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A glass ball of radius \(2.00 \mathrm{~cm}\) sits at the bottom of a container of milk that has a density of \(1.03 \mathrm{~g} / \mathrm{cm}^{3} .\) The normal force on the ball from the container's lower surface has magnitude \(9.48 \times 10^{-2} \mathrm{~N}\). What is the mass of the ball?

Short Answer

Expert verified
The mass of the ball is approximately 0.0442 kg.

Step by step solution

01

Determine the Buoyant Force

Since the ball is in a fluid, it experiences a buoyant force as per Archimedes' Principle. This buoyant force is equal to the weight of the fluid displaced by the ball. The volume of the ball is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] Substituting the radius, \(r = 2.00 \text{ cm}\), we find:\[ V = \frac{4}{3} \pi (2.00)^3 \approx 33.51 \text{ cm}^3 \]The weight of the milk displaced (buoyant force \(F_b\)) is:\[ F_b = V \times \text{Density of milk} \times g \]\[ F_b = 33.51 \times 1.03 \times 9.8 \approx 0.338 \text{ N} \]
02

Apply Newton's Third Law

Apply Newton's third law of motion, which relates the forces acting on the ball. The apparent weight \(W'\) of the ball is the actual weight minus the buoyant force. The normal force \(F_n\) is given to be \(9.48 \times 10^{-2} \text{ N}\). Therefore:\[ W' = F_n = W - F_b \] Where \(W\) is the actual weight of the ball. We rearrange this to solve for \(W\):\[ W = W' + F_b \]\[ W = 9.48 \times 10^{-2} + 0.338 \approx 0.433 \text{ N} \]
03

Calculate the Mass of the Ball

The actual weight \(W\) is equal to the mass \(m\) of the ball times the acceleration due to gravity \(g\). Therefore,\[ W = m \times g \]Solving for mass, we get:\[ m = \frac{W}{g} \]\[ m = \frac{0.433}{9.8} \approx 0.0442 \text{ kg} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Buoyant Force
When an object is submerged in a fluid, such as a glass ball in milk, it experiences an upward force called the buoyant force. This force is a result of the pressure difference at different depths of the fluid. According to Archimedes' Principle, the buoyant force is equal to the weight of the fluid that the object displaces once submerged.
Understanding this concept helps explain why some objects float while others sink. For the glass ball, the buoyant force can be calculated by multiplying the displaced fluid's volume, its density, and the gravitational acceleration.
  • Volume of displaced fluid: You need to know the volume of the object submerged, here it’s the volume of the glass ball.
  • Density of the fluid: Look at the fluid in which the object is submerged (e.g., milk with density of 1.03 g/cm³).
  • Gravitational acceleration: Typically, this is approximately 9.8 m/s² on Earth's surface.
Adding these up provides the magnitude of the buoyant force in newtons.
Density of Fluid
Density is a key concept when analyzing the interaction between an object and a fluid. It is defined as the mass per unit volume of a substance. In our example of the glass ball in milk, density explains how much mass of milk occupies a given volume. A fluid's density directly affects the buoyant force experienced by a submerged object.
If a fluid is denser, it will exert a stronger buoyant force as it can support more weight over the same volume. Therefore, knowing the milk's density is crucial for calculating the buoyant force acting on the submerged glass ball. The formula for density is given by
  • Density (\(\rho \)) = Mass/Volume
This is critical when comparing the immersive behavior of objects in different fluids.
Newton's Third Law
Newton's Third Law states that for every action, there is an equal and opposite reaction. In the context of our topic, when the glass ball exerts a force downward on the milk (due to gravity), the milk exerts an equal force upward, which includes the normal force and the buoyant force.
This upward force counteracts the weight of the ball, making it seem less heavy, and this perceived weight is called the apparent weight. Therefore, understanding Newton's Third Law is important when calculating the forces acting on submerged objects and helps us determine:
  • Apparent weight (\( W' \)) of the object
  • The actual weight (\( W \)) minus the buoyant force (\( F_b \))
These insights are essential for calculating the normal force the container's surface feels from the ball.
Volume of a Sphere
The volume of a sphere is an important geometric formula that is essential when dealing with objects submerged in a fluid. It allows us to determine the extent of fluid displaced by the object, which in turn is used to calculate the buoyant force. The formula for the volume (\( V \)) of a sphere is:
  • \[ V = \frac{4}{3} \pi r^3 \]
This equation shows us that the volume of a sphere depends on its radius raised to the third power, multiplied by a constant (\( \frac{4}{3} \pi \)).
For the glass ball example, using the radius of 2.00 cm, we can calculate its exact volume, which is crucial for further buoyancy and force calculations—it tells us how much space the sphere takes up in the milk.

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Most popular questions from this chapter

Two streams merge to form a river. One stream has a width of \(8.2 \mathrm{~m}\), depth of \(3.4 \mathrm{~m}\), and current speed of \(2.3 \mathrm{~m} / \mathrm{s}\). The other stream is \(6.8 \mathrm{~m}\) wide and \(3.2 \mathrm{~m}\) deep, and flows at \(2.6 \mathrm{~m} / \mathrm{s}\). If the river has width \(10.5 \mathrm{~m}\) and speed \(2.9 \mathrm{~m} / \mathrm{s}\), what is its depth?

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. \(14-50\) ); the cross-sectional area \(A\) of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed \(V\) and then through a narrow "throat" of cross- sectional area \(a\) with speed \(v .\) A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid's speed is accompanied by a change \(\Delta p\) in the fluid's pressure, which causes a height difference \(h\) of the liquid in the two arms of the manometer. (Here \(\Delta p\) means pressure in the throat minus pressure in the pipe.) (a) By applying Bernoulli's equation and the equation of continuity to points 1 and 2 in Fig. \(14-50\), show that $$ V=\sqrt{\frac{2 a^{2} \Delta p}{\rho\left(a^{2}-A^{2}\right)}} $$ where \(\rho\) is the density of the fluid. (b) Suppose that the fluid is fresh water, that the cross-sectional areas are \(64 \mathrm{~cm}^{2}\) in the pipe and 32 \(\mathrm{cm}^{2}\) in the throat, and that the pressure is \(55 \mathrm{kPa}\) in the pipe and 41 \(\mathrm{kPa}\) in the throat. What is the rate of water flow in cubic meters per second?

A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density the same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of \(1.08 \mathrm{~g} / \mathrm{cm}^{3}\). To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is \(60.0 \mathrm{~cm}\), and the density of iron is \(7.87 \mathrm{~g} / \mathrm{cm}^{3} .\) Find the inner diameter.

Suppose that two tanks, 1 and 2, each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth \(h\) below the liquid surface, but the hole in tank 1 has half the cross- sectional area of the hole in tank \(2 .\) (a) What is the ratio \(\rho_{1} / \rho_{2}\) of the densities of the liquids if the mass flow rate is the same for the two holes? (b) What is the ratio \(R_{V 1} / R_{V 2}\) of the volume flow rates from the two tanks? (c) At one instant, the liquid in tank 1 is \(12.0 \mathrm{~cm}\) above the hole. If the tanks are to have equal volume flow rates, what height above the hole must the liquid in tank 2 be just then?

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