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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:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

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