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Chapter 9: Problem 42

An object weighing \(300 \mathrm{~N}\) in air is immersed in water after being tied to a string connected to a balance. The scale now reads \(265 \mathrm{~N}\). Immersed in oil, the object appears to weigh \(275 \mathrm{~N}\). Find (a) the density of the object and (b) the density of the oil.

Short Answer

Expert verified
The density of the object is approximately \(8577.038 kg/m^3\) and the density of the oil is approximately \(699.159 kg/m^3\).

Step by step solution

01

Find Effective Weights

First, determine the effective weights of the object in different fluids by subtracting the weight of the object in the fluid from the weight of the object in air. The effective weight in water is \(265 N - 300 N = - 35 N\) and in oil is \(275 N - 300 N = - 25 N\). The negative sign indicates that the buoyant force is acting upwards.
02

Finding the Volume

The volume of the object can be found using the effective weight in water and the density of water, \(1000 kg/m^3\). Use the formula of buoyant force, \(F_b = \rho_{fluid} \cdot V \cdot g\), where \(F_b = -35 N\), \(g = 9.8 m/s^2\), and solve for \(V\): \(V = - F_b / (\rho_{fluid}\cdot g) = -(-35N) / (1000 kg/m^3 \cdot 9.8 m/s^2) = 0.00357 m^3 \).
03

Finding the Density of the Object

Next, find the density of the object, denoted as \(蟻_{object}\), using the formula \(蟻 = m/v\), where \(m\) is the mass and \(v\) is the volume. As weight \(W = m \cdot g\), the mass \(m = W/g = 300N / 9.8 m/s^2 = 30.612 kg\). Substituting the values into the density formula gives: \(蟻_{object} = m / v = 30.612 kg / 0.00357 m^3 = 8577.038 kg/m^3\).
04

Finding the Density of the Oil

Considering the buoyancy in oil, the density of the oil can be calculated by rearranging the formula of buoyant force to get \(蟻_{fluid} = - F_b / (V \cdot g)\). Substituting the known values gives: \(\rho_{oil} = -(-25N) / (0.00357 m^3 \cdot 9.8 m/s^2) = 699.159 kg/m^3\).

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Most popular questions from this chapter

A large balloon of mass \(226 \mathrm{~kg}\) is filled with helium gas until its volume is \(325 \mathrm{~m}^{3}\). Assume the density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and the density of helium is \(0.179 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Draw a force diagram for the balloon. (b) Calculate the buoyant force acting on the balloon. (c) Find the net force on the balloon and determine whether the balloon will rise or fall after it is released. (d) What maximum additional mass can the balloon support in equilibrium? (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? (f) What limits the height to which the balloon can rise?

A hypodermic syringe contains a medicine with the density of water (Fig. P9.47). The barrel of the syringe has a cross-sectional area of \(2.50 \times 10^{-5} \mathrm{~m}^{2}\). In the absence of a force on the plunger, the pressure everywhere is \(1.00 \mathrm{~atm}\). A force \(\overrightarrow{\mathbf{F}}\) of magnitude \(2.00 \mathrm{~N}\) is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle. Assume the pressure in the needle remains equal to \(1.00 \mathrm{~atm}\) and that the syringe is horizontal.

Old Faithful geyser in Yellowstone Park erupts at approximately 1-hour intervals, and the height of the fountain reaches \(40.0 \mathrm{~m}\) (Fig. P9.57). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c) What is the pressure (above atmospheric pressure) in the heated underground chamber \(175 \mathrm{~m}\) below the vent? You may assume the chamber is large compared with the geyser vent.

A hypodermic needle is \(3.0 \mathrm{~cm}\) in length and \(0.30 \mathrm{~mm}\) in diameter. What pressure difference between the input and output of the needle is required so that the flow rate of water through it will be \(1 \mathrm{~g} / \mathrm{s}\) ? (Use \(1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\) as the viscosity of water.)

The true weight of an object can be measured in a vacuum, where buoyant forces are absent. A measurement in air, however, is disturbed by buoyant forces. An object of volume \(V\) is weighed in air on an equal-arm balance with the use of counterweights of density \(\rho\). Representing the density of air as \(\rho_{\text {air }}\) and the balance reading as \(F_{g}^{\prime}\), show that the true weight \(F_{g}\) is $$ F_{g}=F_{g}^{\prime}+\left(V-\frac{F_{g}^{\prime}}{\rho g}\right) \rho_{\text {air }} g $$
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