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2 years ago

What are the three types of muscle fibers?

2 answers:

7 0

Within skeletal muscle there are three types of fiber.Type one (I or red/slow, slow twitch fibres), type two A (IIa or red fast, fast oxidative fibres) and type two B (IIb or white fast, fast glycolytic fibres). Each fiber types has different qualities in the way they perform and how quickly they fatigue.

I hope this helps:)

uysha [10]2 years ago

7 0

Within skeletal muscle there are three types of fiber.Type one (I or red/slow, slow twitch fibres), type two A (IIa or red fast, fast oxidative fibres) and type two B (IIb or white fast, fast glycolytic fibres). Each fiber types has different qualities in the way they perform and how quickly they fatigue.

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We divide the electromagnetic spectrum into six major categories of light, listed below. Rank these forms of light from left to

makkiz [27]

electromagnetic spectrum is consisting of many frequency range which is from gamma rays to radio waves

they are of various wavelength and different energy levels

minimum wavelength will occurs at Gamma rays

and maximum wavelength at Radio waves

the list of increasing order of wavelength is as following

Gamma rays < X rays < Ultraviolet < Visible Light < Infrared Waves < Radio Waves

so least to maximum order is

1. Gamma rays

2. X rays

3 Ultraviolet

4 Visible light

5 Infrared waves

6 Radio waves

5 0

2 years ago

Why does the evaporation of sweat cool down your skin?

Ne4ueva [31]

So sweat helps cool you down two ways. First, it makes your skin feel cooler when it's wet. And when it evaporates it removes some heat. But sweat will only evaporatoin an environment where there isn't much water in the air.

6 0

2 years ago

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Help ASAP please (: The waves with the MOST energy have

tino4ka555 [31]

Answer: C

high; large

Explanation:

The wave energy is related to its amplitude and frequency.

The wave energy is proportional to the amplitude of the wave. So, wave with the most energy will have high amplitude.

Also, frequency is related to wave energy. The larger the frequency, the more the energy of the wave.

Therefore, The waves with the MOST energy have high amplitudes and large

frequencies.

5 0

2 years ago

A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh

ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

θ = 273.4 º

b) Kf = 37% of Ko

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved.
Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

$p_{o1x} = 2.00 kg*m/s (1)$

$p_{o1y} = 0 (2)$

We can do the same for the particle moving along the positive y-axis:

$p_{o2x} = 0 (3)$

$p_{o2y} = 4.00 kg*m/s (4)$

Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

$p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)$

$p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s (6)$

Now, the total initial momentum, along these directions, must be equal to the total final momentum.
We can write the equation for the x- axis as follows:

$p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x} (7)$

We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

$p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)$

Now, we can repeat exactly the same process for the y- axis, as follows:

$p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y} (9)$

We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

$p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)$

Since we have the x- and y- components of the final momentum of the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

$p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} } = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)$

We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)$

The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

$\frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)$

So, the final kinetic energy has lost a 37% of the initial one.

6 0

2 years ago

The relationship among mass force and acceleration is explained by

Anettt [7]

Newton's second law of motion. F = m a .

7 0

2 years ago

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