Question

3. Suppose you have a rock with a mass of 305 g. When attached to a spring scale and lowered completely into water the spring scale reads Fe 2.10 N. Note that this reading is significantly lower than the rock's weight Fo Mg (3 N), because of the buoyant force helping to support it. From this information, calculate the following: (a) Fe: the magnitude of the buoyant force exerted on the rock. Use N2L in conjunction with the FBD shown above. (b) Use Fa pVg to calculate the volume V of the rock in m. The density of water is p 1000. kg/m2 (precise to 4 SF) (c) Convert the volume V from m' to cm. Be careful: 1 m 100 cm, but 1 m2 isn't equal to 100 cm'! (d) Calculate the rock's density in g/cm For the practice exercises, use 4 SF as you work, and then round to 3 SF after all calculations are done. Check: in the units we're primarily using, the density of water is p = 1.000 g/cm', and anything denser than water will sink. The bowling ball and the rock will sink in water, while the can of peanuts will float (since there's a lot of air trapped inside, lowering the average density). Compare the densities you obtain to the density of water

Answer 1

Solm - mass of the rock, m=305 g Density of water, e = 1000 kg/m weight of rock inside water is: - Fw = 2.10N. (a.) magnitude of Bluoyant force on the rock, FB (Mg - Fw) = (0.305x9:81-21) , Fe = 0.89205 N . (b.) volume of the rock 18, ve (Fol = (0.89205 / egy 7000x981). , V = 9.09327217* 10-5 (c.) volume in cm3 is : va 9.03327217 10 8 10 cm 0 v= 90.9327217 cm. (1) Density of the rock is, = ( )= 50.9824217) on o= 3.354128132 glan*