<span> For any body to move in a circle it requires the centripetal force (mv^2)/r.
In this case a ball is moving in a vertical circle swung by a mass less cord.
At the top of its arc if we draw its free body diagram and equate the forces in radial
direction to the centripetal force we get it as T +mg =(mv^2)/r
T is tension in cord
m is mass of ball
r is length of cord (radius of the vertical circle)
To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So
minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s
In the second case the speed of ball at top = (2*3.285) =6.57 m/s
Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom
we get velocity at bottom as 9.3m/s.
Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force
T-mg=(mv^2)/r
We get tension in cord T=13.27 N</span>
        
                    
             
        
        
        
Answer:
a)       a = 3.72 m / s², b)    a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
              v = v₀ + at
as part of rest the v₀ = 0
              a = v / t
Let's reduce the magnitudes to the SI system
               v = 115 km / h (1000 m / 1km) (1h / 3600s)
               v = 31.94 m / s
               v₂ = 60 km / h = 16.66 m / s
l
et's calculate
              a = 31.94 / 8.58
              a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
             I = Δp
             F Δt = m v_f - m v₀
             F = 
             F = m [16.66 - 31.94] / 0.815
             F = m (-18.75)
Having the force let's use Newton's second law
             F = m a
             -18.75 m = m a
              a = -18.75 m / s²
 
        
             
        
        
        
The beginning development of a
star is marked by a supernova explosion, with the gases present in the nebula
being forced to scatter. As the star shrinks, radiation of the surface increases
and create pressure on the outside shell to push it away and forming a
planetary nebula or white dwarf.
 
        
             
        
        
        
Answer:
B
Explanation:
Gravitational Energy is the energy of position or place. A rock resting at the top of a hill contains gravitational Potential energy. Hydropower, such as water in a reservoir behind a dam, is an example of gravitational potential energy.