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3 years ago

Explain the relationship between resistance and the velocity of shortening.

1 answer:

7 0

The velocity of shortening refers to the speed of the contraction from the muscle shortening while lifting a load. The relationship between the resistance and velocity of shortening is inverse. The greater the resistance, the shorter the velocity of shortening and the smaller the resistance, the larger the velocity of shortening.

Hopefully this help :)

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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi

valina [46]

Answer:

The answer is

" $x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$ ".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs $\frac{8}{3}$ feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

$W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\$

The source of the hooks law is stable,

$16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\$

Number $\frac{1}{2}$ times the immediate speed, i.e .. Damping force

$\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2} \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\$

The m^2+m+12=0 and m is an auxiliary equation,

$m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \ m2 =\frac{-1 - \sqrt{47i}}{2}$

Therefore, additional feature

$x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]$

Use the form of uncertain coefficients to find a particular solution.

Assume that solution equation,

$x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\$

These values are replaced by equation ( 1):

$\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\$

Going to compare cos3 t and sin 3 t coefficients from both sides,

The cost3 t is 3A + 3B= 20 coefficients

The sin 3 t is 3B -3A = 0 coefficient

The two equations solved:

$3A+3B = 20 \\\frac{3B -3A=0}{}\\6B=20\\B= \frac{20}{6}\\B=\frac{10}{3}\\$

Replace the very first equation with the meaning,

$3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\$

equation is

$x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)$

The ultimate plan for both the equation is therefore

$x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)$

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

$x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\$

$x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t) +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\$

$c_2=\frac{-64\sqrt{47}}{141}$

$x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))$

5 0

3 years ago

A 20kg object acceleration by a force of 200N with coefficient of kineticfriction is 0.4 what is acceleration of the object?

Schach [20]

Answer:

Given - M = 20 kg

k = 0.4

F = 200 N

To find -  acceleration

Solution - 

F= kMA

200 = 0.4 * 20 * acceleration

200 = 8 * a

a = 8/200

a = 0.04 m s²

5 0

2 years ago

Find the sum 5.24 g, 43.261 g, and 7.3458 g. Write your answer with the correct amount of significant figures.

lapo4ka [179]

Answer:

This is how I do it:

5.24 rounded to one significant figure is 5
43.261 rounded to one significant figure is 40.
7.3458 rounded to one significant figure is 7.
5 + 40 + 7 is 52 g

Hope this helps you.

Explanation:

7 0

3 years ago

A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin

Katarina [22]

I’m not sure look it up in the internet!

4 0

2 years ago

In an economy, the demand for labor is given by the equation W = 15 - (1/200) L and the supply of labor is given by the equation

mr_godi [17]

Answer:

the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

Explanation:

The equilibrium wage rate and the equilibrium quantity of labor are found as the point where the equation of demand intercepts the equation of supply, so the equilibrium quantity of labor is:

$W_{Demand} = W_{Supply}$

15 - (1/200) L = 5 + (1/200) L

15 - 5 = (1/200) L + (1/200) L

10 = (2/200) L

(10*200)/2 = L

1000 = L

Then, the equilibrium wage rate is calculated using either the equation of demand for labor or the equation of supply of labor. If we use the equation of demand for labor, we get:

W = 15 - (1/200) L

W = 15 - (1/200) 1000

W = 10

Finally, the equilibrium wage rate is 10 and the equilibrium quantity of labor is 1000 workers

7 0

3 years ago