Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 31

A tank of oil has a mass of 25 slugs. (a) Determine its weight in pounds and in newtons at the Earth's surface. (b) What would be its mass (in slugs) and its weight (in pounds) if located on the moon's surface where the gravitational at raction is approximately one-sixth that at the Earth's surface?

Short Answer

Expert verified
The weight of the oil tank on Earth is 805 lb or 3580 N. Its weight on the Moon would be 134.25 lb, and its mass, on both the Earth and the Moon, is 25 slugs.

Step by step solution

01

Calculate weight on Earth

First, calculate the weight of the oil tank on Earth. The formula to calculate weight is \( W = m × g \) where \( m \) is the mass and \( g \) is the gravitational constant. On Earth, \( g = 32.2 ft/s² \). So, \( W = 25 slugs × 32.2 ft/s² = 805 lb \).
02

Convert pounds to newtons

To convert pounds to newtons, use the conversion factor \( 1 lb = 4.44822 N \). So, \( 805 lb = 805 × 4.44822 N = 3580 N \). That's the weight of the oil tank on Earth in newtons.
03

Calculate mass on the Moon

The mass of the object doesn't change, irrespective of its location. So, the mass of the oil tank on the Moon remains the same as Earth, which is 25 slugs.
04

Calculate weight on the Moon

The gravitational constant on the Moon is approximately one-sixth of that on Earth, i.e., \( g_{moon} = 32.2 ft/s²/6 = 5.37 ft/s² \). So the weight of the oil tank on the Moon would be \( W_{moon} = m × g_{moon} = 25 slugs × 5.37 ft/s² = 134.25 lb \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Often the assumption is made that the flow of a certain fluid can be considered as incompressible flow if the density of the fluid changes by less than \(2 \%\). If air is flowing through a tube such that the air pressure at one section is 9.0 psi and at a downstream section it is 8,6 psi at the same temperature, do you think that this flow could be considered an incompressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressire.

At what atmospheric pressure will water boil at \(35^{\circ} \mathrm{C} ?\) Express your answer in both SI and BG units.

The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed of sound in the fluid. For an airplane flying at velocity \(V\) in air at absolute temperature \(T,\) the Mach number Ma is \\[ \mathrm{Ma}=\frac{V}{\sqrt{k R T}} \\] where \(k\) is a dimensionless constant and \(R\) is the specific gas constant for air. Show that Ma is dimeasionless.

The force, \(F\), of the wind blowing against a building is given by \(F=C_{D} \rho V^{2} A / 2,\) where \(V\) is the wind speed, \(\rho\) the density of the air, \(A\) the cross-sectional area cf the building, and \(C_{D}\) is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.

For a parallel plate arrangement of the type shown in Fig. 1.5 it is found that when the distance between plates is \(2 \mathrm{mm}\), a shearing stress of 150 Pa develops at the upper plate when it is pulled at a velocity of \(1 \mathrm{m} / \mathrm{s}\). Determine the viscosity of the fluid between the plates. Express your answer in SI units.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Thermodynamics

Read Explanation

Wave Optics

Read Explanation

Fundamentals of Physics

Read Explanation

Turning Points in Physics

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free