Calculating Potential Energy on a 1.9° Incline

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To calculate the potential energy (PE1) of a glider on a 1.9° incline, the formula PE1 = mgh is used, where m is the mass, g is the acceleration due to gravity, and h is the height. The glider weighs 0.190 kg, and the height is given as 3.6 cm, which should be converted to meters for proper unit consistency. The calculation performed was PE1 = 0.190 x 9.8 x 0.036, which is correct if the height is adjusted to meters. Ensuring unit accuracy is crucial for obtaining the correct potential energy value. The approach is fundamentally sound, but attention to unit conversion is necessary.
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In a lab we set up an air track and gave it a 1.9 degree incline. we then slid a glider down the track. the glider passed through two photogates, which calculated two velicities.
the glider weighed .190kg
the first velocity V1 was 0.524m/s
the second velocity V2 was 1.06m/s
h=3.6cm

Now i am supposed to calculate potential energy PE1 which is given by the equation
PE1=mgh,
is was wonder if i did this right, i did PE1=0.190x9.8x3.6
if not, what is the approach?
 
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the equation is basically correct. Check the units - (height)
 
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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