40.REASONING the Ice with the Bear on It Is Floating, So That the Upward-Acting Buoyant

40.REASONING The ice with the bear on it is floating, so that the upward-acting buoyant force balances the downward-acting weight Wice of the ice and weight Wbear of the bear. The magnitude FB of the buoyant force is the weight of the displaced water, according to Archimedes’ principle. Thus, we have , the expression with which we will obtain Wbear. We can express each of the weights and Wice as mass times the magnitude of the acceleration due to gravity (Equation4.5) and then relate the mass to the density and the displaced volume by using Equation 11.1.

SOLUTION Since the ice with the bear on it is floating, the upward-acting buoyant force FB balances the downward-acting weight Wice of the ice and the weight Wbear of the bear. The buoyant force has a magnitude that equals the weight of the displaced water, as stated by Archimedes’ principle.Thus, we have

(1)

In Equation (1), we can use Equation 4.5 to express the weights and Wice as mass m times the magnitude g of the acceleration due to gravity. Then, the each mass can be expressed as (Equation 11.1). With these substitutions, Equation (1) becomes

(2)

When the heaviest possible bear is on the ice, the ice is just below the water surface and displaces a volume of water that is . Substituting this result into Equation (2), we find that

41.REASONING The buoyant force exerted on the balloon by the air must be equal in magnitude to the weight of the balloon and its contents (load and hydrogen). The magnitude of the buoyant force is given by . Therefore,

where, since the balloon is spherical, . Making this substitution for V and solving for r, we obtain

SOLUTION Direct substitution of the data given in the problem yields

43.REASONING Since the duck is in equilibrium, its downward-acting weight is balanced by the upward-acting buoyant force. According to Archimedes’ principle, the magnitude of the buoyant force is equal to the weight of the water displaced by the duck. Setting the weight of the duck equal to the magnitude of the buoyant force will allow us to find the average density of the duck.

SOLUTION Since the weight Wduck of the duck is balanced by the magnitude FB of the buoyant force, we have that Wduck = FB. The duck’s weight is Wduck = mg = (duckVduck)g, where duck is the average density of the duck and Vduck is its volume. The magnitude of the buoyant force, on the other hand, equals the weight of the water displaced by the duck, or
FB = mwaterg, where mwater is the mass of the displaced water. But , since one-quarter of the duck’s volume is beneath the water. Thus,

Solving this equation for the average density of the duck (and taking the density of water from Table 11.1) gives

61.REASONING AND SOLUTION

a. Using Equation 11.12, the form of Bernoulli's equation with , we have

b. The pressure inside the roof is greater than the pressure on the outside. Therefore, there is a net outward force on the roof. If the wind speed is sufficiently high, some roofs are "blown outward."

62.REASONING We apply Bernoulli’s equation as follows:

SOLUTION The vaccine’s surface in the reservoir is stationary during the inoculation, so that vS=0 m/s. The vertical height between the vaccine’s surface in reservoir and the opening can be ignored, so yS=yO. With these simplifications Bernoulli’s equation becomes

Solving for the speed at the opening gives