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lara [203]
4 years ago
12

A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 10-27

kg. Determine the density of the proton.
Physics
2 answers:
Step2247 [10]4 years ago
8 0
For the answer to the question above asking to d<span>etermine the density of the proton. 
</span>Density is mass over volume. 

The volume of a sphere is 4πr³/3. r is half the diameter. 

So the density would be 2.3×10¹⁷ kg/m³. 
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
mr_godi [17]4 years ago
6 0
<h3>The density of the proton is 2.3 x 10²⁰ g/m³.</h3><h3 /><h3>FURTHER EXPLANATION</h3>

The density of a substance is equal to the ratio of its mass and volume.

Density = \frac{Mass}{Volume}

To get the density of the proton, the following steps must be completed:

1. Determine the volume of the spherical proton.

2. Calculate the ratio of the mass and the volume of the proton.

<u>STEP 1:</u> To determine the volume, the equation for the volume of a sphere must be known.

V \ = \frac{4 \pi r^{3}}{3}\\

where r is 1/2 the diameter of the sphere.

Plugging in the values to get the volume in cubic meters:

V \ = \frac{4 \pi (1.2 \times 10^{-15} \ m)^3}{3}\\\\\\\\boxed {V = 7.23456 \times 10^{-45} \ m^3\\}

STEP 2: Once the volume is known, the density can be calculated by getting the ratio of the mass (in grams) and the volume in cubic meters.

Density \ = 1.67 \times 10^{-27} \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1}{7.23456 \times 10^{-45} \ m^3}\\\\\\\boxed {Density \ = 2.3084 \times 10^{20} \frac{g}{m^3}\\}

Since the least number of significant figures in the given is 2, the final answer must also only have two significant figures.

Therefore,

\boxed {\boxed {Density = 2.3 \times 10^{20} \frac{g}{m^3}}} \

<h3>LEARN MORE</h3>
  1. Dimensional Analysis brainly.com/question/1594497
  2. Stoichiometry brainly.com/question/4867681
  3. Significant Figures brainly.com/question/1566507

<em>keywords: unit conversion, dimensional analysis, density</em>

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Answer:

Because of the speed of the sound.

Explanation:

The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].

Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.

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The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>
  • Range = U²× sin(2θ)/g
  • U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
  • U=√{Range×g/sin(2θ)}
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<h3>What's the expression of time of flight in projectile motion?</h3>
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Answer: Yes

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Explanation:

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