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s2008m [1.1K]
2 years ago
5

What is equilibrium?

Physics
1 answer:
Lunna [17]2 years ago
4 0

Answer:

In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.

Explanation:

Equilibrium is defined as a state of balance or a stable situation where opposing forces cancel each other out and where no changes are occurring. ... An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.

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2. A ball tied to a pole by a rope swings in a circular path with a centripetal acceleration of 2.7 m/s. If the ball has a
Helga [31]

Answer: The diameter of the circular path is 2.96m

Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.

Centripetal acceleration = 2.7m/s^2

Tangential speed = 2.0m/s

Radius = 2.0^2 / 2.7 = 4/2.7

= 1.48m

Diameter = radius*2

= 1.48*2 = 2.96m.

3 0
3 years ago
To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.070-kg ball from the end of a wire. The w
mote1985 [20]

Answer:

g = 1.64m/s²

Explanation:

1.5m in 0.078s

V = 15 / 0.078

= 19.23m/s

Tension = mg

μ = 3.10 × 10⁻⁴

T = V²μ

mg =  V²μ

g =  V²μ / m

g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)

g = 1.64m/s²

7 0
3 years ago
Which statement is a characteristic of a concave lens
Free_Kalibri [48]

<em>"A concave lens is thinner at the center than it is at the edges."</em>

If this isn't on the list of choices, that's tough.  We can't help you choose the best one if we don't know what any of them is.

5 0
3 years ago
We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth
Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = \frac{m}{m+M} \ v_o

for when n balls have collided

          v = \frac{m}{n \ m + M}  v₀

Now let's analyze the case of the bouncing ball (elastic)

     

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = \frac{m-M}{m + M}   v₀

we substitute in equation 1 to find v₂

            \frac{M}{m}  v₂ = v₀ -  \frac{m-M}{m+M}   v₀

            v₂ = \frac{m}{M}  ( 1 - \frac{m-M}{m+M} ) \ v_o

            v₂ = \frac{m}{M}  ( \frac{2M}{m+M} ) \ \ v_o

            v₂ = \frac{2m}{m +M}  \ v_o  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

8 0
2 years ago
A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t
Furkat [3]

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


3 0
3 years ago
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