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dexar [7]
2 years ago
7

Which of the following accurately describes the way in which a muscle moves?

Physics
1 answer:
Vaselesa [24]2 years ago
7 0
<h3><u>Answer</u>;</h3>

B. When actin filaments are pulled toward the center of the sarcomere, the fiber shortens.

<h3><u>Explanation;</u></h3>
  • <em><u>The events of muscle fiber shortening occurs with in the sacromeres in the fibers. </u></em>
  • <em><u>Contraction of striated muscle fibers takes place as the sacromeres shorten as myosin heads pull on the actin filaments.</u></em>
  • <em><u>Filament movement starts at the region or zone where thin and thick filaments overlap. </u></em>
  • <em><u>Myofibril contains many sacromeres along its length and thuse myofibrils and muscle cells contract as the sacromeres contract.</u></em>
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What is the magnitude of δv12 if the bulb is removed from the socket (i.e. the circuit is not closed)?
andreev551 [17]

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current , R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

4 0
3 years ago
Read 2 more answers
True or false the refraction of a wave is how many wavelengths pass a fixed point each second
Ksenya-84 [330]
False. That description fits the wave's 'frequency'. 
It has nothing to do with refraction.
8 0
3 years ago
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The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t&gt;0 , where is measured in seconds and
8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at 0.

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>
  • A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s

The velocity is positive at when the time of motion is as follows;

0.

The total distance traveled in the first 10 seconds is calculated as follows;

2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

Learn more about motion of particles here: brainly.com/question/11066673

4 0
2 years ago
A DC generator is the same as a(n)
nikdorinn [45]

Answer:

A DC generator is the same as a(n)

C. armature

7 0
3 years ago
A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw
mezya [45]

Answer:

The mass is  m  = 3.45408 kg

Explanation:

From the question we are told that

    The extension  of the rod is from , \   x_1 = 0 \to x_2 = 13.0

     The area is  A =  8.0 cm^2

      The density increase as follows from  \  \rho_1 =2.5 g/cm^2 \to  \rho_2= 19.0 g/cm^3

    The equation  \rho =  B + Cx

at  x_1= 0  \rho_1 =2.5 g/cm^2

So

      2.5  =  B  + 0

=>  B =2.5

    So at x_2 = 13.0 ,  \rho_2= 19.0 g/cm^3

So

            19.0 = 2.5 + C(13)

       =>   C = 1.27

Now  

       m  =  8   \int\limits^{13}_{0} {2.5 + 1.27x} \, dx

      m  =  8   [{2.5 +\frac{ 1.27x^2}{2} } ]\left  | 13} \atop {0}} \right.

      m  =  8   [{2.5 +\frac{ 1.27(13)^2}{2} } ]

      m  = 3454.08 g

        m  = 3.45408 kg

         

8 0
3 years ago
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