& this one...
4. A person drops 740 g rock originally held 6.5 ft above the ground. What is its velocity the moment before it hits the gound. (Assume rock doesn't gain/lose energy)... (Hint: You will need to calculate the P.E of the rock before it's dropped!
Last one! Thankyou for the help thus far...
calculate the KE in kJ of 9.100 x 10^5 atoms of silver moving at 1800. km/hr. ... the atoms part is what I'm confused about.
3 answers
it's a fraction of a mole
... 9.1E5 / 6.02E23
multiply the fraction by the molar mass
... 9.1E5 / 6.02E23
multiply the fraction by the molar mass
m g h = 1/2 m v^2 ... v = √(2 g h)
1 answer