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3 years ago

What does stressing an Equilibrium system mean? How is stress Applied?

2 answers:

7 0

Stressing an equilibrium simply means that the physical properties in which already exists are balanced. Stress can be applied by either changing the pressure or the volume or temperature.

Kruka [31]3 years ago

3 0

Answer:

Explanation:

As per the law of Mass of Action, Rate of forward reaction is equal to the rate of reverse reaction.

In such condition,a system is stable and attains a phase known as dynamic equilibrium.

Consider an equation having some reactants and product to be on a balanced load or sea saw.

If stress is applied, the balance will be distributed and the sea-saw will bend towards the stress factor applied side.

The stress can be applied in different way :

By changing concentration

By changing temperature

By altering pressure or volume

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A convex lens of focal length 35 cm produces a magnified image 2.5 times the size of the object. What is the object distance if

Zanzabum

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

$m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm$

$v=-2.5\times 21=-52.5\ cm$

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

3 0

3 years ago

Which of the following changes would decrease the coefficient of friction needed for this ride?

posledela

Friction occurs between two contacting surfaces. The coefficient of friction is very much dependent on the roughness of these surfaces. Some of the many ways in which the coefficient can be lessened or decreased are to lubricate the surface or make it shiny by eliminating the spikes which caused the roughness.

7 0

3 years ago

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

frez [133]

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

$F_D = c_D A \frac{\rho V^2}{2}$

Where,

F_D = Drag Force

$c_D$ = Drag coefficient

A = Area

$\rho$ = Density

V = Velocity

Our values are given by,

$c_D = 0.5$ (That is proper of a cone-shape)

$A = 9m^2$

$\rho = 1.2Kg/m^3$

$V = 6.5m/s$

Part A ) Replacing our values,

$F_D = 0.5*9*\frac{1.2*6.5^2}{2}$

$F_D = 114.075N$

Part B ) To find the torque we apply the equation as follow,

$\tau = F*d$

$\tau = (114.075N)(7)$

$\tau = 798.525N.m$

3 0

3 years ago

Suppose that the velocity (in meters per second) of a sky diver falling near the Earth's surface is given by the following expon

Mice21 [21]

Answer:

4.7 s

Explanation:

The complete question is presented in the attached image to this solution.

v(t) = 61 - 61e⁻⁰•²⁶ᵗ

At what time will v(t) = 43 m/s?

We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.

43 = 61 - 61e⁻⁰•²⁶ᵗ

- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18

e⁻⁰•²⁶ᵗ = (18/61) = 0.2951

In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205

-0.26t = -1.2205

t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.

Hope this Helps!!!

6 0

2 years ago

If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w

uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

$V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2$

V₂ is the final volume

$V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL$

So, the final volume of the sample is 200 mL.

5 0

2 years ago