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1 year ago

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One description of the potential energy of a diatomic molecule is given by the Lennard-Jones potential,U =A/r¹² - B/r⁶ where A a

nd B are constants and r is the separation distance between the atoms. For the H₂ molecule, take A= 0.124 × 10⁻¹²⁰ eV . m¹² and B = 1.488 × 10⁻⁶⁰eV . m⁶. = Find (a) the separation distance r₀ at which the energy of the molecule is a minimum

1 answer:

Andrei [34K]1 year ago

3 0

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A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid

Sergeeva-Olga [200]

Answer:

$8.37*10^5$ rpm

Explanation:

Given that rotational kinetic energy = $4.66*10^9J$

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = $\frac{1}{2} mr ^2$

Rotational K.E is illustrated as $(K.E)_{rt} = \frac{1}{2} I \omega^2$

$\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }$

$\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }$

$\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }$

$\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }$

$\omega = 87636.04$

$\omega = 8.76*10^4 rad/s$

Since 1 rpm = $\frac{2 \pi}{60} rad/s$

$\omega = 8.76*10^4(\frac{60}{2 \pi})$

$\omega = 836518.38$

$\omega = 8.37 *10^5 rpm$

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2 years ago

The record for the highest speed achieved in alaboratory for a

Ira Lisetskai [31]

Answer:

0.000234 seconds

Explanation:

Since the row is 0.15m, its radius of rotation must be 0.15 / 2 = 0.075 m

We can start by calculating the angular speed of the rod:

$\omega = \frac{v}{r} = \frac{2010}{0.075} = 26800 rad/s$

Since one revolution equals to 2π rad. The speed in revolution per second must be

26800 / 2π = 4265 revolution/s

The number of seconds per revolution, or period, is the inverse:

1/4265 = 0.000234 seconds

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2 years ago

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Aleksandr [31]

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

(c) From [2] we can solve for a:

$\displaystyle a= 2\frac{x-v_ot}{t^2}$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

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2 years ago

According to Charles' law, as the temperature of a given gas at constant pressure is increased, the volume will also

Alenkasestr [34]

A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion. ... The equation shows that, as absolute temperature increases, the volume of the gas also increases in proportion.

i hope it will help

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3 years ago

Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons

dybincka [34]

<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N

3 0

2 years ago

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