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BlackZzzverrR [31]
2 years ago
12

Why is it important to assess flexibility both before and during a workout regimen? a. to assess flexibility progress b. to keep

muscles warm c. to establish flexibility baselines d. none of the above please select the best answer from the choices provided. a b c d
Physics
2 answers:
Inga [223]2 years ago
8 0

Answer:

A

Explanation:

lidiya [134]2 years ago
5 0

It important to assess flexibility both before and during a workout regimen because to assess flexibility progress.

<h3>What is a flexibility exercise?</h3>

Flexibility exercises, as the name implies, help to increase the stretch of the muscles, improving the range of motion. More flexibility is also important to help prevent possible post-workout soreness.

In this case we have that the reason to continue doing stretching before and during the exercises is about  to assess flexibility progress.

See more about flexibility at brainly.com/question/15395713

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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
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Stella [2.4K]

Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

F = \frac{kq_1q_2}{r^2}

here we know that

q_1 = +8.4\mu C

q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

so it is separated by 80 cm distance

3 0
3 years ago
Read 2 more answers
A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume
olchik [2.2K]

A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume increases to 30.0 ml, at constant temperature is 0.25atm.

On constant temperature, the pressure and volume relation become constant before and after the change in quantitities have occurred.

According to Boyle's Law,

P₁V₁ = P₂V₂

where, P₁ is pressure exerted by the gas initially

V₁ is the volume of gas initially

P₂ is pressure exerted by the gas finally

V₂ is the volume of gas finally

Given,

P₁ = 1.5 atm

V₁ = 5 ml

V₂ = 30 ml

P₂ =?

On substituting the given values in the above equation:

P₁V₁ = P₂V₂

1.5 atm × 5 ml = P₂ × 30 ml

P₂ = 0.25 atm

Hence, pressure exerted by the gas is 0.25atm.

Learn more about Boyle's Law here, brainly.com/question/1437490

#SPJ4

8 0
2 years ago
The moon's surface becomes hot during the long lunar day because the sun transfers heat to the moon
In-s [12.5K]
Radiation would be the answer (:
7 0
3 years ago
When liquid water gets into cracks of rock and freezes, it __ and ___.
hoa [83]
It expands and pushes the crack further aprt
3 0
2 years ago
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