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TiliK225 [7]
4 years ago
10

Several years from now you have graduated with an engineering/physics degree and have been hired by a nanoengineering firm as an

intern. You have been assigned to work under a top engineer from the company. Their current project is to design a microscopic oscillator as a time keeping device. The engineering design involves placing a negative charge at the center of a very small positively charged metal ring. Your boss claims that the negative charge will undergo simple harmonic motion if displaced away from the center of the ring. Furthermore, they claim they can change the period (timing) of oscillation by adjusting the amount of charge on the ring. The first task they give you is to check the validity of their design.
a. Consider a charge −???? located a small distance z above the center of a positively charged ring with total charge +Q and radius R. Write an expression for the net force exerted on the charge −???? due to the ring of charge. What is the magnitude of the force on the charge −???? if it is at the location z = 0?

b. Use the binomial expansion as a mathematical tool to understand the motion of the charged particle under the assumption that it is only displaced a small distance z ≪ ???? away from the center of the ring. Using the binomial approximation, simplify your expression from part (a). You should have two terms in your expression for the net force.

c. Suppose that z =1.0 nanometer and R = 1.0 millimeter. Using these values, determine the relative strength of the force that each term in your expression contributes to the total force (calculate the ratio of z/R that appears in each term). Is one term negligible (i.e. is the value of one term very close to zero)?

d. Using results from part (c) show that the relevant force on charge −???? simplifies to a Hook’s law restoring force: ???? = −????x, where k is now a more complicated constant made up of several constants from this problem. What is k?

e. Using Newton’s 2nd Law, write down the differential equation of motion for the particle as it undergoes oscillations due to this Hook’s law force. Determine the period T of oscillations charge −???? makes about the equilibrium point z = 0 (this should not involve any calculations, see equations 15.32 and 15.37 from your book). Asses the validity of your solution by analyzing the physical units to make sure that your solution has units of time.
Physics
1 answer:
nadya68 [22]4 years ago
5 0

Answer:

See attached handwritten document for answer

Explanation:

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Katyanochek1 [597]

Answer:

When the temperature increases

Explanation:

5 0
3 years ago
In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le
expeople1 [14]

Answer:

The  value  is   m \approx   310

Explanation:

From the question we are told that

     The  focal length of the objective is  f_o =  1.0 \ cm

    The  focal length of the eyepiece is  f_e  =  2.0 \  cm

    The  tube length is  L  =  25 \  cm

Generally the magnitude of the overall magnification is mathematically represented as

            m =  m_o  *  m_e

Where  m_o is the objective magnification which is mathematically represented as

        m_o  =  \frac{L}{f_o }

=>      m_o  =  \frac{25}{1 }

=>      m_o  =  25

m_e is the eyepiece magnification which is mathematically evaluated as

     m_e  =  \frac{L }{f_e }

     m_e  =  \frac{25 }{ 2}

      m_e  =  12.5 \  cm

So

    m =  25 * 12.5

     m \approx   310

6 0
3 years ago
In Fig., block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.50 m and then collides with stationary
Evgesh-ka [11]

Answer:

Explanation:

Given

initial height h=2.5 m

m_2=2m_1

coefficient of static friction \mu =0.5

When collision is elastic respective velocities after collision is

v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}

v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}

where u_1 and u_2=initial velocities of object

v_1 and v_2 final velocities of object

u_1=\sqrt{2\times 9.8\times 2.5}

u_1=7 m/s

v_2=\frac{0+2m_1\times 7}{m_1+2m_1}

v_2=\frac{14}{3} m/s

using v^2-u^2=2 as

0-(4.67)^2=2\times (-0.5\times 9.8)\times s

s=2.22\ m

(b)Completely Inelastic

In Completely Inelastic objects stick with each other

m_1u_1=(m_1+m_2)v

v=\frac{u_1}{3}=\frac{7}{3} m/s

using v^2-u^2=2 as

0-(2.33)^2=2\times (-0.5\times 9.8)\times s

s=0.55\ m                          

7 0
4 years ago
A car travels on a banked circular curve at constant speed. All four of the car's tires roll without slipping. The car moves at
kvv77 [185]

Answer:

b. The horizontal component of the normal force acts toward the center of the circle.

Explanation:

When a car on a banked turn move with just the right speed so that the frictional force is zero, a component of the normal force -which is always perpendicular to the surface- acts as the centripetal force on the car. This is because the normal force now has a horizontal component, and this component supplies the centripetal force.

7 0
3 years ago
What conditions are required for a solar eclipse?
IgorLugansk [536]
The phase of the Moon must be new, and the nodes of the Moon's orbit must be nearly aligned with Earth and the Sun.
5 0
3 years ago
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