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Kitty [74]
3 years ago
11

Acceleration of a body is 10m/s² , what does it mean?​

Physics
2 answers:
GREYUIT [131]3 years ago
8 0
20 m/s^2 means that every second the velocity increases by 10 m/s. ie, 0,10,20,30,40 etc
Ganezh [65]3 years ago
7 0

Answer:

It is actually an answer to a question. It has to do with acceleration due to gravity. To understand it more, solve work, energy and time. Search it online. God will help u

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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.

4 0
3 years ago
The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use
saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

M = \frac{Nl}{f_ef_0}

Where

N = Near point

l = distance between the object lens and eye lens

f_0= Focal length

f_e= Focal of eyepiece

Given that the minimum distance at which the eye is able to focus is about 25cm we have that N = 25cm

Replacing,

M = \frac{25*10}{3*5}

M = 16.67\approx 17\\

Therefore the correct answer is C.

3 0
3 years ago
Which formula correctly expresses the property density?
Olenka [21]

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

6 0
2 years ago
A lightbulb has a power of 100 W and is used for 4 hours. A microwave has a power of 1200 W and is used for 5 minutes. How much
ladessa [460]

Answer:

a. E=1440KJ

b. E=360KJ

c. E=1.8 J

Explanation:

I have the power (Watts) is expressed as Energy (Joules) / Time (seconds), also I have to1hour * \frac{60min}{1hour}*\frac{60s}{1min}\\1h=3600s

so, for the lightbulb

100W= \frac{Energy}{3600s*4}\\Energy=100W*14400s\\ E=1440000J =1440KJ

Analogously, for the microwave

1200W= \frac{Energy}{60s*5}\\Energy=1200W*300s\\ E=360000J =360KJ

Now, I have to express the efficiency as Heat energy / power * 100 so1.8= \frac{Heat Energy}{100W}*100\\Heat Energy=1.8J

Done

4 0
3 years ago
A pendulum Bob released from some initial height such that the speed of the bob at the bottom of the swing is 1.9m/s. What is th
Dmitriy789 [7]

Answer:

h = 18.4 cm

Explanation:

Given that,

The speed of the bob at the bottom of the swing is 1.9m/s.

We need to find the initial height of the bob. Let it is h.

We can find it using the conservation of energy i.e.

mgh=\dfrac{1}{2}mv^2

Where

v is speed of the bob

So,

h=\dfrac{v^2}{2g}\\\\h=\dfrac{(1.9)^2}{2\times 9.8}\\\\h= 0.184\ m

or

h = 18.4 cm

So, the initial height of the bob is 18.4 cm.

8 0
2 years ago
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