/*! 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 76 An Extraterrestrial Volcano The ... [FREE SOLUTION] | 91Ó°ÊÓ

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An Extraterrestrial Volcano The first active volcano observed outside the Earth was discovered in 1979 on lo, one of the moons of Jupiter. The volcano was observed to be ejecting material to a height of about \(2.00 \times 10^{5} \mathrm{m}\). Given that the acceleration of gravity on lo is \(1.80 \mathrm{m} / \mathrm{s}^{2}\), find the initial velocity of the ciected material.

Short Answer

Expert verified
The initial velocity is approximately 848 m/s.

Step by step solution

01

Understand the Variables

The problem provides the maximum height (\[ h = 2.00 \times 10^{5} \text{ m} \]) reached by the volcanic material and the acceleration due to gravity on Io (\[ g = 1.80 \text{ m/s}^{2} \]). We need to find the initial velocity (\[ v_0 \]) of the material.
02

Apply the Kinematic Equation

Use the kinematic equation for motion under uniform acceleration, which is:\[ v^2 = v_0^2 + 2gh \]Since the final velocity \( v \) at the maximum height is 0 (the material stops rising before it falls back down), this simplifies to:\[ 0 = v_0^2 - 2gh \]
03

Solve for Initial Velocity

Rearrange the equation to solve for the initial velocity \( v_0 \):\[ v_0^2 = 2gh \]Take the square root to find \( v_0 \):\[ v_0 = \sqrt{2gh} \]
04

Calculate the Initial Velocity

Substitute the known values into the equation:\[ v_0 = \sqrt{2 \times 1.80 \times 2.00 \times 10^5} \]Simplify inside the square root:\[ 2 \times 1.80 \times 2.00 \times 10^5 = 7.20 \times 10^5 \]Now, calculate:\[ v_0 = \sqrt{7.20 \times 10^5} \approx 848 \text{ m/s} \]

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Key Concepts

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

Acceleration Due to Gravity
Acceleration due to gravity is a fundamental concept in kinematics, which describes how objects fall or are projected in a gravitational field. On Earth, this acceleration is approximately
  • 9.81 m/s²
, but it varies on other celestial bodies. For instance, on Io, a moon of Jupiter, gravity is much weaker, at 1.80 m/s².
This implies that objects fall at a slower rate compared to Earth. Gravity on Io affects how high volcanic material can reach when ejected.
Understanding the local acceleration due to gravity is crucial when predicting the behavior of objects, such as determining how high they might rise or how fast they will fall. This is because it is directly involved in calculating the force exerted on the object and influences all kinematic calculations.
In equations, it is denoted as \( g \) and is used to determine other parameters like velocity and time when an object is in free fall or at the point of being projected.
Initial Velocity
Initial velocity, often represented by \( v_0 \), is the speed an object starts with when observed or projected onto a path. It is a crucial factor in determining the subsequent motion of an object under uniform acceleration.
In the context of Euclidean or extraterrestrial volcanos, like those on Io, determining the initial velocity of ejected materials helps in ascertaining how far or high they can travel.
If you know the initial velocity and the acceleration acting on an object (like gravity), you can forecast its journey using kinematic equations. In the provided exercise, the initial velocity was calculated by understanding how high the volcanic material traveled before coming back down.
Hence, initial velocity is not just a starting speed, but it dictates the trajectory and potential energy of an object in motion, as defined by its ability to counteract gravitational forces and inertia.
Kinematic Equations
Kinematic equations are essential tools in physics that describe the motion of objects under uniform acceleration. They allow you to calculate variables such as velocity, displacement, and time without needing complex calculus.
These equations arise out of Newton's laws of motion, and for an object moving in a straight line with constant acceleration, they are used in various forms:
  • \( v = v_0 + at \)
  • \( s = v_0 t + \frac{1}{2} a t^2 \)
  • \( v^2 = v_0^2 + 2as \)
Where, \( v \) is the final velocity, \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement. For the projectile motion on Io, we used the kinematic equation
\[v^2 = v_0^2 + 2gh\]assuming the final velocity at the top of the projectile is 0. This allows us to solve for \( v_0 \) and understand the needs of extraterrestrial projectile mechanics.
Kinematic equations enable predictions about motion in a gravity field and are prerequisite knowledge for exploring beyond Earth's atmosphere or even understanding everyday phenomena like throwing a ball.

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Most popular questions from this chapter

You drive your car in a straight line at \(15 \mathrm{m} / \mathrm{s}\) for \(10 \mathrm{kilometers},\) then at \(25 \mathrm{m} / \mathrm{s}\) for another \(10 \mathrm{kilome}^{-}\) ters. (a) Is your average speed for the entire trip more than, less than, or equal to \(20 \mathrm{m} / \mathrm{s} ?\) (b) Choose the best explanation from among the following: I. More time is spent at \(15 \mathrm{m} / \mathrm{s}\) than at \(25 \mathrm{m} / \mathrm{s}\). II. The average of \(15 \mathrm{m} / \mathrm{s}\) and \(25 \mathrm{m} / \mathrm{s}\) is \(20 \mathrm{m} / \mathrm{s}\) III. Less time is spent at \(15 \mathrm{m} / \mathrm{s}\) than at \(25 \mathrm{m} / \mathrm{s}\).

Two cars drive on a straight highway. At time \(t=0,\) car 1 passes mile marker 0 traveling due east with a speed of \(20.0 \mathrm{m} / \mathrm{s} .\) At the same time, car 2 is \(1.0 \mathrm{km}\) east of mile marker 0 traveling at \(30.0 \mathrm{m} / \mathrm{s}\) due west. Car 1 is speeding up with an acceleration of magnitude \(2.5 \mathrm{m} / \mathrm{s}^{2}\), and car 2 is slowing down with an acceleration of magnitude \(3.2 \mathrm{m} / \mathrm{s}^{2}\) (a) Write \(x\) -versust equations of motion for both cars, taking east as the positive direction. (b) At what time do the cars pass next to one another?

The position of a particle as a function of time is given by \(x=(-2.00 \mathrm{m} / \mathrm{s}) t+\left(3.00 \mathrm{m} / \mathrm{s}^{3}\right) t^{3}\) (a) Plot \(x\) versus \(t\) for time from \(t=0\) to \(t=1,00 \mathrm{s}\) (b) Find the average velocity of the particle from \(t=0.150 \mathrm{s}\) to \(t=0.250 \mathrm{s},\) (c) Find the average velocity from \(t=0.190 \mathrm{s}\) to \(t=0.210 \mathrm{s}\) (d) Do you expect the instantaneous velocity at \(t=0.200 \mathrm{s}\) to be closer to \(-1.62 \mathrm{m} / \mathrm{s},\) or \(-1.66 \mathrm{m} / \mathrm{s}\) ? Explain.

IP Standing at the edge of a cliff \(32.5 \mathrm{m}\) high, you drop a ball. Later, you throw a second ball downward with an initial speed of \(11.0 \mathrm{m} / \mathrm{s}\). (a) Which ball has the greater increase in speed when it reaches the base of the cliff, or do both balls speed up by the same amount? (b) Verify your answer to part (a) with a calculation.

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