/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 A 15-kg rock is dropped from res... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 47

A 15-kg rock is dropped from rest on the earth and reaches the ground in 1.75 \(\mathrm{s}.\) When it is dropped from the same height on Saturn's satellite Enceladus, it reaches the ground in 18.6 \(\mathrm{s}.\) What is the acceleration due to gravity on Enceladus?

Short Answer

Expert verified
The acceleration due to gravity on Enceladus is 尘/蝉虏.

Step by step solution

01

Understand the Problem

We need to calculate the acceleration due to gravity on Enceladus, given the time it takes for a rock to fall. We know the time it takes to fall on Earth and the height from which the rock is dropped is the same.
02

Use the Formula for Free Fall on Earth

On Earth, the time taken for an object to fall from rest is given by the formula \( t = \sqrt{\frac{2h}{g}} \), where \( g \) is the acceleration due to gravity \( 9.81 \, \mathrm{m/s^2} \), \( t \) is the time \(1.75 \, \mathrm{s}\), and \( h \) is the height. Rearrange to find \( h = \frac{1}{2}gt^2 \).
03

Calculate the Falling Height on Earth

Using the formula from Step 2, substitute \( g = 9.81 \, \mathrm{m/s^2} \) and \( t = 1.75 \, \mathrm{s} \):\[ h = \frac{1}{2} \times 9.81 \, \mathrm{m/s^2} \times (1.75 \, \mathrm{s})^2 \]Calculate \( h \) to get the height in meters.
04

Apply the Free Fall Formula on Enceladus

On Enceladus, the rock takes \( 18.6 \, \mathrm{s} \) to fall from the same height \( h \). Use the formula: \( t = \sqrt{\frac{2h}{g_{enc}}} \). Rearrange to find \( g_{enc} = \frac{2h}{t^2} \).
05

Calculate the Acceleration on Enceladus

Substitute the calculated \( h \) from Step 3 into the formula from Step 4 and solve for \( g_{enc} \) using \( t = 18.6 \, \mathrm{s}\):\[ g_{enc} = \frac{2h}{(18.6 \, \mathrm{s})^2} \].
06

Simplify and Solve for \( g_{enc} \)

Substitute the calculated height \( h \) from Step 3 into the formula and compute:\[ g_{enc} = \frac{2 \times }{(18.6)^2} \]This will give the acceleration due to gravity on Enceladus in \( \mathrm{m/s^2} \).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Free Fall
In physics, free fall refers to the motion of an object under the influence of gravitational force alone, without any resistance, such as air resistance, acting on it. This idealized situation helps simplify calculations in physics problems. When an object is in free fall:
  • It is only acted upon by gravity.
  • The initial velocity is often zero when it starts from rest.
  • The object's motion can be described using the kinematic equations of motion.
For instance, the formula to calculate the time, \( t \, \) it takes for an object to fall from a certain height \( h \, \) when dropped from rest is \( t = \sqrt{\frac{2h}{g}} \, \). This equation represents the balance between the gravitational pull and the inertia of the object.
Enceladus
Enceladus is one of Saturn's intriguing moons, known for its icy surface and geysers that eject water into space. In physics problems like the one given, Enceladus captures interest due to its lower gravity compared to Earth.
  • The gravitational acceleration on Enceladus, denoted as \( g_{enc} \, \), is much less than that on Earth, affecting the free fall time of objects.
  • When analyzing free fall on Enceladus, the same principles apply, yet the different gravitational force alters the results significantly.
Such variations in gravity across celestial bodies are crucial for astronomers and physicists, as they can influence the potential for human exploration and habitation.
Physics Problem Solving
Solving physics problems involves a step-by-step approach to understanding and applying scientific principles. Here's how to approach such problems:
  • Understand the problem: Identify what is known and what needs to be found.
  • Choose the right formulas: Apply equations that are relevant to the scenario, such as the kinematic equations for free fall.
  • Perform calculations: Replace variables with known values and solve for unknown quantities.
  • Check your work: Ensure that answers are reasonable and units are consistent.
In the original problem, calculating the gravitational acceleration on Enceladus involves:
  1. Determining the height from Earth's free fall data.
  2. Using that height to find the acceleration due to gravity on Enceladus.
This systematic approach helps in tackling any physics problem with confidence and precision.
Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause such movement. It uses equations to describe how an object moves through space and time.
  • Key equations in kinematics relate to displacement, velocity, acceleration, and time.
  • Kinematic equations can solve complex problems related to the motion of objects under various conditions.
In free fall scenarios, like the one involving Enceladus, kinematics simplifies the motion description by assuming only gravitational forces come into play. The equation \( h = \frac{1}{2}gt^2 \, \) is an example of using kinematic principles to calculate height from Earth's common values of acceleration and time.Learning how to use these equations effectively can aid in solving not only textbook exercises but also real-world physical phenomena.

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

An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?

Two cars start 200 \(\mathrm{m}\) apart and drive toward each other at a steady 10 \(\mathrm{m} / \mathrm{s} .\) On the front of one of them, an energetic grasshopper jumps back and forth between the cars (he has strong legs!) with a constant horizontal velocity of 15 \(\mathrm{m} / \mathrm{s}\) relative to the ground. The insect jumps the instant he lands, so he spends no time resting on either car. What total distance does the grasshopper travel before the cars hit?

Trip Home. You normally drive on the freeway between San Diego and Los Angeles at an average speed of 105 \(\mathrm{km} / \mathrm{h}\) (65 \(\mathrm{mi} / \mathrm{h} ),\) and the trip takes 2 \(\mathrm{h}\) and 20 min. On a Friday afternoon, however, heavy traffic slows you down and you drive the same distance at an average speed of only 70 \(\mathrm{km} / \mathrm{h}(43 \mathrm{mi} / \mathrm{h})\) How much longer does the trip take?

On a 20 -mile bike ride, you ride the first 10 miles at an average speed of 8 \(\mathrm{mi} / \mathrm{h} .\) What must your average speed over the next 10 miles be to have your average speed for the total 20 miles be (a) 4 \(\mathrm{mi} / \mathrm{h} ?\) (b) 12 \(\mathrm{mi} / \mathrm{h} ?\) (c) Given this average speed for the first 10 miles, can you possibly attain an average speed of 16 \(\mathrm{mi} / \mathrm{h}\) for the total 20 -mile ride? Explain.

The acceleration of a bus is given by \(a_{x}(t)=\alpha t\) , where \(\alpha=1.2 \mathrm{m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{s}\) is 5.0 \(\mathrm{m} / \mathrm{s},\) what is its velocity at time \(t=2.0 \mathrm{s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{s}\) is \(6.0 \mathrm{m},\) what is its position at time \(t=2.0 \mathrm{s} ?(\mathrm{c}) \operatorname{Sketch} a_{x}-t, v_{x}-t,\) and \(x\) -t graphs for the motion.
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Physics of Motion

Read Explanation

Nuclear Physics

Read Explanation

Turning Points in Physics

Read Explanation

Thermodynamics

Read Explanation

Radiation

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.