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sergejj [24]
3 years ago
9

A 3.0-kg block being pulled across a table by a horizontal

Physics
1 answer:
zlopas [31]3 years ago
6 0
Acceleration = 75/3 = 25 m/s^2
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A student needs to prepare 100.0 mL of vitamin C solution according to the directions in Part 1 of the experiment. S/he weighs o
Rudiy27

Answer:

4.99 mg of vitamin C are in the beaker.

Explanation:

Given that,

Weight of vitamin = 0.0499 g

Molar mass = 176.124 g/mol

Weight of water = 100.0 ml

We need to calculate the mg of vitamin C in the beaker

We dissolve 0.0499 g vitamin C in water to from 100.0 ml solution.

100 ml solution contain 49.9 mg vitamin C

Now, we take 10 ml of this vitamin C solution in breaker

Since, 100 ml solution =49.9 mg vitamin C

Therefore,

10\ ml\ solution =\dfrac{10\times49.9}{100}

10\ ml\ solution =4.99\ mg

Hence, 4.99 mg of vitamin C are in the beaker.

4 0
3 years ago
Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en
Blizzard [7]

Answer:

c. 2 m/s

Explanation:

The speed of a wave is given by:

v=f \lambda

where

v is the speed of the wave

f is the frequency

\lambda is the wavelength

For the wave in this problem, we have

f = 4 Hz is the frequency

\lambda = 0.5 m is the wavelength

So, the speed of the wave is

v=(4 Hz)(0.5 m)=2 m/s

5 0
4 years ago
Please help I cannot fail!
alekssr [168]
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4 0
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
4 years ago
Use examples to explain how the geosphere interacts with two other of Earth's spheres. Explain the interaction for each using co
Orlov [11]

The geosphere interacts with the hydrosphere when water causes rock to erode. The atmosphere provides the geosphere with heat and energy for erosion, and the geosphere reflects the sun's energy back into the atmosphere.

7 0
3 years ago
Read 2 more answers
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