Recent content by TkoT

my solution: a) F(upward)=Fb +Fw =(1.3+1.5)X9.8 =27.44N total Mass = 2.8kg b) Volume increased = π(0.2/2)^2 x 1.5/100 =4.7x10-4 m^3 T+Fb =mg T=mg-Fb T=2.2x9.8 -1000 x 4.7x10-4 x 9.8 T=17.4N T is the apparent weight of the rock, so the mass of the rock in the water...

thank you for your reply. for the second turning point, is it because of the external force so it is two times larger? if not external force, is the second turning point equal to the first one? i attached the answer

For the first part, the mass sits at rest on the spring, so it is at the equilibrium position and thus mg = kd So, d = mg/k For the second part, I assume the uncompressed spring position is 0. When the mass at rest at the top. Its KE and PE is 0. When the mass at distance D, the question said...

thank you for your help

typo,why do the exhaust velocity relative to the rocket u=v+deltav-U? should be "v" not "u". i get the information from this website: http://hyperphysics.phy-astr.gsu.edu/hbase/rocket.html#c1 thank u for your reply