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sleet_krkn [62]
3 years ago
5

Sandra wanted to see how fast she could kick a soccer ball. She recorded her results below. Unfortunately, she forgot to record

her time during her third kick. Fill in the chart below for Sandra

Physics
1 answer:
alexira [117]3 years ago
5 0
Kick | Distance (M) | Time (s) | Average Speed

1. 55. 5.0. 11

2. 50. 5.0. 10

3. 20. 3.0. 10
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Which idea is associated with Copernicus? Select one: a. The orbits of the planets are circles. b. The orbits of the planets are
Schach [20]

Answer:

d. The earth rotates around the sun

Explanation:

  • Nicolas Copernicus is considered the first person to give the theory of heliocentric, or Sun-centered system of our planetary system.
  • In the heliocentric system, it is considered that the sun is stationary and the earth revolves around the sun.
  • He stated that the sun is at the center of the universe and the earth spins on its axis once daily and revolves around the sun in one year.
  • And today we know it is correct that earth rotates on its own axis and also revolves around the sun.

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Which of the following best characterizes the field of physics
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A teacher wants to perform a classroom demonstration that illustrates both chemical and physical changes. Which would be the bes
anygoal [31]

Answer

D) burning a candle

Explanation

When burning a candle no new substance is form.

We have both physical and chemical change occuring.

Physical part: Melting of the solid wax and evaporation of the liquid forms the physical change.

Chemical part: burning of the wax vapour forms the chemical change.

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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.

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3 years ago
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Answer:

Scientific method

Explanation:

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