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Evgesh-ka [11]
3 years ago
6

What can destroy stuff yet does not have any mass

Physics
1 answer:
Artist 52 [7]3 years ago
3 0
If energy could<span> be created or </span>destroyed<span>, all of our ideas of how the world works ... Historically, of course </span>not<span> all the forms of energy were known to begin with. ... too messy or complicated to make sense, we </span>would have<span> had to give up the law. ... </span>can<span> be converted into rest </span>mass<span> and back again (particle physicists </span>do<span> this </span><span>every )</span>
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Why is physics important for high school students to learn?
Bond [772]

It is important for people to learn physics so that they can better understand the basic laws of the universe, and most importantly, so we can look at things with common sense, for you my friend i recommend watching <em>mythbusters</em>, they teach physics and make learning fun they teach you to look at the world with common sense. hope i helped!!! have a good day.


7 0
3 years ago
Of all of the types of forest biomes, tropical rain forests contain the most biodiversity, even though they do not have the most
BlackZzzverrR [31]
False. They do tend to have fertile soil do to the animal fecal matter.
3 0
3 years ago
Read 2 more answers
A force is applied to an ideal spring (initially in its equilibrium position) and does 1.9 JJ of work stretching it 2.2 cmcm . H
Verdich [7]

Answer:

W=16.58J

Explanation:

initial information we have

work: W=1.9J

stretched distance: x=2.2cm=0.022m

from this, we can find the value of the constant of the spring k, with the equation for work in a spring:

W=\frac{1}{2} kx^2

substituting known values:

1.9J=\frac{1}{2}k(0.022)^2\\

and clearing for k:

k=\frac{2(1.9J)}{0.022^2} \\k=7,851.24

and now we want to know how much work is done when we stretch the spring a distance of 6.5cm from equilibrium, so now x is:

x=6.5cm=0.065m

and using the same formula for work, with the value of k that we just found:

W=\frac{1}{2} kx^2

W=\frac{1}{2}(7851.24)(0.065)^2\\W=16.58J

5 0
3 years ago
Read 2 more answers
According to Newton's Second Law of Motion, an object will accelerate if you apply what kind of force? Question 1 options: Frict
Bumek [7]

An unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

<h3>What does Newton's Second Law of Motion state?</h3>

It states that the force applied to the object is equal to the product of mass and acceleration.

F = ma

  • An object will accelerate when the net force applied on the object is more than zero or unbalanced.
  • The acceleration is the change in the direction or speed of the object. To achieve acceleration the force must be greater in a direction.

  • When force is greater in one the object move in that direction which is known as acceleration.

Therefore, an unbalanced force is required to accelerate an object according to Newton's Second Law of Motion.

Learn more about Newton's Second Law of Motion.:

brainly.com/question/25810165

5 0
2 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
3 years ago
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