Physics, 18.03.2020 00:28 itsyagirlgona21

The sense of equilibrium responds to movements of the head. It is divided into two types--static and dynamic equilibrium. Describe and give an example of these two types of equilibrium. Select all the statements that accurately describe equilibrium.

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Answer from: sandrathompson7250

nodding of head ,yes- static equilibrium

nodding of head, no- dynamic equilibrium.

Explanation:

static equilibrium monitors head position when body is not moving .

dynamic equilibrium monitors the angular or rotational movements of the head when body moves.

Answer from: Thania3902

Static equilibrium is simply described as one in which all forces acting on an object are balanced which implies that there is no resultant forces. And the object is not in motion releative to the reference plane.i.e the object does not have a velocity.

While

Dynamic equilibrium is type of equilibrium in which all the forces acting on an object are balanced but the object it moving, i.e there exist a veleocity.

Explanation:

Equilibrium simply means, forces are in balance. i.e the net force should be zero.

It can further explained forces acting downward and acting upward, and forces acting right and acting left should be equal in magnitude.

1) dynamic equilibrium can be described as the one in which the reactants are converted to products and products are converted to reactants at an equal and constant rate. Equilibrium is the state of equal, opposite rates, not equal concentrations.

Examples of dynamic equilibrium are; a)a car moving at constant velocity, b)a fan rotating at uniform speed

2)Static equilibrium is the one in which the rate of forward process is equal to reverse process.It is usually regarded as MECHANICAL EQUILIBRIUM because it occurs when all particles in the reaction are at rest and there is no motion between reactants and products.

Static equilibrium can be noticed when there is a steady-state in a system that is the dynamic forces are not acting on the potential energies of the reverse and forward reactions.

An example of static equilibrium is when graphite turn into diamond, and it is illustrated as

C(diamond) ⇌ C(graphite)

Answer from: Quest

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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