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Simora [160]
2 years ago
5

An oxygen atom at a particular site within a DNA molecule can be made to execute simple harmonic motion when illuminated by infr

ared light. The oxygen atom is bound with a spring-like chemical bond to a phosphorus atom, which is rigidly attached to the DNA backbone. The oscillation of the oxygen atom occurs with frequency fO=3.7×1013Hz.
Physics
1 answer:
lukranit [14]2 years ago
4 0

Answer:

The frequency of the sulphur is 2.6×10^13 Hz.

We find the frequency of the sulphur because sulphur is just below the oxygen atom in the periodic table. So, we find it by using the frequency of the oxygen.

Explanation:

As, the frequency in terms of spring constant and mass is :

f = (1/2π) × √(k/m)

k is the spring constant and m is the mass.

Square on both sides of the equation and it becomes then,

f^2 = (1/4π^2) × (k/m)

f^2×m = k/4π^2

Now, we will solve it with oxygen and sulphur because the oxygen atom is chemically replaced with a sulphur atom, the spring constant of the bond is unchanged so, the relation between fo and fs is:

<h3>fo^2×mo = fs^2×ms</h3>

As, fo is the frequency of the oxygen and fs is the frequency of the sulphur atom and mo is the mass of the oxygen and ms is the mass of the sulphur.

So,

fs^2 = (fo^2×mo) / ms

fs = fo×√(mo/ms)

As, we also know the mass of the oxygen and sulphur which is 16amu and 32amu:

fs = (3.7×10^13) × √(16/32)

fs = (3.7×10^13) × 1/√2

fs = 2.6×10^13 Hz.

Therefore, the frequency is: 2.6×10^13 Hz.

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Aleks04 [339]
W=Fd. Force is not given so we solve for it. F=ma, m=5kg, a=2m/s^2, F=10N. Distance is not given so we solve for it, x=.5a(t^2)=.5(2)(7x7)=49m. F=10N, d=49m, W=490J.
5 0
2 years ago
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A wave is propagating from left to right in a medium. The particles in the medium are also vibrating from left to right. What ki
Anna71 [15]
Based on the direction of propagation compared to direction of vibration, waves are classified into:
1- Transverse waves: The direction of propagation of the wave is perpendicular to the direction of vibration of the medium particles.
2- Longitudinal waves: The direction of propagation of the wave is the same as the direction of vibration of the medium particles.

For the question we have here, since the direction of the wave is the same as the direction of vibration of particles, therefore, this wave is a longitudinal wave
6 0
2 years ago
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
A technician is troubleshooting a problem. The technician tests the theory and determines the theory is confirmed. Which of the
vladimir1956 [14]

Answer:

b)  Document lessons learned.

Explanation:

First he should do documentation

then C

then D

then A

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