We have that for the Questions
a the gravitational potential energy at a relative to B
b. the particle's kinetic energy at B
c the particle's speed at B
d. the potential energy and kinetic energy at C
it can be said that
a)P.E=0.63210Js
b)K.E=0.63210J
c)v=2.420m/s
d)K.E=0.5mv^2
P.E=mgh
From the question we are told
215 g particle is released from rest at point A inside a smooth hemispherical bowl of radius 30.0 cm, as shown at right
a)
Generally the equation for the gravitational potential energy is mathematically given as
P.E=mgh
P.E=0.215x9.8 x 0.3
P.E=0.63210J
b)
Generally the equation for the kinetic energy is mathematically given as
K.E=0.5mv^2
Where
P.E=K.E at point B
Therefore
K.E=0.63J
c)
Generally the equation for the speed is mathematically given as
[tex]v=\sqrt{\frac{0.63}{0.5m}}[/tex]
v=2.420m/s
d)
The potential energy and kinetic energy at C
Will be given by the equation
K.E=0.5mv^2
P.E=mgh
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