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Anvisha [2.4K]
3 years ago
8

The drawing shows a tire of radius R on a moving car

Physics
1 answer:
masha68 [24]3 years ago
7 0

Answer:

 H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

          Em₀ = K = 1/2 m v²

Final. In the sought height

         Em_{f} = U = mg h

Energy is conserved

        Em₀ =  Em_{f}

        ½ m v² = m g h

        v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

        Em₀ = K = ½ I w²

Final. Heights sought

        Emf = U = m g R

        Em₀ =  Em_{f}

       ½ I w² = m g R

The moment of inertial of a cylinder is

       I = I_{cm} + ½ m R²

      I= ½ I_{cm}  + ½ m R²

Linear and rotational speed are related

       v = w / R

       w = v / R

We replace

      ½ I_{cm}  w² + ½ m R² w²  = m g R

moment of inertia of the center of mass      

      I_{cm}  = ½ m R²

     ½ ½ m R² (v²/R²) + ½ m v² = m gR

    m v² ( ¼ + ½ ) = m g R

   

       v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

        2 g H = 4/3 g R

         H / R = 2/3

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Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 &lt; x a. 283 eV <br> b. 339 eV <br> c.
denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113   eV  

d. 226 eV        

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

E_{nx,ny,nz = ( n_x^2 + n_y^2 + n_z^2 ) × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( n_x = n_y = n_z = 1 )

so

( n_x^2 + n_y^2 + n_z^2 ) ×  π²h"² / 2ml²

since h" = h/2π

( n_x^2 + n_y^2 + n_z^2 ) × π²h² / (2π)²2ml²

so we substitute

E_{111 = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

E_{111 = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]    

E_{111 = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

E_{111 = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

E_{111 = 3 × [ 37.645578 ]

E_{111 = 112.9 ≈ 113 eV

E_{211 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{211 = ( 1² + 1² + 2² ) × [ 37.645578 ]

E_{211 = 6 × [ 37.645578 ]

E_{211 = 225.87 ≈ 226 eV

E_{221 = ( n_x^2 + n_y^2 + n_z^2 )  × π²h² / (2π)²2ml²

we substitute

E_{221 = ( 2² + 2² + 1² ) × [ 37.645578 ]

E_{211 = 9 × [ 37.645578 ]

E_{211 = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

8 0
2 years ago
A conveyer belt moves a 40 kg box at a velocity of 2 m/s. What is the kinetic energy of the box while it is on the conveyor belt
ankoles [38]
Kinwtic energy = (40kg*2*2)/2=40*2=80J
4 0
3 years ago
Read 2 more answers
A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f
jonny [76]

Answer:

<em>Entropy Change = 0.559 Times</em>

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

5 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
Assign a positive velocity to the red box and negative velocity to the blue box. Are they moving in opposite direction?
attashe74 [19]

Answer:

<em>Yes, they are moving in opposite direction one to the other.</em>

Explanation:

Velocity is a vector quantity, which means that it has both magnitude and direction. The magnitude shows the size of the velocity, and the direction shows which way it is moving in reference to a chosen reference direction. If the red box is assigned a positive velocity, and the blue box is assigned a negative velocity, as indicated in the question, then it means that the red box, and the blue box, both move in opposite direction to the other.

4 0
3 years ago
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