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nikklg [1K]
2 years ago
13

What role does actin and myosin play in the process of muscle contraction?

Physics
1 answer:
Margarita [4]2 years ago
4 0

Answer:

Muscle contraction thus results from an interaction between the actin and myosin filaments that generates their movement relative to one another. The molecular basis for this interaction is the binding of myosin to actin filaments, allowing myosin to function as a motor that drives filament sliding.

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It took a bulldozer 62,000 J of work to move a rock 30 m. It took 5 minutes. How much force did the bulldozer have to apply?
NeX [460]

Answer:

A (2066,6 N)

Explanation:

Use the Work formula

62.000J = F . 30

62.000/30 = 2066,6 N

The amout of time it took to move the rock doesn´t matter at all.

It is called a distraction variable, We don´t need it to solve the problem it is there just to confuse.

5 0
3 years ago
A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16
erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle \phi = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

V =\pi D( \frac{N}{60} )

V = 3.14 (0.034) \frac{(1650)}{60}

V = 2.93 \frac{m}{s}

Form factor for the pinion gear is

Y = 0.303

Now

K_{v} = \frac{6.1 +0.303}{6.1} = 1.049

Force on gear tooth

F = \frac{P}{V}

F = \frac{1200}{2.93}

F = 408.73 N

Now the bending stress is given by the formula

\sigma = \frac{K_{v} F}{m b y}

\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}

\sigma = 35.38 M pa

This is the value of bending stress on the pinion

8 0
3 years ago
NEES HELP!!!!
OLga [1]

Answer:

Doppler Theory

Explanation:

it's a theory regarding the change in wave frequency during the relative motion between a wave source and its observer.

6 0
2 years ago
Order the layers, with the oldest at the bottom and most recent at the top.
givi [52]

Answer:

adbce i think

Explanation:

7 0
3 years ago
Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0
3 years ago
Read 2 more answers
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