A 71 kg girl dives off a raft 2.34 square meters floating in a freshwater lake. By how much does the raft rise in cm?
Write your answer correct to two decimal places.
71 kg = volume displaced * 1000 kg/m^3
so
V = 71/1000 m^3
area * delta h = volume = 71/1000
2.34 * delta h = 71/1000
delta h = 30.34 /1000 in meters
= 30.34/10 in cm
= 3.03 cm
i think that's wrong !!!
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To calculate the rise of the raft, we need to apply the principle of buoyancy. The buoyant force acting on an object is equal to the weight of the liquid displaced by the object.
First, let's find the weight of the girl using the equation W = mg, where m is the mass of the girl and g is the acceleration due to gravity.
W = (71 kg)(9.8 m/s^2) = 696.8 N
Next, we calculate the volume of water displaced by the raft using the formula V = A × h, where A is the area of the raft and h is the height the raft is lifted.
V = (2.34 m^2)(h)
According to Archimedes' principle, the weight of the water displaced by the raft is equal to the weight of the girl. Therefore, we can write the equation as:
(2.34 m^2)(h) × (density of water) × (acceleration due to gravity) = 696.8 N
The density of water is approximately 1000 kg/m^3, and the acceleration due to gravity is 9.8 m/s^2. Substituting these values into the equation, we have:
(2.34 m^2)(h)(1000 kg/m^3)(9.8 m/s^2) = 696.8 N
Now, we can solve for h by rearranging the equation:
h = 696.8 N / [(2.34 m^2)(1000 kg/m^3)(9.8 m/s^2)]
h ≈ 0.031 m
Finally, to convert the height from meters to centimeters, we multiply by 100:
h ≈ 0.031 m × 100 cm/m
h ≈ 3.1 cm
Therefore, the raft rises approximately 3.1 cm.