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Simora [160]
3 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]3 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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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi
tigry1 [53]

Answer:

a

The  radial acceleration is  a_c  = 0.9574 m/s^2

b

The horizontal Tension is  T_x  = 0.3294 i  \ N

The vertical Tension is  T_y  =3.3712 j   \ N

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

   The length of the string is  L =  10.7 \ cm  =  0.107 \ m

     The mass of the bob is  m = 0.344 \  kg

     The angle made  by the string is  \theta  =  5.58^o

The centripetal force acting on the bob is mathematically represented as

         F  =  \frac{mv^2}{r}

Now From the diagram we see that this force is equivalent to

     F  =  Tsin \theta where T is the tension on the rope  and v is the linear velocity  

     So

          Tsin \theta  =   \frac{mv^2}{r}

Now the downward normal force acting on the bob is  mathematically represented as

          Tcos \theta = mg

So

       \frac{Tsin \ttheta }{Tcos \theta }  =  \frac{\frac{mv^2}{r} }{mg}

=>    tan \theta  =  \frac{v^2}{rg}

=>   g tan \theta  = \frac{v^2}{r}

The centripetal acceleration which the same as the radial acceleration  of the bob is mathematically represented as

      a_c  =  \frac{v^2}{r}

=>  a_c  = gtan \theta

substituting values

     a_c  =  9.8  *  tan (5.58)

     a_c  = 0.9574 m/s^2

The horizontal component is mathematically represented as

     T_x  = Tsin \theta = ma_c

substituting value

   T_x  = 0.344 *  0.9574

    T_x  = 0.3294 \ N

The vertical component of  tension is  

    T_y  =  T \ cos \theta  = mg

substituting value

     T_ y  =  0.344 * 9.8

      T_ y  = 3.2712 \ N

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is  

         

       T  = T_x i  + T_y  j

substituting value  

      T  = [(0.3294) i  + (3.3712)j ] \  N

         

3 0
3 years ago
During the fission reaction shown, how did the target nucleus change ?
Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0
3 years ago
The phenomenon of water sticking to a surface, such as a window pane or
Korolek [52]

Answer:

Because of the Cohesion

4 0
2 years ago
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In an intergalactic competition, spaceship pilots compete to see who can cover the distance between two asteroids in the short-
pogonyaev

Answer:

a)  truc is C,  b) correct result is the B

Explanation:

As the speed of the competition is very high, for the judges the speed is

           v = d / t

           v = 3 109 m / 20

           v = 1.5 108 m / s

This is half the speed of light. For these high speeds we must use the relations of special relativity.

For the time          t = to γ

For distance         L = Lo / γ

                            γ = √ (1-v2 / c2)

Own time and distance (to and Lo) corresponds to the observer who is not moving the judges in this case

Let's look for the range value

                     γ = 1 / √ (1 - (1.5 / 3) 2) = 1 / 0.866 = 1.15

The time              t = 20 1.15 = 23 s

The distance       L = 3 10 9 /1.15 = 2.60 109 m

From these results we see that time increases and the distance is shorter.

Let's review the claims

A) False. It's the opposite

B) False

C) True. It is according to the result found

D) False.

In the nuclear fusion process, we will also use the special relativity that has a relationship between energy and mass

         ΔE = c² Δm

As in the process energy is released, for the law of conservation of the mass of energy to be fulfilled, the total mass of the products, He atom, must be reduced.

Therefore the correct result is the B

4 0
3 years ago
Can I also get help on this??
Xelga [282]

Answer:

25

Explanation:

8 0
2 years ago
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