<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:
c
Explanation:
a vector quantity has both magnitude and direction
Gravitational potential energy = mass × gravity × height
Ep = (4)(9.81)(3)
Energy = 117.72 Joules
= 1.2x10^2 Joules
Conservation of momentum: total momentum before = total momentum after
Momentum = mass x velocity
So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32
Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32
Therefore β = 12.8 m/s
Explanation:
Substances with their density less than that of water which is 1 g/cm³ will float on it whiles those greater than that of water will sink into the water
From the question the density of the object is greater than that of water so the object will sink into the water
Hope this helps you
