It is more difficult to hold a piece of aluminum or a piece of cork completely submerged in water if they have the same weight of 0.5 kg. what forces should be applied for this?
m1 = m2 = 0.5 kilograms is the mass of a piece of aluminum and a cork;
ro1 = 2700 kg / m3 is the density of aluminum;
ro2 = 200 kg / m3 is the density of the plug;
g = 10 m / s2 – free fall acceleration;
ro = 1000 kg / m3 – water density.
It is required to determine the buoyancy forces acting on a piece of aluminum and cork in water.
Let’s find the buoyancy force acting on a piece of aluminum:
A1 = V1 * ro * g = m1 * ro * g / ro1 = 0.5 * 1000 * 10/2700 = 5000/2700 = 1.9 Newtons.
Let’s find the buoyancy force acting on the plug:
A2 = V2 * ro * g = m2 * ro * g / ro2 = 0.5 * 1000 * 10/200 = 5000/200 = 25 Newtons.
Since A2> A1, it will be more difficult to keep the plug.
Answer: it will be more difficult to keep the plug. To hold a piece of aluminum, you need to apply a force of 1.9 Newtons, and a cork – 25 Newtons.
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