5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Water flows to the right through a horizontal pipe. Treat the water as an ideal fluid of density 1000 kg/m³. Consider a small cube-shaped parcel of the water with edges 0.020 m long. At one instant, the gauge pressure on the parcel's upstream (left) face is 6.0 × 10² Pa and the gauge pressure on its downstream (right) face is 4.0 × 10² Pa. What is the magnitude of the parcel's horizontal acceleration at that instant?
Answer and reasoning
A3.0 × 10¹ m/s² A student who counts only the push from the water behind the parcel uses 0.24 N as the net force and gets 30 m/s². The water ahead of the parcel pushes back on its downstream face with 0.16 N, so the net force is only 0.080 N.
B1.0 × 10¹ m/s²Correct Treat the parcel as an object and apply Newton's second law. The water behind pushes it forward with (6.0 × 10²)(4.0 × 10⁻⁴) = 0.24 N and the water ahead pushes it back with (4.0 × 10²)(4.0 × 10⁻⁴) = 0.16 N, so Fnet = 0.080 N. Its mass is ρV = 8.0 × 10⁻³ kg, so a = 0.080/(8.0 × 10⁻³) = 10 m/s².
C5.0 × 10¹ m/s² A student who adds the two pressure forces as magnitudes gets 0.24 + 0.16 = 0.40 N and 50 m/s². The forces on the two faces point in opposite directions, so they partly cancel: Fnet = 0.24 − 0.16 = 0.080 N.
D2.5 × 10⁴ m/s² A student who treats the pressure difference as if it were a force divides 2.0 × 10² Pa by the mass and gets 2.5 × 10⁴. Pressure is force per unit area; the difference in force on the two faces is (2.0 × 10² Pa)(4.0 × 10⁻⁴ m²) = 0.080 N.
Working Face area A = (0.020 m)² = 4.0 × 10⁻⁴ m²; volume V = (0.020 m)³ = 8.0 × 10⁻⁶ m³; mass m = ρV = (1000 kg/m³)(8.0 × 10⁻⁶ m³) = 8.0 × 10⁻³ kg. Horizontal forces on the parcel: from the water behind, (6.0 × 10² Pa)(4.0 × 10⁻⁴ m²) = 0.24 N to the right; from the water ahead, (4.0 × 10² Pa)(4.0 × 10⁻⁴ m²) = 0.16 N to the left. Fnet = 0.24 N − 0.16 N = 0.080 N to the right. a = Fnet/m = 0.080 N / 8.0 × 10⁻³ kg = 10 m/s² = 1.0 × 10¹ m/s². (Atmospheric pressure adds equal and opposite forces on the two faces, so using gauge pressures gives the same net force.)
A glass of water rests on a level table. The water in the glass is treated as a system. Which statement correctly explains why the center of mass of the water does not accelerate?
Answer and reasoning
AThe external forces on it, from Earth, the glass and the air, add to zero.Correct Only external forces can change the motion of a system's center of mass. Earth pulls the water down, the air pushes down on its surface, and the glass pushes up on it (and inward on its sides). These external forces add to zero, so the center of mass does not accelerate.
BThe forces that its molecules exert on each other hold the water up. A student who thinks internal forces can support a system picks this. The forces the molecules exert on each other come in equal and opposite pairs and add to zero for the system; they cannot hold the water up. Only the glass, an object outside the system, can.
CNo forces are exerted on the water, since the water is at rest. A student who thinks nothing acts on an object at rest picks this. Earth, the glass and the air all exert forces on the water; being at rest means only that these forces add to zero.
DThe glass pushes up with a force equal to the water's weight; air exerts none. A student who thinks still air exerts no force picks this. The atmosphere presses down on the water's surface with a pressure of about 1.0 × 10⁵ Pa, a force far larger than the water's weight for an ordinary glass, so the glass must push up with the weight plus that force.
A small metal block is released from rest under water and sinks toward the bottom of a deep tank. Which statement correctly describes the buoyant force exerted on the block by the water while the block is sinking?
Answer and reasoning
AIt is zero, since the block is sinking through the water. A student who thinks only floating objects are pushed up by a fluid picks this. The block displaces water, so the water exerts an upward buoyant force ρVg on it; the block sinks only because its weight is larger.
BIt points downward, as the water above pushes it down. A student who thinks the water above an object pushes it down overall picks this. The water does push down on the top of the block, but it pushes up harder on the deeper bottom face, so the net force from the water is upward.
CIt points upward and is equal to the block's weight. A student who applies the floating condition to every object picks this. If the buoyant force equaled the weight, the net force on the block would be zero and, released from rest, it would stay at rest instead of sinking.
DIt is upward and less than the block's weight.Correct The buoyant force is the net upward force the water exerts on any object immersed in it, whether the object floats or sinks. The block sinks because its weight is greater than the buoyant force, so the buoyant force is upward and smaller than the weight.
The diagram shows a solid cube held at rest under water with its top face horizontal. The arrows show the forces exerted by the water on the cube's faces, and the cube's dimensions and the gauge pressures in the water at its top and bottom faces are labeled. What is the magnitude of the buoyant force exerted on the cube?
Answer and reasoning
A3.0 × 10¹ N A student who thinks the water pushes on a submerged object only from below takes the 30 N push on the bottom face as the buoyant force. The water above also pushes down on the top face, with 20 N; the buoyant force is the difference, 10 N.
B2.0 × 10¹ N A student who thinks the water above an object pushes it down overall takes the 20 N push on the top face as the buoyant force. That push is real, but the push on the deeper bottom face is larger, so the net force from the water is 30 N − 20 N = 10 N, upward.
C1.0 × 10¹ NCorrect The buoyant force is the net force of all the water's pushes on the cube. The pushes on the side faces cancel. The bottom face is pushed up with (3.0 × 10³)(0.010) = 30 N and the top face down with (2.0 × 10³)(0.010) = 20 N, so the net force is 10 N upward, the same as ρVg.
D1.0 × 10³ N A student who treats pressure as force reads the pressure difference, 1.0 × 10³ Pa, as a force in newtons. Pressure must be multiplied by the area it acts on: (1.0 × 10³ Pa)(0.010 m²) = 10 N.
Working Face area A = (0.10 m)² = 0.010 m². The horizontal forces on opposite side faces cancel. Upward force on the bottom face: (3.0 × 10³ Pa)(0.010 m²) = 30 N. Downward force on the top face: (2.0 × 10³ Pa)(0.010 m²) = 20 N. Fb = 30 N − 20 N = 10 N = 1.0 × 10¹ N, upward. Check with Fb = ρVg = (1000 kg/m³)(1.0 × 10⁻³ m³)(10 m/s²) = 10 N. (Atmospheric pressure adds the same force to the top and bottom faces and cancels.)
Two solid balls of equal volume, one made of lead and one of aluminum, hang from strings so that each is completely under water at the same depth. Lead is denser than aluminum. How does the buoyant force on the lead ball compare with the buoyant force on the aluminum ball?
Answer and reasoning
AGreater on the lead ball, since it has the greater mass A student who thinks a heavier object displaces more water picks this. A completely submerged object displaces its own volume of water, whatever its mass, so the two equal-volume balls displace equal amounts.
BGreater on the aluminum ball, since it is the less dense A student who thinks a less dense object is pushed up more picks this. Aluminum would rise more easily than lead only because its weight is smaller; the buoyant force, ρVg, is the same for both balls.
CZero on both, since both would sink if released A student who thinks sinking objects are not buoyed up picks this. Each ball displaces water and has an upward buoyant force on it; each would sink only because its weight is greater than that force.
DEqual, since the balls displace equal volumes of waterCorrect The buoyant force is the weight of the water displaced, ρVg, where V is the volume of water displaced. Both balls are completely submerged and have equal volumes, so they displace equal volumes of water and have equal buoyant forces, whatever their masses.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
8.3.A.1 Fluid parcel Fix
Fluid parcel
A small portion of a fluid, such as a tiny cube of water, treated as a single object. Newton's laws apply to it exactly as to any other object: its velocity changes only if the net force exerted on it by the surrounding fluid and by Earth is not zero.
Net force on a fluid parcel
The vector sum of the forces the surrounding fluid exerts on all faces of a parcel, plus the parcel's weight. Along a horizontal pipe the net horizontal force is (Pbehind − Pahead)A, where A is the area of a face, so it points from higher toward lower pressure. Unit: N.
Conditions for a change in a fluid's velocity
A parcel of fluid speeds up, slows down or changes direction only when the net force on it is not zero. In horizontal flow it speeds up where the pressure falls along the direction of flow, slows down where the pressure rises, and keeps a constant velocity where the pressure is uniform.
Students often think Pressure and force are the same quantity, so a pressure (or a pressure difference) in pascals can be used as a force in newtons. In fact No. Pressure is force per unit area, P = F⊥/A. To find the force a fluid exerts on a surface, multiply the pressure by the area of the surface; a pressure difference of 200 Pa across a face of 4.0 × 10⁻⁴ m² gives a force difference of only 0.080 N.
Students often think The forces on an object can be added as plain numbers, whatever their directions, so opposite forces add up to a larger total. In fact No. Forces are vectors. Forces in opposite directions partly cancel, so the net force is found by subtracting the smaller from the larger (or by adding them with signs), not by adding their magnitudes.
8.3.A.2 Fluid as a system Fix
Fluid as a system
A chosen body of fluid treated as a system. The forces its particles exert on each other are internal and cancel in pairs (Newton's third law), so only external forces, such as its weight and the forces from a container and the surrounding air, can change the motion of its center of mass.
Macroscopic behavior of a fluid
Properties observed for the fluid as a whole, such as the pressure it exerts on a surface or the way it flows or stays at rest. They result from the many interactions among its particles, which transmit forces through the fluid, combined with the external forces exerted on it.
Students often think The forces between the particles of a fluid can hold the fluid up, or set it moving as a whole, without any help from outside. In fact No. Forces between parts of a system are internal: by Newton's third law they come in equal and opposite pairs, so their sum is zero. Only external forces, such as the fluid's weight and the forces from its container and the air, change the motion of its center of mass.
Students often think An object, or a body of fluid, that is at rest has no forces exerted on it. In fact No. An object or fluid at rest can have many forces exerted on it; at rest means only that they add to zero. Water in a glass has its weight and forces from the glass and the air exerted on it.
8.3.B.1 Buoyant force (Fb) Fix
Buoyant force (Fb)
The net upward force that a surrounding fluid exerts on an object that is partly or completely immersed in it. It acts on floating and sinking objects alike, in liquids and in gases. Unit: N.
Scale reading for a submerged object
For an object hanging at rest from a spring scale while immersed in a fluid, the scale's upward pull plus the buoyant force balances the object's weight, so the reading is Fg − Fb. The reading is less than the weight; the object's weight itself does not change.
Students often think An object that sinks has no buoyant force exerted on it; only floating objects are pushed up by the fluid. In fact Yes. Every object immersed in a fluid displaces some fluid and has an upward buoyant force exerted on it, ρVg. An object sinks because its weight is greater than the buoyant force, not because the buoyant force is missing.
Students often think The water above a submerged object pushes it down, so the net force from the water is downward, and more water above means a larger downward push. In fact No. The water does push down on the object's top surface, but it pushes up harder on the bottom surface, which is deeper. The net force from the water, the buoyant force, is upward.
8.3.B.2 Origin of the buoyant force Fix
Origin of the buoyant force
The sum of the forces exerted by the fluid's particles on every part of an object's surface. Horizontal forces on opposite sides cancel. Pressure increases with depth, so the fluid pushes up on the lower surface harder than it pushes down on the upper surface; for a block with horizontal faces of area A, Fb = (Pbottom − Ptop)A.
Buoyant force and depth
For a completely submerged object in an incompressible fluid, going deeper increases the pressure on the top and bottom surfaces by the same amount, so their difference, and therefore the buoyant force, does not depend on depth.
Students often think A fluid pushes on an object or a parcel from one side only (from behind in a flow, or from below for a submerged object), so the push from the fluid on the other side can be left out. In fact No. A fluid pushes on every surface it touches, perpendicular to that surface. A parcel in a pipe is pushed forward by the fluid behind it and backward by the fluid ahead of it; a submerged object is pushed up on its bottom and down on its top. The net force is the difference.
Students often think The buoyant force on a submerged object increases with depth, because the pressure increases with depth. In fact No. Going deeper increases the pressure on the top and bottom of the object by the same amount, so the difference, and the buoyant force, is unchanged. For an incompressible fluid and a rigid object, the buoyant force is the same at every depth.
8.3.B.3 Displaced fluid Fix
Displaced fluid
The fluid whose place is taken by an object: a volume equal to the whole volume of a completely submerged object, or to only the submerged part of a floating object. Its volume V is measured in m³.
Archimedes' principle (Fb = ρVg)
The magnitude of the buoyant force equals the weight of the fluid displaced by the object: Fb = ρVg, where ρ is the density of the fluid (not of the object) and V is the volume of fluid displaced. Unit: N.
Floating condition
An object floating at rest has Fb = Fg, so it sinks into the fluid until the weight of the fluid it displaces equals its own weight. An object whose weight is greater than ρVg for its whole volume sinks.
Students often think A heavier object displaces more fluid, so it has a greater buoyant force exerted on it than a lighter object of the same size. In fact No. The buoyant force depends on the volume of fluid displaced, not on the object's mass. Two completely submerged objects of equal volume have equal buoyant forces, whatever they are made of.
Students often think A lighter or less dense object has a greater buoyant force exerted on it than a heavier object of the same size, because the fluid pushes it up more easily. In fact No. When both are completely submerged, objects of equal volume displace equal volumes of fluid and have equal buoyant forces. The less dense object is more likely to float only because its weight is smaller.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
Water flows steadily in the +x direction along a horizontal pipe that is completely filled. Treat the water as an ideal fluid. The graph shows the pressure P in the water as a function of position x along the pipe. In which interval does a small parcel of the water slow down as it moves along the pipe?
Answer and reasoning
AFrom x = 0 m to x = 1 m A student who thinks water moves faster where its pressure is higher picks this interval, where the pressure falls. A pressure that falls along the flow means the water behind a parcel pushes it harder than the water ahead pushes back, so the net force is forward and the parcel speeds up.
BFrom x = 1 m to x = 2 m A student who thinks moving water slows down whenever nothing pushes it along picks this interval, where the pressure is uniform. With equal pressures behind and ahead, the net force on a parcel is zero, so by Newton's first law it keeps a constant velocity.
CFrom x = 2 m to x = 3 mCorrect A parcel slows down only if the net force on it points backward, against its motion. From x = 2 m to x = 3 m the pressure rises along the direction of flow, so the water ahead of a parcel pushes back on it harder than the water behind pushes it forward. The net force is in the −x direction, and the parcel slows.
DFrom x = 3 m to x = 4 m A student who judges the force on a parcel by how high the pressure is, rather than by how it changes along the pipe, picks the interval where the pressure is highest. Here the pressure is uniform, so the water behind and ahead of a parcel push on it equally hard, the net force is zero and its speed does not change.
Working Treat a parcel as an object. Its net horizontal force is (Pbehind − Pahead)A, which points from higher toward lower pressure. From x = 0 to 1 m, P falls along the flow: net force in +x, the parcel speeds up. From 1 m to 2 m and from 3 m to 4 m, P is uniform: net force zero, constant speed. From 2 m to 3 m, P rises along the flow: net force in −x, opposite to the motion, so the parcel slows down.
A foam ball of volume V would float, but it is held at rest completely under water by a light string tied to the bottom of a tank. The density of the foam is one-fourth of the density ρ of water. Which expression gives the tension in the string?
Answer and reasoning
A0.75ρVgCorrect The ball is at rest, so the upward buoyant force balances its weight plus the downward pull of the string: ρVg = 0.25ρVg + FT, which gives FT = 0.75ρVg.
B1.00ρVg A student who thinks gravity does not act on objects under water sets the tension equal to the buoyant force, ρVg. The ball's weight, 0.25ρVg, still acts and does part of the job of holding the ball down, so the string's pull is smaller than the buoyant force.
C1.25ρVg A student who adds the magnitudes of the buoyant force and the weight gets 1.25ρVg. The buoyant force is upward and the weight is downward; the string supplies only the difference, ρVg − 0.25ρVg.
D0.25ρVg A student who takes the tension in a string to be the weight of the object tied to it picks 0.25ρVg. Here the string holds the ball down against the buoyant force, so its tension comes from the balance of all three forces, not from the weight alone.
Working Forces on the ball: buoyant force Fb = ρVg upward; weight Fg = (ρ/4)Vg downward; string tension FT downward (the string pulls the ball toward the bottom of the tank). The ball is at rest, so Fb = Fg + FT and FT = ρVg − 0.25ρVg = 0.75ρVg.
A sealed, rigid steel can sinks slowly through a deep freshwater lake. Treat the water as incompressible. How does the buoyant force on the can when it is 5.0 m below the surface compare with the buoyant force on it when it is 1.0 m below the surface?
Answer and reasoning
AGreater at 5.0 m, as the water pressure is greater at greater depth. A student who thinks the buoyant force grows with depth picks this. The pressure is greater at 5.0 m, but it is greater on the top of the can by the same amount as on the bottom, so the difference, which is the buoyant force, does not change.
BThe same, as the pressures on its top and bottom rise by equal amounts.Correct The buoyant force is the difference between the upward push on the can's bottom and the downward push on its top. Going 4.0 m deeper adds ρg(4.0 m) to the pressure on both surfaces, so the difference, and the buoyant force, is unchanged. Equivalently, the can displaces the same volume at both depths.
CSmaller at 5.0 m, as more water above the can pushes it down. A student who thinks the water above an object pushes it down overall picks this. The extra water above does push harder on the top, but it pushes harder on the bottom by exactly the same amount, so the net upward force is unchanged.
DZero at both depths, as a can that is sinking has no buoyant force. A student who thinks sinking objects are not buoyed up picks this. The can displaces water, so an upward buoyant force ρVg acts on it at every depth; it sinks because its weight is greater than that force.
A rock hangs from a spring scale. The scale reads 12 N when the rock hangs at rest in air and 8.0 N when the rock hangs at rest completely under water. (Density of water = 1000 kg/m³; use g = 10 m/s².) What is the volume of the rock?
Answer and reasoning
A8.0 × 10⁻⁴ m³ A student who takes the scale reading under water to be the buoyant force uses 8.0 N and gets 8.0 × 10⁻⁴ m³. The reading is the scale's upward pull; the buoyant force is the drop in the reading, 12 N − 8.0 N = 4.0 N.
B1.2 × 10⁻³ m³ A student who thinks the buoyant force equals the object's weight uses 12 N and gets 1.2 × 10⁻³ m³. That holds only for a floating object; the rock still needs 8.0 N from the scale, so the buoyant force is only 4.0 N.
C4.0 × 10⁻⁴ m³Correct Under water the rock is at rest, so Fb + 8.0 N = 12 N and Fb = 4.0 N. The buoyant force is the weight of the displaced water, ρVg = 4.0 N, so V = 4.0/(1000 × 10) = 4.0 × 10⁻⁴ m³. The rock is completely submerged, so this is its volume.
D4.0 × 10⁻³ m³ A student who takes the buoyant force to be the mass of the displaced water, ρV, leaves out g and gets V = 4.0/1000 = 4.0 × 10⁻³ m³. The buoyant force is the weight of the displaced water, ρVg.
Working In air the reading equals the weight: Fg = 12 N. Under water the rock is at rest, so Fb + 8.0 N = 12 N and Fb = 4.0 N. Fb = ρVg gives V = Fb/(ρg) = 4.0 N / ((1000 kg/m³)(10 m/s²)) = 4.0 × 10⁻⁴ m³. The rock is completely submerged, so this is the rock's volume.
A block 4.0 cm tall hangs from a spring scale and is lowered slowly, at constant speed, into a deep tank of water, with its bottom face horizontal. Resistance from the water is negligible. The graph shows the scale reading F as a function of the depth d of the block's bottom face below the water surface. Which statement about the buoyant force on the block is supported by the graph?
Answer and reasoning
AIt keeps growing with depth for as long as the block keeps on going deeper into the water. A student who thinks the buoyant force grows with depth picks this. The graph is flat beyond d = 4.0 cm: once the block is completely under water it displaces the same volume at every depth, so the buoyant force stays 2.0 N.
BIt is greatest just as the block touches the water, and then it becomes smaller. A student who takes the scale reading to be the buoyant force sees the graph fall and picks this. The reading is the scale's pull; the buoyant force is the drop in the reading from 6.0 N, which grows from 0 to 2.0 N.
CIt equals the block's weight, 6.0 N, once the block is completely under water. A student who thinks the buoyant force equals the object's weight picks this. Once the block is under water the scale still reads 4.0 N, so the buoyant force is 6.0 N − 4.0 N = 2.0 N, less than the weight.
DIt grows as the block enters the water, then stays constant once the block is submerged.Correct The block moves at constant velocity, so at every depth Fb + F = 6.0 N and Fb = 6.0 N − F. From d = 0 to 4.0 cm the reading falls, so the buoyant force grows from 0 to 2.0 N as the block displaces more water. Beyond 4.0 cm the reading stays 4.0 N, so the buoyant force stays 2.0 N at every greater depth.
A beaker of water rests on a scale. A solid block of mass M, made of a material whose density ρb is greater than the density ρ of water, hangs from a string. The block is lowered into the water until it is completely submerged, and it then hangs at rest without touching the beaker; no water spills. By how much does the magnitude of the force exerted on the beaker by the scale increase?
Answer and reasoning
AρMg/ρbCorrect The block's volume is V = M/ρb, so the water exerts a buoyant force ρVg = ρMg/ρb upward on it. By Newton's third law the block pushes down on the water with the same magnitude, and the beaker–water system stays at rest, so the scale must push up harder by ρMg/ρb.
BMg A student who takes the buoyant force to be equal to the block's weight gets a reaction force Mg on the water. The block is denser than water and does not float; the buoyant force is the weight of the water displaced, ρMg/ρb, which is less than Mg, and the string supports the rest.
CMg − ρMg/ρb A student who thinks the block presses on the water with its apparent weight picks this. The apparent weight, Mg − ρMg/ρb, is what the string supports. The water pushes up on the block with the buoyant force, so the block pushes down on the water with that force, ρMg/ρb.
D0 A student who takes the string's tension to be the block's full weight concludes that the string holds the block up and nothing is added to the scale. The water exerts an upward buoyant force on the block, so the tension is less than Mg, and the block pushes down on the water with the reaction to that force, ρMg/ρb.
Working Volume of block: V = M/ρb. Buoyant force on block: Fb = ρVg = ρMg/ρb, upward. By Newton's third law the block exerts ρMg/ρb downward on the water. The beaker–water system stays at rest, so the scale's upward force increases by ρMg/ρb. (The string supports the rest, Mg − ρMg/ρb.)
Solid ball X, of radius r, hangs at rest from a string, completely submerged in water. Solid ball Y has radius 2r and is made of a material whose density is half the density of ball X's material. Ball Y hangs at rest from a string, completely submerged in an oil whose density is 0.80 times the density of water. The buoyant force exerted on ball Y is how many times the buoyant force exerted on ball X?
Answer and reasoning
A6.4Correct The buoyant force is the weight of the fluid displaced, ρVg. Each ball is completely submerged, so it displaces its own volume, which scales with r³: doubling the radius gives 8 times the volume. The oil is 0.80 times as dense as water: 0.80 × 8 = 6.4. The balls' own densities do not enter.
B4.0 A student who takes the buoyant force to equal the object's weight picks this: Y has 8 times the volume at half the density, so 4 times X's weight. The buoyant force is the weight of the fluid displaced, which depends on the fluid's density and the displaced volume: 0.80 × 8 = 6.4.
C8.0 A student who thinks the buoyant force depends only on the displaced volume, whatever the fluid, picks this: 8 times the volume gives 8.0. Fb = ρVg uses the fluid's density, and the oil is only 0.80 times as dense as water: 0.80 × 8 = 6.4.
D1.6 A student who thinks doubling the radius doubles the volume picks this: 0.80 × 2 = 1.6. Volume scales with the cube of the radius, so the displaced volume is 2³ = 8 times as large: 0.80 × 8 = 6.4.
Working Fb = ρfluid V g, the weight of fluid displaced. A completely submerged ball displaces its own volume, V = (4/3)πr³, so doubling the radius multiplies V by 2³ = 8. The fluid density changes by 0.80. Ratio = 0.80 × 8 = 6.4. The density of each ball's material does not enter the buoyant force.
Compiled from the AP Physics 1 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
