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seropon [69]
3 years ago
8

In the Bohr model of the hydrogen atom, an electron in the 7th excited state moves at a speed of 4.47 104 m/s in a circular path

having a radius of 2.59 10-9 m. What is the effective current associated with this orbiting electron
Physics
2 answers:
natima [27]3 years ago
8 0

Answer:

4.4 x 10^-7 A

Explanation:

charge on an electron, q = 1.6 x 10^-19 C

velocity, v = 4.47 x 10^4 m/s

radius, r = 2.59 x 10^-9 m

Let the effective current is i.

According to the definition of current, the rate of flow of charge through the conductor is called current.

i = q / T

where, T is the time period of rotation.

T = \frac{2\pi r }{v}

T = \frac{2\times 3.14\times 2.59\times 10^{-9} }{4.47\times 10^{4}}

T = 3.64 x 10^-13 s

So, current

i=\frac{1.6\times 10^{-19}}{3.64\times 10^{-13}}

i = 4.4 x 10^-7 A

tekilochka [14]3 years ago
7 0

Answer:

I = 4.38 x 10⁻⁷ A

Explanation:

Given,

speed = 4.47 x 10⁴ m/s

radius of the circular path, r = 2.59 x 10⁻⁹ m

effective current = ?

The current represented by the orbiting electron is equal to

I = q \times f

f is the frequency

q is charge of electron

we know,

f = \dfrac{v}{2\pi r}

f = \dfrac{4.47\times 10^{4}}{2\pi\times 2.59\times 10^{-9}}

f = 2.74 x 10¹² Hz

now,

I = 1.6 \times 10^{-19}\times 2.74\times 10^{12}

I = 4.38 x 10⁻⁷ A

Hence, the effective current associated with the orbiting electron is equal to I = 4.38 x 10⁻⁷ A

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An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-
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Answer:

  1. Power requirement <u>P</u> for the banner is found to be  30.62 W
  2. Power requirement <u>P</u> for the solid flat plate is found to be 653.225 W
  3. Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The <em>Cd </em>for banner was given, whereas the <em>Cd </em>for a flat plate is generally found to be around <em><u>1.28</u></em><em> </em>which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
  4. Power requirement <u>P</u> for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

  1. v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
  2. The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
  3. The power can be determined this equation: P = F . v, where F represents <em>the drag-force</em> that we will need to determine and v represents the<em> velocity of the airplane</em>
  4. The equation to determine drag-force is: F = 1/2 * d *  C_d * A

In the drag-force equation Cd represents the c<em>o-efficient of drag</em> and A represents the <em>frontal area of the banner/plate/balloon (the object being towed)</em>

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

<u>Part a)</u> We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = <em>1/2 * 0.06* 1.225 * 20</em> = 0.735 N. Now to find the power P we will use P = F . v i.e.<em> 0.735 * 41.66</em> = <u><em>30.62 W</em></u>

<em></em>

<u>Part b) </u>For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = <em>1/2 * 1.28 * 1.225 * 20</em> = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = <u><em>653.225 W</em></u>

<u><em></em></u>

<u>Part c)</u> The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

<u>Part d)</u> The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : <em>pi* r^2 (r = d / 2).</em> Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = <u><em>40.08 W</em></u>

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Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is  T =8391.6 N

Explanation:

From the question we are told that

       The strain on the strain on the head is \Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m

         The contact area is  A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2  

Looking at the first diagram

           At  600 MPa of stress

               The strain is  0.3mm/mm

          At   450 MPa of stress

                 The strain is   0.0015 mm/mm

 To find the stress at  \Delta l we use the interpolation method

            \frac{\sigma_{\Delta l} -  \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l }  - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}

Substituting values

              \frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}

            \sigma _{\Delta l} -450 = 49.50

             \sigma _{\Delta l} = 499.50 MPa

Generally the force on each head is mathematically represented as

              F = \sigma_{\Delta l} * A

Substituting values

             F = 499.50*10^{6} * 2.8*10^{-6}

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Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

              T = 6 * F

                  = 6* 1398.6

              T =8391.6 N

                 

     

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