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Anuta_ua [19.1K]
4 years ago
12

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

Physics
2 answers:
horsena [70]4 years ago
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]4 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 student uses the right-hand rule as shown.
m_a_m_a [10]

Answer: O:right

Explanation: In this case you place your finger on the current, and your fingers should curl showing the way, you must use your right hand in this case, otherwise that would mean the fingers on your left would bend way back, and snap off, (Not really lol,  just saying)

5 0
3 years ago
an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2
Degger [83]
60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N
7 0
3 years ago
Read 2 more answers
What is the acceleration of an object with a constant velocity
valina [46]

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: <span>a=<span><span>Δv</span>t</span></span>

Where a=acceleration, <span>Δv</span>=change in velocity and t=time

6 0
3 years ago
Read 2 more answers
A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat
postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let 'F_A' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

F_A × 10.0 m - W × 4.0 m = 0

∴ F_A × 10.0 m = W × 4.0 m = 600 N × 4.0 m

F_A × 10.0 m = 600 N × 4.0 m

∴  F_A = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', F_A = 240 N.

8 0
3 years ago
A zero-order reaction has a constant rate of 2.30×10−4 M/s. If after 80.0 seconds the concentration has dropped to 1.50×10−2 M,
dolphi86 [110]

Answer:

Initial concentration of the reactant = 3.34 × 10^(-2)M

Explanation:

Rate of reaction = 2.30×10−4 M/s,

Time of reaction = 80s

Final concentration = 1.50×10−2 M

Initial concentration = Rate of reaction × Time of reaction + Final concentration

= 2.30×10−4 M/s × 80s + 1.50×10−2 M = 3.34 × 10^(-2)M

Initial concentration = 3.34 × 10^(-2)M

6 0
3 years ago
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