1. What is its volume?
Express your answer using two significant figures.
2. The ingot is suspended from a rope and totally immersed in water. What is the tension in the rope (the apparent weight of the ingot in water)?
Express your answer using two significant figures.
2007-12-06 04:11:58 · 3 answers · asked by Bugz 3 in Science & Mathematics ➔ Physics
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Weight of aluminum ingot(in air) =W=72 N
Density of aluminum = d =2700 kg/m^3
Mass = m =weight (W) /acceleration of gravity(g)
m=W/g
Volume= v = mass (m) / density(d)=m/d
Volume= v = W/gd = 72/9.8*2700
Volume= v = 0.0027 m^3
Volume of aluminum ingot is 0.0027 m^3
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the tension in the rope (the apparent weight of the ingot in water = T = wt of ingot in air - buoyant force(upthrust)
the tension in the rope = T = wt in air - wt of equal volume of water
T =W - vpg
where p is density of water , p = 1000 kg/m^3
T =72 -0.0027*1000*9.8=72 - 26.46=45.54 N
The tension in the rope (the apparent weight of the ingot in water) is 45.54 N
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2007-12-06 05:21:33 · answer #1 · answered by ukmudgal 6 · 6⤊ 0⤋
Aluminum Ingot
2016-11-07 05:35:43 · answer #2 · answered by javoronkov 4 · 0⤊ 1⤋
Information is missing. 72 WHAT? Pounds? Ounces? And what do you mean by "In air?" Floating? Thrown in the air? Or merely stated "dry" that is, out of water.
2007-12-06 04:20:35 · answer #3 · answered by Marvinator 7 · 0⤊ 1⤋