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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?
Physics
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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A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast
Neko [114]

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

8 0
2 years ago
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Metals are used in many products because of the characteristic properties that most metals have. Which product requires the high
labwork [276]
<span>The answer is mirrors. Mirrors are made by applying a metal thin layer on the back surface of a transparent substrat, typically glass. The metal layer in the antiquity was bronze, mercury and later silver whose luster gave the reflective property to the mirror.</span>
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2 years ago
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A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick to
daser333 [38]

Answer:

Magnitude of impulse, |J| = 4 kg-m/s                                                                                

Explanation:

It is given that,

Mass of cart 1, m_1=2\ kg

Mass of cart 2, m_2=4\ kg  

Initial speed of cart 1, u_1=3\ m/s          

Initial speed of cart 2, u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

m_1u_1+m_2u_2=(m_1+m_2)V

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}        

V=\dfrac{2\times 3}{(2+4)}

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:

J=F\times t=m(V-u)

J=m(V-u)

J=2\times (1-3)    

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

8 0
2 years ago
A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?
guajiro [1.7K]

The correct answer to the question is-  2.2\ \mu C

CALCULATION:

As per the question, the electric field generated by the source charge is 1236  N/C at a distance of 4 m.

Hence , electric field  E =  1236 N/C.

The distance of the point R = 4m

We are asked to calculate the charge possessed by the source.

The electric field produced by a source charge of Q at a distance R is calculated as -

                    Electric field E = \frac{1}{4\pi \epsilon_{0}}\frac{Q}{R^2}

Here, \epsilon_{0} is called the absolute permittivity of the free space.

Hence, the charge of source is calculated as -

                                         Q = E\times 4\pi \epsilon_{0}\times R^2

                                            = 1236\times \frac{1}{9\times 10^9}\times (4)^2\ Coulomb

                                            = 2197.33\times 10^{-9}\ C

                                             = 2.19733\times 10^{-6}\ C

                                             = 2.2\ \mu C

Hence, the charge of source is 2.2\ \mu C

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2 years ago
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A bicyclist of mass 68 kg rides in a circle at a speed of 3.9 m/s. If the radius of the circle is 6.5 m, what is the centripetal
ASHA 777 [7]
Data:
Centripetal Force = ? (Newton)
m (mass) = 68 Kg
s (speed) = 3.9 m/s
R (radius) = 6.5 m

Formula:
F_{centripetal\:force} =  \frac{m*s^2}{R}

Solving:
F_{centripetal\:force} = \frac{m*s^2}{R}
F_{centripetal\:force} = \frac{68*3.9^2}{6.5}
F_{centripetal\:force} = \frac{68*15.21}{6.5}
F_{centripetal\:force} = \frac{1034.28}{6.5}
\boxed{\boxed{F_{centripetal\:force} = 159.12\:N}}
Answer:
<span>B.159 N</span>
3 0
2 years ago
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