In a lecture demonstration, an object is suspended from a spring scale which reads 8n when the object is in air. the object is then lowered into a beaker of water, and when the object is fully submerged in the water, the scale reads 6n. a) find the density of the object. b) during the entire demonstration, the beaker is standing on a kitchen scale which reads 20n before the object is lowered into the water. what does this scale read when the object is fully submerged? c) compare the sum of the two scale readings before and after the object is submerged.

a) density of the object is 3995.01, b) the weight scale reads 22N c) the sum individually will be the same with when added together.

Step-by-step explanation:

The weight of the object in air is 8N,

and weight = Mass * acceleration due to gravity = m * 9.81

8/9.81 = 0.815,

upthrust( force acting on the body from the liquid impeding the immersion) on the body when fully submerged = weight in air - weight in water = 8N - 6N =2N

Upthrust = weight of water displaced = 2N = mass * acceleration

2/9.81 = 0.204kg

density of water(1000kg/m^3) = mass of water / volume of water

volume of water displaced = 0.204/1000 = 0.000204m^3 (204cm^3)

volume of water displaced = volume of the solid

density of solid = mass/ volume = 0.815/0.000204 = 3995.01kg/m^3

b) when fully submerge in water the the scale experience according to newton third law of motion ( equal and opposite reaction of forces) additional 2N push so that total weight with the fully submerge solid = 20N + upthrust = 20N + 2N =22N

c) the of two scale reading is before (8N + 20N = 28N) and after (6N + 22N = 28) since there is no loss of matter; the demonstration was in equilibrium.

i think the answer is fried chicken ; >

step-by-step explanation:

step-by-step explanation:

bob the banana