1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sav [38]
3 years ago
12

Question: How does the total amount of energy present before an energy transformation compare to the total amount of energy pres

ent afterward?
A: The energy before is equal to the energy afterward.

B: The mass before is equal to the mass afterward.

C: The energy before is greater than the energy afterward.

D: The energy afterward is greater than the energy before.
Physics
2 answers:
sasho [114]3 years ago
7 0

Answer: A

Explanation: i took the test

juin [17]3 years ago
5 0
The correct option is: (A) <span>The energy before is equal to the energy afterward.

Explanation:
According to the law of conservation of Energy, energy can neither be created nor destroyed; it can only be transformed into one form or another. It means that the total initial energy must be equal to the total final energy of the system. By considering this law, we can infer that the energy before is equal to the energy afterward.</span>
You might be interested in
As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz
Eddi Din [679]

a)E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = \frac{1}{2}kx² +  \frac{1}{2}mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) <em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude  is 'A'

\omega=\sqrt{\frac{k}{m}} is the angular frequency of the motion

t is the time

\phi is the phase (we can take \phi=0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K=\frac{1}{2}mv^2=0

E=\frac{1}{2}kA^2\\ -->(1)

<em>A=</em>\sqrt{\frac{2E}{k}}<em> </em>

c)v_{max}=\omega A<u></u>

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E=\frac{1}{2}mv_{max}^2

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2

and solving for v_{max}we find an expression for the maximum speed:

v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A

<h2><u></u>v_{max}=\omega A<u></u></h2>
4 0
3 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
Arisa [49]

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

              = 1/2 (6 X 15)

              = 1/2 (90)

              = 45 meters

Hence, the object traveled 45 meters.

4 0
3 years ago
A star is a ___ that emits energy produced by nuclear reactions in its interior.
grigory [225]

A star is a large ball of gas that emits energy produced by nuclear reactions in the star's interior. Much of this energy is emitted as electromagnetic radiation, including visible light. Light emitted by stars enables other objects in the universe to be seen by reflection.

5 0
3 years ago
A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves
KIM [24]

Answer:A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

Explanation:

A piece of driftwood moves up and down as water waves pass beneath it. However, it does not move toward the shore with the waves. What does this demonstrate about the propagation of waves through a medium?

A) Waves transmit energy but not matter as they progress through a medium.

B) Waves transmit matter but not energy as they progress through a medium.

C) Waves do not transmit matter or energy as they progress through a medium.

D) Waves transmit energy as well as matter as they progress through a medium.

4 0
3 years ago
Energy Conservation With Conservative Forces: If a spring-operated gun can shoot a pellet to a maximum height of 100 m on Earth,
crimeas [40]

Answer:

h' = 603.08 m

Explanation:

First, we will calculate the initial velocity of the pellet on the surface of Earth by using third equation of motion:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity on the surface of earth = - 9.8 m/s² (negative sign due to upward motion)

h = height of pellet = 100 m

Vf = final velocity of pellet = 0 m/s (since, pellet will momentarily stop at highest point)

Vi = Initial Velocity of Pellet = ?

Therefore,

(2)(-9.8 m/s²)(100 m) = (0 m/s)² - Vi²

Vi = √(1960 m²/s²)

Vi = 44.27 m/s

Now, we use this equation at the surface of moon with same initial velocity:

2g'h' = Vf² - Vi²

where,

g' = acceleration due to gravity on the surface of moon = 1.625 m/s²

h' = maximum height gained by pellet on moon = ?

Therefore,

2(1.625 m/s²)h' = (44.27 m/s)² - (0 m/s)²

h' = (1960 m²/s²)/(3.25 m/s²)

<u>h' = 603.08 m</u>

4 0
3 years ago
Other questions:
  • A 9.0-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introd
    8·1 answer
  • what is the acceleration of a bowling ball that starts at rest and moves 300m down the gutter in 22.4 sec
    10·1 answer
  • Which of the following is correctly describes a radio wave
    8·2 answers
  • A driver in a car traveling at a speed of 26.8 m/s sees a deer 100 m away on the road. Calculate the minimum constant accelerati
    11·1 answer
  • A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0
    6·1 answer
  • What is the least possible initial kinetic energy kmin the oxygen atom could have and still excite the cesium atom?
    9·1 answer
  • A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu
    11·1 answer
  • A man wishes to travel due north in order to cross a river 5 kilometers wide flowing due East at 3 kilometers per hour . if he c
    9·1 answer
  • PLEASE HELP 100 POINTS!!!!!
    5·2 answers
  • In terms of energy transfer, how is a rise in
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!