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3 years ago

Why are slow twitch muscles more beneficiall than fast twitch mucles for cardiorespiratory fitness

Physics

2 answers:

Dmitriy789 [7]3 years ago

4 0

Answer:

otherwise known as option D

Explanation:

Lady_Fox [76]3 years ago

4 0

Answer:

D Slow-twitch muscles are able to use oxygen more efficiently than fast-twitch muscles.

is the answer

Explanation:

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What scenario involves kinetic Energy transform into potential energy

Vinil7 [7]

The object which is projected upwards. The kinetic energy imparted as its launch is converted into potential energy as an object Rises until, at the peak of its motion, all the kinetic energy has been converted to additional potential energy

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3 years ago

What is the independent variable? What is the dependent variable? Please help!

e-lub [12.9K]

graphic too wide.

indep variable is one you control

dep var is what happens

7 0

3 years ago

Date: 12-3-21<br> 4.<br> The momentum of a 5-kilogram object moving at<br> 6 meters per second is

ASHA 777 [7]

Answer

30 kg . m/sec

Explanation:

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2 years ago

Read 2 more answers

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh

Goshia [24]

Answer:

a) $R_s = 0.092$

b) $R_p = 0.085$

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

$R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2$

using snell's law to calculate θ t

$sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}$

$cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}$

a) $R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2$

$R_s = 0.092$

b) $R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2$

$R_p = 0.085$

3 0

3 years ago