Energy Dissipated: Calculate Rate in Joules/sec

  • Thread starter lexia
  • Start date
  • Tags
    Energy
In summary: Last edited by a moderator: May 8, 2017Hello @lexia, in summary, a measuring instrument with a mass of 2.2 kg is attached to a balloon for atmospheric studies. It is released from the balloon and falls to Earth, with the acceleration at any instant before the parachute opens being given by a= g exp ^(-t/R), where g is the acceleration due to gravity, e is the base of natural logarithms, and t is a time constant. The instrument's heat due to air resistance is the primary concern, and the rate of energy dissipation before the parachute opens can be expressed as a function of t, with t = 0 at release.
  • #1
lexia
1
1
New user has been reminded to show their work on schoolwork questions
Homework Statement
A 2.2-kg measuring instrument is mounted on a balloon by your scientific team for atmospheric studies. At the top of its flight, the instrument is released from the balloon and falls most of the way back to Earth before a parachute opens. You are told that the magnitude of the acceleration at any instant t before the parachute opens is given by a= g exp ^(-t/R), where g is the acceleration due to gravity, e is the base of natural logarithms, and t is a time constant that depends on the shape of the instrument and in this case is 5.68 s. Your primary concern is how much the instrument heats up as it falls, due to air resistance. At what rate, in joules per second, is energy dissipated before the parachute opens? Express your answer as a function of t, where t = 0 at release. (Hint: Integrate the acceleration to calculate speed and displacement.)
Relevant Equations
Kinetic energy = (m(V^2))/2
A 2.2-kg measuring instrument is mounted on a balloon by your scientific team for atmospheric studies. At the top of its flight, the instrument is released from the balloon and falls most of the way back to Earth before a parachute opens. You are told that the magnitude of the acceleration at any instant t before the parachute opens is given by a= g exp ^(-t/R), where g is the acceleration due to gravity, e is the base of natural logarithms, and t is a time constant that depends on the shape of the instrument and in this case is 5.68 s. Your primary concern is how much the instrument heats up as it falls, due to air resistance. At what rate, in joules per second, is energy dissipated before the parachute opens? Express your answer as a function of t, where t = 0 at release. (Hint: Integrate the acceleration to calculate speed and displacement.)
 
Physics news on Phys.org
  • #2

1. What is energy dissipation?

Energy dissipation is the process of converting energy from one form to another, usually resulting in the loss of some of the original energy. This can occur through various mechanisms such as friction, heat transfer, or electrical resistance.

2. Why is it important to calculate the rate of energy dissipation?

Calculating the rate of energy dissipation allows us to understand how quickly energy is being lost in a system. This can be important in determining the efficiency of a process or identifying potential sources of energy loss.

3. How is energy dissipation measured?

Energy dissipation is typically measured in joules per second (J/s), also known as watts. This unit represents the rate at which energy is being dissipated or transferred.

4. What factors affect the rate of energy dissipation?

The rate of energy dissipation can be affected by various factors such as the type of material, the surface area of contact, the speed of movement, and the temperature. These factors can impact the amount of friction or resistance present, which in turn affects the rate of energy dissipation.

5. How can we reduce energy dissipation?

There are several ways to reduce energy dissipation, such as using lubricants to reduce friction, insulating materials to minimize heat transfer, or using more efficient equipment. Understanding the sources of energy dissipation and implementing strategies to reduce it can lead to improved efficiency and cost savings.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
4K
Replies
20
Views
678
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
6K
Back
Top