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Dafna11 [192]
1 year ago
5

64) Compare skeletal, smooth, and cardiac muscles as to their body location, microscopic anatomy, regulation of contraction, spe

ed of contraction, and rhythmicity.
65) What is the effect of aging on skeletal muscles?

66) Discuss the role of the myosin heads in sliding filament theory.
Physics
1 answer:
storchak [24]1 year ago
5 0
The answers are as follows:
64. SKELETAL MUSCLES
Body location: it is usually attached to the bone or to the skin.
Microscopic anatomy: it is made up of very long, cylindrical multinucleated cells which are striated.
Regulation of contraction: the nervous system controls the voluntary contraction of the skeletal muscles.
Speed of contraction: the speed of contraction ranges from slow to fast.
Rhythmicity: the skeletal muscle is arrhythmic.
SMOOTH MUSCLES
Body location: found in the wall of hollow visceral organs [not including those of the heart].
Microscopic anatomy: made up of single fusiform, uninucleated cells that are without striation.
Regulation of contraction: smooth muscles undergo involuntary contractions which are controlled by the nervous system and hormones.
Speed of contraction: very slow. it is the slowest of the three muscles.
Rhythmicity: rhythmic.
CARDIAC MUSCLES
Body location: located in the wall of the heart.
Microscopic anatomy: it is composed of branching chains of cells, that are uninucleated; they are striated and posses intercalated discs.
Regulation of contraction: Undergo involuntary contractions, which are controlled by nervous system, heart pacemarker and hormones.
Speed of contraction: slow.
Rhythmicity: rhythmic.

65. Aging brings about gradual loss in muscle functions. As one grows older, there are usually age related alterations in the skeletal muscle functions. The factors that affect the rate of muscle loss are sex and level of muscle activity. Loss of muscle mass also occurs as one grows older.

66. The sliding filament theory states that, during contraction the thin filaments slide past the thick filaments and the sacomere shortens.
During contraction, the myosin head attaches to the myosin binding site on the actin filament. Using energy from ATP, the myosin head move toward the center of the sacomere, attaching and detaching several times. As a result of this, the thin actin filament is pulled toward the center of the sacomere. This leads to the shorten of the muscle cells.
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Answer:

T=C*P*V

Explanation:

It is said that a variable - let's call 'y' -, is proportional to another - let's call it 'x' - if x and y are multiplicatively connected to a constant 'C'. It means that their product (x*y) can be always equaled to the constant 'C' or their division (\frac{x}{y}) can be always equaled to 'C'. The first case is the case of the inverse proportionality: It is said that x and y are inversely proportional if

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The second case is the case of the direct proportionality: It is said that x and y are directly proportional if

\frac{x}{y} =C : x is directly proportional to y.

or

\frac{y}{x} =C : y is directly proportional to x.

Always that any text does not specify about directly or inversely proportionality, it is assumed to mean directly automatically.

For our case, we are said that the temperature T is proportional to the pressure P and the volume V (we assume that it means directly); it is a double proportionality but follows the same rules:

If T were just proportional to P, we would have:

\frac{T}{P} =C

If T were just proportional to V, we would have:

\frac{T}{V} =C

As T is proportional to both P and V, the right equation is:

\frac{T}{P*V}=C

In order to isolate the temperature, let's multiply (P*V) at each side of the equation:

\frac{T}{P*V}*(P*V)=C*(P*V)\\T=C*P*V

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2 years ago
(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang
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4m at 1.2 radian means 4m at 68.75^0

4m at 68.75^0 can be written as 4*cos68.75^0i+4*sin68.75^0j = 1.45i+3.73j

6m at -210^0 can be written as 6*cos(-210)^0i+6*sin(-210)^0j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = \sqrt{1.274^2+0.6^2} = 1.98 m

c) Angle,  tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0

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2 years ago
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled
SVETLANKA909090 [29]

Answer:

The  value is  v =  -0.04 \  m/s

Explanation:

From the question we are told that

   The  mass  of the block is  m  =  2.0 \ kg

   The  force constant  of the spring is  k  =  590 \ N/m

   The amplitude  is  A =  + 0.080

   The  time consider is  t =  0.10 \  s

Generally the angular velocity of this  block is mathematically represented as

      w =  \sqrt{\frac{k}{m} }

=>   w =  \sqrt{\frac{590}{2} }

=>   w = 17.18 \  rad/s

Given that the block undergoes simple harmonic motion the velocity is mathematically represented as  

         v  =  -A w sin (w* t )

=>       v  = -0.080 * 17.18 sin (17.18* 0.10 )

=>       v =  -0.04 \  m/s

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1
notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is  F_{net}=  3.22*10^{-5} \ J

Explanation:

From the question we are told that

    The third charge is  q_3 =  55 nC =  55 *10^{-9} C

    The position of the third charge is  x = -1.220 \ m

     The first charge is q_1 =  -16 nC  =  -16 *10^{-9} \ C

     The position of the first charge is x_1 =  -1.650m

      The second charge is  q_2 =  32 nC  =  32 *10^{-9} C

      The position of the second charge is  x_2 =   0  \ m  

The distance between the first and the third charge is

      d_{1-3} =  -1.650 -(-1.220)

     d_{1-3} = -0.43 \ m

The force exerted on the third charge by the first is  

     F_{1-3} =  \frac{k  q_1 q_3}{d_{1-3}^2}

Where k is the coulomb's constant with a value  9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

substituting values

      F_{1-3} =  \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}

       F_{1-3} = 4.28 *10^{-5} \ N

 The distance between the second and the third charge is      

  d_{2-3} =  0- (-1.22)

   d_{2-3} =1.220 \ m

The force exerted on the third charge by the first is mathematically evaluated as

       F_{2-3} =  \frac{k  q_2 q_3}{d_{2-3}^2}

substituting values

       F_{2-3} =  \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}

       F_{2-3} =  1.06*10^{-5} N

The net force is

      F_{net} =  F_{1-3} -F_{2-3}

substituting values

    F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}

    F_{net}=  3.22*10^{-5} \ J

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Molodets [167]

Answer:

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Explanation:

Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.

Based on the aforementioned, let's analyze the statements in order of importance

b) True. Since the moon is material evaporated from Earth, its compassion is similar

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f) False. The moon's rotation and translation are equal has no relation to its formation phase

c) false. The amount of vaporized material on the moon is large

Therefore, the order of importance must be

         b e a f c

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