Several years from now you have graduated with an engineering/physics degree from OSU and have been hired by a nanoengineering f
irm 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 of 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.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?
1 answer:
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Answer:
(a). The coefficient of permeability is .
(b). The seepage velocity is 0.0330 cm/s.
(c). The discharge velocity during the test is 0.0187 cm/s.
Explanation:
Given that,
Length = 175 mm
Diameter = 175 mm
Time = 90 sec
Volume= 405 cm³
We need to calculate the discharge
Using formula of discharge
Insert the values into the formula
(a). We will compute the coefficient of permeability
Applying the formula for permeability
Where, Q=discharge
l = length
A = cross-sectional area
h=constant head that causes flow
Plugging the value into the formula
(c). To ascertain the discharge velocity during the testing phase
Utilizing the discharge velocity formula
Substituting the values into the equation
The discharge velocity during the test measures 0.0187 cm/s.
Thus, (a). The coefficient of permeability is .
(b). The seepage velocity computes at 0.0330 cm/s.
(c). The observed discharge velocity during the test equals 0.0187 cm/s.
To start, we first need to determine the kinetic energy of the penny before it strikes the ground. This is calculated using the formula where m equals 5.25 g, which is 0.00525 kg for the penny's mass, and v equals 3.27 m/s for its speed. Replacing the values into the equation provides: When the penny lands, all this kinetic energy transforms into internal energy for both the penny and the ground. If half of this energy goes into the penny's internal energy, the change is determined by a specific formula where m is the penny's mass, Cs is its specific heat capacity (2.03 J/gC), and 
, the change in temperature. To find the last element, the equation will be solved.
Answer:
The number of photons emitted each second is
Explanation:
Let 'n' stand for the quantity of photons released by the bulb.
Provided Information:
The bulb radiates energy at a rate of 100 J per second (E).
Wavelength of emitted light is (λ) = 525 nm =
The energy of a photon is calculated by:
Where,
Now, if we have 'n' photons, the total energy is equivalent to the energy of a single photon multiplied by the count of photons. Thus,
To express in terms of 'n', we find:
Insert the provided values and solve for 'n'. The resulting calculation yields
Consequently,
photons are discharged every second.