<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.
<span>ρwAd1g = Mg</span>
ρw<span>Ad2g = (M + m) g</span>
<span>d2∕d1 = (M + m)/g</span>
m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>
This means that Bubba’s mass is 120 kg.
Answer:
Explanation:
As we know that the pressure inside the liquid level is given as
here we have
h = 10.9 km
also we know that
now we have
Answer:
Around 44.01g.
Explanation:
One mole of carbon dioxide molecules has a mass of 44.01g, while one mole of sodium sulfide formula units has a mass of 78.04g.
