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Lunna [17]
3 years ago
12

The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen

atom orbits the nucleus at a distance of 5.3 × 10-11 m. What is the ratio of the density of the hydrogen nucleus to the density of the complete hydrogen atom?
Physics
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

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Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 20 cm apart. The sound
Sophie [7]

Answer:

a. Wavelength = λ = 20 cm

b. Next distance of maximum intensity will be 40 cm

Explanation:

a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.

Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.

This we get the equation:

(nλ + λ/2) - nλ = 30-20

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b. at what distance, sound intensity will be maximum again.

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When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

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So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

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