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3 years ago

What does it mean for a system to be in dynamic equilibrium?

Physics

1 answer:

Ainat [17]3 years ago

6 0

Answer:

in chemistry, and in physics, a dynamic equilibrium exists once a reversible reaction occurs.

Explanation:

Substances transition between the reactants and products at equal rates, meaning there is no net change. Reactants and products are formed at such a rate that the concentration of neither changes :)

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The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

Andrews [41]

Explanation:

It is given that,

Mass of the runner, m = 70 kg

Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, $A=110\ mm^2=0.00011\ m^2$

Part A,

Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, $Y=0.15\times 10^{10}\ N/m^2$

The formula of the Young modulus is given by :

$Y=\dfrac{F/A}{\dfrac{\Delta L}{L}}$

$0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}$

$\Delta L=0.0049\ m$

Part B,

The fraction of the tendon's length does this correspond is given by :

$\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}$

$\dfrac{\Delta L}{L}=0.0326$

Hence, this is the required solution.

6 0

3 years ago

A phone cord is 2.28 m long. The cord has a mass of 0.2 kg. A transverse wave pulse is produced by plucking one end of the taut

Debora [2.8K]

The characteristics of the speed of the traveling waves allows to find the result for the tension in the string is:

T = 10 N

The speed of a wave on a string is given by the relationship.

v = $\sqrt{\frac{T}{\mu } }$

Where v es the velocty, t is the tension ang μ is the lineal density.

They indicate that the length of the string is L = 2.28 m and the pulse makes 4 trips in a time of t = 0.849 s, since the speed of the pulse in the string is constant, we can use the uniform motion ratio, where the distance traveled e 4 L

v = $\frac{d}{t}$

v = $\frac{4 L}{t}$

v = $\frac{4 \ 2.28 }{0.849}$

v = 10.7 m / s

Let's find the linear density of the string, which is the length of the mass divided by its mass.

μ = $\frac{m}{L}$

$\mu = \frac{0.2}{2.28}$

μ = 8.77 10⁻² kg / m

The tension is:

T = v² μ

Let's calculate

T = 10.7² 8.77 10⁻²

T = 1 0 N

In conclusion using the characteristics of the velocity of the traveling waves we can find the result for the tension in the string is:

T = 10 N

Learn more here: brainly.com/question/12545155

7 0

3 years ago

3–101 It is estimated that 90 percent of an iceberg’s volume is below the surface, while only 10 percent is visible above the su

Artist 52 [7]

Answer:

The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

Explanation:

Given that,

Volume of seawater = 90 % of Volume of iceberg

Volume of visible surface = 10 %

Density of seawater = 1025 kg/m³

We need to calculate the density of the iceberg

Using equilibrium condition

$W_{I}=F_{B}$

$\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g$

Put the value into the formula

$\rho_{I}\times V_{I}\times g=\rho_{w}\times0.9V_{I}\times g$

$\rho_{I}=1025\times0.9$

$\rho_{I}=922.5kg/m^3$

We need to calculate the density of the iceberg

Using equilibrium condition

$W_{I}=F_{B}$

$\rho_{I}\times V_{I}\times g= \rho_{w}\timesV_{w}\times g$

Put the value into the formula

$\rho_{I}\times V_{I}\times g=\rho_{w}\times0.1V_{I}\times g$

$\rho_{I}=1025\times0.1$

$\rho_{I}=102.5\ kg/m^3$

Hence, The density of iceberg upper water and under water are 102.5 kg/m³ and 922.5 kg/m³

5 0

3 years ago

What distance separates Earth and the Sun? A 300,000 km per second B Eight light-minutes C 149 million light-seconds D One light

lina2011 [118]

Answer:

B Eight light-minutes

Explanation:

In the case when the distance separated earth and the sun so here we orbit the sun for a 150 million km distance and the light moves would be 300,000 kilometers per second

Now divide this

= 150 million ÷ 300,000 kilometers per second

= 500 seconds

This 500 seconds represent 8 minutes and 20 seconds

Hence, option B is correct

5 0

3 years ago

Read 2 more answers

Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022k

Taya2010 [7]

Answer: $g=1.97 m/s^2$