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amid [387]
3 years ago
5

What would happen to the marbles speed if we had longer tables

Physics
2 answers:
Zanzabum3 years ago
7 0

Answer:

The marble will start to decrease in speed

Explanation:

The longer the table is the more the marbles speed would decrease because marbles can only be fast for a certain amount of time

egoroff_w [7]3 years ago
5 0

If we had longer tables and if we did the ramp marble experiment  the speed would increase and the marble would  roll longer  . .

Hope I helped -

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Describes the relationship between the free energy change, the reaction quotient, and the equilibrium constant.
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<u>Explanation:</u>

Reaction quotient is defined as the ratio of the concentration of the products and reactants of a reaction at any point of time with respect to some unit. It is represented by the symbol <em>Q</em>.

The ratio of the concentration of products and reactants of a reaction in equilibrium with respect to some unit is said to be equilibrium constant expression. It is represented by the symbol <em>K</em>.

The relationship between Gibbs free energy change and reaction quotient of the reaction is:

\Delta G=\Delta G^o+RT ln Q           ......(1)

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\Delta G = Gibbs free energy change

\Delta G^o = Standard Gibbs free energy change

R = Gas constant

T = Temperature

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\Delta G^o=-RT ln Q           ...(2)

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What do all hydrogen atoms and ions have in common?
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A bullet is shot from shoulder height (1.50m)at an angle 60.0°above the horizontalwith an initial speed of 400m/s.Neglect air re
gizmo_the_mogwai [7]

Answer:

a)   t₁ = 70,717 s , b)  x = 14143.4 m , c) θ = -60.02º

Explanation:

We will work on this problem using kinematics,

a) time to reach the ground, let's look for the initial speed

          v_{oy} = v₀ sin.60

          v₀ₓ = v₀ cos 60

          v_{oy} = 400 sin. 60 = 346.41 m / s

          v₀ₓ = 400 cos 60 = 200 m / s

Let's use the projectile launch equation

           y = y₀ + v_{oy} t - ½ g t²

Upon reaching the ground y = 0

            0 = 1.5+ 346.41 t - ½ 9.8 t²

             4.9 t² - 346.41 t - 1.5 = 0

             t² - 70.696 t - 0.306 = 0

We solve the second degree equation

            t = [70.696 ±√ (70.696² + 4 0.306)] / 2

           

            t₁ = 70,717 s

            t₂ = -0.0085 s

We take the positive time

b) let's use the equation

              x = v₀ₓ t

              x = 200 70,717

              x = 14143.4 m

c) to find the angle let's look for the velocity at the point of impact, the velocity on the x-axis is constant

             vₓ = v₀ₓ = 200 m / s

The speed on the y axis

            v_{y} = v_{oy}  - g t

            v_{y} = 346.41 - 9.8 70.717

            v_{y} = -346.62 m / s

Let's use trigonometry

           tan θ = v_{y} / vₓ

           θ = tan⁻¹ v_{y} / vₓ

           θ = tan⁻¹ (-346.62 / 200)

           θ = -60.02º

6 0
3 years ago
Read 2 more answers
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