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ExtremeBDS [4]
3 years ago
5

In the Bohr model of the hydrogen atom, an electron orbits a proton in a circular orbit of radius 0.53 X 10^-10 m. What is elect

ric potential at the electron's orbit due to the proton? What is the Potential Energy of the electron? What is the total energy of the electron in its orbit? What is the ionization energy? Express your results in electron volts.
Physics
1 answer:
maxonik [38]3 years ago
7 0

Answer:

a) 27.2 V

b)27.2 V

Explanation:

Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C

Radius = r = 0.53×10⁻¹⁰ m

Electric Potential = V = k q/r

k = 9 ×10⁹ N m²/C² = Coulomb's constant.

V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V

b) Potential Energy of the electron =  k q × q / r

   = [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,

since 1 electron volt = (1.6 × 10⁻¹⁹)joules

   = 27.2 eV

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A girl throws a rock horizontally at 10 m/s from the top of a building, 22 m above street level. Assuming free fall conditions a
MAXImum [283]

Answer:21.18 m

Explanation:

Given

initial speed u=10 m/s

height of building h=22 m

time taken to complete 22 m

h=ut+\frac{1}{2}at^2

initial vertical velocity =0

22=\frac{1}{2}gt^2

t=\sqrt{\frac{22\times 2}{g}}

t=2.11 s

Horizontal Distance moved

R=u_x\times t

R=10\times 2.11

R=21.18 m

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3 years ago
What is the largest planet in our galaxy
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In our Solar System, Jupiter is the largest planet we have. it has the surface area of 23.71 billion mi^2. it beats all the other planets in both mass and volume.
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2 years ago
True or false the potential energy of a freely object increases as it begins to fall
Eva8 [605]
False, as an object falls its potential energy turns into kinetic energy thus decreasing the potential energy.
3 0
3 years ago
I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

7 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
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