Apparent Weight EK Physics Q
Started by Bond03
apparent weight = weight - buoyant force
1.5 * 10 = (2 * 10) - (0.5 * 10)
buoyant force = density fluid * volume fluid displaced * acceleration due to gravity
(0.5 * 10) = 5 * volume fluid displaced * 10
5 / 50 = 0.10 = volume fluid displaced = volume object
specific gravity = density = mass / volume
2 / 0.10 = 20
1.5 * 10 = (2 * 10) - (0.5 * 10)
buoyant force = density fluid * volume fluid displaced * acceleration due to gravity
(0.5 * 10) = 5 * volume fluid displaced * 10
5 / 50 = 0.10 = volume fluid displaced = volume object
specific gravity = density = mass / volume
2 / 0.10 = 20
From EK
"Apparent loss of mass can obtained simply by dividing apparent loss of weight by g.
So when they say the object experiences an apparent loss of mass of 0.5 kg, that means it has apparent loss of weight of 5 N. Which means that 5 N is the fraction of the object's weight that the fluid IS supporting, which means the bouyant force on the object is 5N.
So:
p*g*V = 5 N
So the problem with your equation is that it states the bouyant force, pgV, supports the entire weight of the object, mg ---> which would mean the object is floating. But the "submerged" object is not floating, and the bouyant force on it is less than its weight. That net downward force on it (after you subtract the 5 N bouyant force from the weight) is what keeps it sunk.
From this revised equation, you can calculate V = 1 x 10^-4 m^3
Then you can take the 2 kg / 1 x 10^-4 = 2 x 10^4 = 20,000 kg/m^3 = 20 SG."
"Apparent loss of mass can obtained simply by dividing apparent loss of weight by g.
So when they say the object experiences an apparent loss of mass of 0.5 kg, that means it has apparent loss of weight of 5 N. Which means that 5 N is the fraction of the object's weight that the fluid IS supporting, which means the bouyant force on the object is 5N.
So:
p*g*V = 5 N
So the problem with your equation is that it states the bouyant force, pgV, supports the entire weight of the object, mg ---> which would mean the object is floating. But the "submerged" object is not floating, and the bouyant force on it is less than its weight. That net downward force on it (after you subtract the 5 N bouyant force from the weight) is what keeps it sunk.
From this revised equation, you can calculate V = 1 x 10^-4 m^3
Then you can take the 2 kg / 1 x 10^-4 = 2 x 10^4 = 20,000 kg/m^3 = 20 SG."
Ok. Use proportions. mass is all you need. it's mass is 2 kg. That means outside the Mg=pVg
Now, the object is submerged, meaning the b force is pfluid*V*G=
Now, since submerged, the V in water is equal to V outside water so we can cancel everything.
Everything cancels and since the ratio is 2 to .5, the object has 4 times the specific density.
No math is required like the poster above me. That is too slow on the test.
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For the love of all that is holy, thank you for simplifying that.
