# 1. Lower in water 2. Higher in water

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1. Lower in water 2. Higher in water
Compared to an empty ship, will a ship loaded with a cargo of Styrofoam float lower in water or higher in water? Ch 13-1 Thanks to Milo Patterson and Steve Hewitt. 1. Lower in water 2. Higher in water

1. Lower in water 2. Higher in water
Compared to an empty ship, will a ship loaded with a cargo of Styrofoam float lower in water or higher in water? Ch 13-1 Thanks to Milo Patterson and Steve Hewitt. Answer: 1 The ship loaded with Styrofoam will float lower in water. A ship will float highest when its weight is least—that is, when it is empty. Loading any cargo will increase its weight and make it float lower in the water. Whether the cargo is a ton of Styrofoam or a ton of iron, the water displacement will be the same. 1. Lower in water 2. Higher in water

Everybody knows that “water seeks its own level,” but very few people know why water seeks its own level. The reason has most to do with Ch 13-2 1. atmospheric pressure. 2. water pressure depending on depth. 3. water’s density.

Everybody knows that “water seeks its own level,” but very few people know why water seeks its own level. The reason has most to do with Ch 13-2 Answer: 2 Water pressure depends on depth, so only at equal depths of water will the pressure be equal. Consider the U-tube. If water is at rest where each X is, the pressures must be equal—otherwise a flow would occur from the region of higher to the region of lower pressure until the pressures equalize. For this to happen, the depths below the surfaces must be equal. This is true whatever the density of water or whether or not there is atmospheric pressure. 1. atmospheric pressure 2. water pressure depending on depth. 3. water’s density

The density of the block of wood floating in water is
1. greater than the density of water. 2. equal to the density of water. 3. less than half that of water. 4. more than half the density of water. 5. … not enough information is given. Ch 13-4

The density of the block of wood floating in water is
1. greater than the density of water. 2. equal to the density of water. 3. less than half that of water. 4. more than half the density of water. 5. … not enough information is given. Ch 13-4 Answer: 4 A very-low-density object, like an inflated balloon, floats high on the water, and a denser object, like a piece of hardwood, floats lower into the water. An object half as dense as water floats halfway into the water (because it weighs as much as half its volume of water). Wood that floats 3/4 submerged, is 3/4 as dense as water—like the block in question—more than half the density of water. The density of the block compared to the density of water is the same as the fraction of the block below the water line.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 its volume submerged. If she were instead in a lunar module parked on the Moon, the ice in the same soft drink would float with 1. less than 9/10 submerged. 2. 9/10 submerged. 3. more than 9/10 submerged. Ch 13-5 Thanks to Josip Slisko.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 its volume submerged. If she were instead in a lunar module parked on the Moon, the ice in the same soft drink would float with 1. less than 9/10 submerged. 2. 9/10 submerged. 3. more than 9/10 submerged. Ch 13-5 Thanks to Josip Slisko. Answer: 2 How much a floating object extends below and above the liquid depends on the weight of the object and the weight of the displaced fluid, both of which are proportional to g. Lower g or increase it; the proportion floating above and below is unchanged.

1. Rise 2. Fall 3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged? Ch 13-6 1. Rise 2. Fall 3. Remain unchanged

1. Rise 2. Fall 3. Remain unchanged
Consider a boat loaded with scrap iron in a swimming pool. If the iron is thrown overboard into the pool, will the water level at the edge of the pool rise, fall, or remain unchanged? Ch 13-6 Answer: 2 The water level at the side of the pool will fall because the iron will displace less water submerged than when floating. When floating it displaces its weight of water (a lot!)—when submerged it displaces only its volume (less because iron is more dense than water). 1. Rise 2. Fall 3. Remain unchanged

1. half. 2. the same. 3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be Ch 13-8 1. half. 2. the same. 3. twice. 4. more than twice.

1. half. 2. the same. 3. twice. 4. more than twice.
The weight of the stand and suspended solid iron ball is equal to the weight of the container of water as shown above. When the ball is lowered into the water the balance is upset. The amount of weight that must be added to the left side to restore balance, compared to the weight of water displaced by the ball, would be Ch 13-8 Answer: 3 The additional weight that must be put on the left side to restore balance will equal twice the buoyant force, that is, twice the weight of water displaced by the submerged ball. Why twice? Because what the right side gains due to submersion and the heightened water level, the left side loses. (For example, if each side initially weighs 10 N and the right side gains 2 N to become 12 N, the left side loses 2 N to become 8 N. So an additional weight of 4 N, not 2 N, is required on the left side to restore balance.) 1. half. 2. the same. 3. twice. 4. more than twice.

1. doesn’t change. 2. shows an increase. 3. shows a decrease.
Gently push down on the pan of the scale and the display shows an increase in force. Likewise if you do the same on the rim of the beaker. But what if you immerse your fingertip in the water, without touching the beaker? Then the scale reading 1. doesn’t change. 2. shows an increase. 3. shows a decrease. Ch 13-9 Thanks to Peter Hopkinson.

1. doesn’t change. 2. shows an increase. 3. shows a decrease.
Gently push down on the pan of the scale and the display shows an increase in force. Likewise if you do the same on the rim of the beaker. But what if you immerse your fingertip in the water. without touching the beaker? Then the scale reading 1. doesn’t change. 2. shows an increase. 3. shows a decrease. Ch 13-9 Thanks to Peter Hopkinson. Answer: 2 Consider the system of beaker and water, resting in equilibrium on the scale. Before your finger is introduced, the only downward force on the system is the weight of both beaker and water. The scale supplies an equal upward force, the normal force shown by the scale reading. Now add another downward force—that of your finger poking into the water. How big is this force? It is as big as the buoyant force on your finger—equal to the weight of water displaced. This additional downward force on the system increases the normal force supplied by the scale. Hence the scale reading increases.

Consider a solid brass cube and a solid brass sphere that have equal surface areas. When both are completely submerged in water, the one experiencing the greater buoyant force is the 1. cube. 2. sphere. 3. … both the same. 4. … not enough information to say. Ch 13-11 Thanks to Hasan Fakhruddin.

Consider a solid brass cube and a solid brass sphere that have equal surface areas. When both are completely submerged in water, the one experiencing the greater buoyant force is the 1. cube. 2. sphere. 3. … both the same. 4. … not enough information to say. Ch 13-11 Thanks to Hasan Fakhruddin. Answer: 2 A sphere confines the largest possible volume within a given surface area. Buoyant force is equal to the weight of displaced water, and the greater-volume sphere displaces a greater volume, and hence greater weight, of water than the cube. So the sphere experiences the greater buoyant force.

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