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amid [387]
2 years ago
5

The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, U (r) = ARe(R-01.5 — 1)2 wh

ere r is the distance between the two atoms and A, R, and S are positive constants with S << R. Sketch this function for 0 < r < cc. Find the equilibrium separation rip , at which U (r) is minimum. Now write r = rox so that x is the displacement from equilibrium, and show that, for small displacements, U has the approximate form U = const kx 2 . That is, Hooke's law applies. What is the force constant k?
Physics
1 answer:
Alex2 years ago
4 0

Answer:

The constant Hooke law is

k = 2 * A / S

Explanation:

To the force constant k Hooke law can use the Morse function as a:

U (r) = A * [ e ⁽ ᵇ ⁻ ⁿ / ˣ ⁾ - 1 ] ² - 1 ]

b = R , n = r₀ , x = s

U (r) = A * [ e ⁽ ᵇ ⁻ ⁿ / ˣ ⁾ - 1 ] e ⁽ ᵇ ⁻ ⁿ / ˣ ⁾ * ( - ¹ /ₓ ) = 0

e ⁽ ᵇ ⁻ ⁿ / ˣ ⁾ = 1

r₀ = R

r = r₀ + x , r = R + x  

U (x) = - A + A / 2 * 2 / s * x²

U (x) = const + ¹/₂ * k * x²

Solve k'

k = 2 * A / S

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You use a slingshot to launch a potato horizontally from the edge of a cliff with speed v0. The acceleration due to gravity is g
Ray Of Light [21]

Answer:

\displaystyle t=\frac{2v_o}{g}

Explanation:

<u>Horizontal Launch</u>

When an object is launched horizontally at a speed vo, it describes a curved called parabola as the speed in the x-direction does not change and the speed in the y-direction increases with time because the gravity makes it return to the ground.

The vertical distance the object (potato) travels downwards is:

\displaystyle y=\frac{gt^2}{2}

The horizontal distance is

x=v_ot

We need to find the time when both distances are equal, thus

\displaystyle \frac{gt^2}{2}=v_ot

Simplifying by t

\displaystyle \frac{gt}{2}=v_o

Solving for t

\displaystyle \boxed{t=\frac{2v_o}{g}}

8 0
2 years ago
You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg
Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

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2 years ago
If Emily throws the ball at an angle of 30∘ below the horizontal with a speed of 12m/s, how far from the base of the dorm should
Hitman42 [59]

Answer:

3.6 m

Explanation:

let x = horizontal distance between emily and allison should be for allison to catch the ball

Find horizontal speed of the ball

vx = 12 sin 30 = 12 x 0.5 = 6 m/s

To find time taken, we will use vertical values of the ball motion

Initial velocity in vertical direction

u = 12 cos 30 = 10.392 m/s

let a = g = 9.8m/s2

Use equation of motion

s = ut +1/2at^2

s = vertical distance = 8

8 = (10.392)t + (1/2)(9.8)t^2

8 = (10.392)t + (4.9)t^2

4.9t^2 + 10.392t - 8 = 0

Using formula of quadratic or calculator, we'll find

t = 0.6 and t = -2.72

We pick t=0.6s since it's not logical time in negative

Assuming no air resistance or external forces, the ball will move 6m/s horizontally. Hence using the formula of speed

speed vx = distance x / time

x = (vx)(t)

  = 6 x 0.6

  = 3.6 m

6 0
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jasenka [17]

Answer:

Daria probably suffers from Entomophobia.

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Use Scoratic it works with any time of subject
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