Conservation of energy of an ice flake

AI Thread Summary
The problem involves calculating the speed of a 1.90 g ice flake released from the edge of a hemispherical bowl with a radius of 28.0 cm. The conservation of energy principle is applied, equating kinetic energy and gravitational potential energy. The initial calculations yielded a velocity of 23.44, but the radius needed conversion from centimeters to meters. After correcting for units, the final speed calculated is 2.344 m/s, which is confirmed as correct. The discussion emphasizes the importance of unit consistency in physics calculations.
lauriecherie
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Homework Statement



A 1.90 g ice flake is released from the ege of a hemispherical bwon whose radius r is 28.0 cm. The flake bowl contact is frictionless. What is the speed of the flake when it reaches the bottom of the bowl.

Homework Equations


I took KINETIC ENERGY = GRAVITATIONAL POTENTION ENERGY. since KINETIC ENERGY = .5*M*V^2, and GRAVITATIONAL POTENTIAL ENERGY = MGH, i canclled the masses and solved for V. That gave me VELOCITY = SQUARE ROOT OF (2*G*R).


The Attempt at a Solution


I got 23.44 for my answer but after looking over my calculations I noticed that the radius is given in cm. Need I convert it to meters? If I do my answer comes out as 2.344 m/s. Is this correct now?
 
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Nevermind. It is correct.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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