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ioda
3 years ago
7

The acceleration of gravity on the moon is about 1.6 m/sec2. An experiment to test gravity compares the time it takes objects to

reach a speed of 10 m/sec after being dropped from rest. How long does an object dropped on the moon have to fall compared to an object dropped on Earth?
a. 1.6 times as long
b. 6.1 times as long
c. 9.8 times as long
d. 15.7 times as long
Physics
1 answer:
wlad13 [49]3 years ago
3 0
I really think that the answer A
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Black_prince [1.1K]

Answer:

The answer to your question is given below

Explanation:

From the question given above, we can see that the wave with a higher frequency has a shorter wavelength while that with a lower frequency has a longer wavelength. This is so because the frequency and wavelength of a wave has inverse relationship. This can further be explained by using the following formula:

Velocity = wavelength x frequency

Divide both side by wavelength

Frequency = Velocity /wavelength

Keeping the velocity constant, we have:

Frequency ∝ 1 / wavelength

From the above illustration, we can see clearly that the frequency and wavelength are in inverse relationship. This implies that the higher the frequency, the shorter the wavelength and the shorter the frequency, the higher the wavelength.

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2 years ago
Where is the planet moving faster?
valentinak56 [21]

Answer:

I looked it up but got perihelion so I don't know if that will help at all but try um.... I don't know try L

Explanation:

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3 years ago
What is the term for the process by which a portion of a glacier breaks off and falls into the water
zhenek [66]

The term for the process by which a portion of a glacier breaks off and falls into the water is called calving.

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3 years ago
When do sunspots disappear?
dedylja [7]

Answer:

In 5 years or so, the sun will be awash in sunspots and more prone to violent bursts of magnetic activity.

Explanation

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3 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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