><u>n</u><u> </u><u>i</u><u>a</u><u> </u><u>2</u><u>/</u><u>3</u><u>V</u><u> </u><u>.</u>