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uranmaximum [27]

2 years ago

8

A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from

the mass of a proton. If the HIHI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring?

1 answer:

vlada-n [284]2 years ago

0 0

Answer:

The force constant of the spring is 317.8 N/m.

Explanation:

Given that,

Frequency $f=7.0\times10^{13}\ Hz$

We need to calculate the reduced mass

Using formula of reduced mass

$\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}$

Where, $m_{H}$ = atomic mass of H

$m_{I}$ = atomic mass of I

Put the value into the formula

$\mu=\dfrac{1\times126.9}{1+126.9}$

$\mu=0.99\ u$

$\mu=0.99\times1.66\times10^{-27}\ Kg$

$\mu=1.643\times10^{-27}\ kg$

We need to calculate the force constant of the spring

Using formula of frequency

$f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}$

$k=f^2\times 4\pi^2\times\mu$

Put the value into the formula

$k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}$

$k=317.8\ N/m$

Hence, The force constant of the spring is 317.8 N/m.

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Waves hitting at an angle and then bending around features of the coast is known as

Pavel [41]

<span>Waves hitting at an angle and then bending around features of the coast is known as Wave refraction
When waves hitting a specific angle, some part of the waves will be closer to the shallow part of the water and some part will be closer to the deeper part of the water, which makes the wave became somehow bent around the shore.</span>

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2 years ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

5 0

2 years ago

A ball is dropped from the top of a cliff. By the time it reaches the ground, all the energy in its gravitational potential ener

Bingel [31]

The ball was dropped from a height 20 meters

Explanation:

The given is

1. A ball is dropped from the top of a cliff

2. By the time it reaches the ground, all the energy in its gravitational

potential energy store has been transferred into its kinetic energy

store, that mean K.E = P.E

3. The ball is travelling at 20 m/s when it hits the ground

4. The gravitational field strength is 10 N/kg

We need to find the height that the ball dropped from it

The ball dropped from the top of a cliff means the initial speed is 0

→ K.E = $\frac{1}{2}m(v^{2}-v_{0}^{2})$

where v is the final speed, $v_{0}$ in the initial speed and m

is the mass

→ v = 20 m/s and $v_{0}$ = 0 m/s

→ K.E = $\frac{1}{2}m(20^{2}-0^{2})$

→ K.E = $\frac{1}{2}m(400)$

→ K.E = 200 m joules ⇒ when the ball hits the ground

→ P.E = m g h

where g is the gravitational field strength, m is the mass and h is

the height

→ g = 10 N/kg

→ P.E = m(10)(h)

→ P.E = 10 m h joules

→ P.E = K.E

→ 10 m h = 200 m

Divide both sides by 10 m

→ h = 20 meters

The ball was dropped from a height 20 meters

Learn more

You can learn more about gravitational potential energy in brainly.com/question/1198647

#LearnwithBrainly

8 0

2 years ago

Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged part

erica [24]

Answer:

Two possible points

<em>x= 0.67 cm to the right of q1</em>

<em>x= 2 cm to the left of q1</em>

Explanation:

<u>Electrostatic Forces</u>

If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude

$\displaystyle f=k\frac{q_1\ q_2}{d^2}$

We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.

$\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}$

$\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}$

$\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}$

$\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}$

$\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}$

Equating

$\displaystyle F_{13}=F_{23}$

$\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}$

Operating and simplifying

$\displaystyle (0.02-x)^2=4x^2$

To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.

$\displaystyle 0.02-x=\pm 2x$

Assuming the positive sign :

$\displaystyle 0.02-x= 2x$

$\displaystyle 3x=0.02$

$\displaystyle x=0.00667\ m$

$x=0.67\ cm$

Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative

$\displaystyle 0.02-x=-2x$

$\displaystyle x=-0.02\ m$

$\displaystyle x=-2\ cm$

The negative sign of x means q3 is located to the left of q1 (assumed in the origin).

5 0

2 years ago

Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A

il63 [147K]

Answer:

The question is incomplete, below is the complete question

"Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?

a. A and B must be displacements.

b. A and B must have equal lengths.

c.A and B must point in opposite direction.

d. A and B must point in the same direction.

e.A and B must be at right angles to each other."

answer:

d. A and B must point in the same direction.

Explanation:

a.false:From vector analysis, all forms of vector with the same unit can be added, and we add vector component by component. hence this defile option (a).

b. false: in addition of vectors, length is not a criteria to consider before carrying out the operation, hence vectors of different lengths can be added

c. false: this is against the claim as vectors in opposite direction can give rise to a new vector of negative value.

d. true: this is true as vector in the same direction add up to give rise to another vector.

e.false: this is not a true assumption

4 0

2 years ago