Answer:
f = 130 Khz
Explanation:
In a circuit driven by a sinusoidal voltage source, there exists a fixed relationship between the amplitudes of the current and the voltage through any circuit element, at any time.
For an inductor, this relationship can be expressed as follows:
VL = IL * XL (1) , which is a generalized form of Ohm's Law.
XL is called the inductive reactance, and is defined as follows:
XL = ω*L = 2*π*f*L, where f is the frequency of the sinusoidal source (in Hz) and L is the value of the inductance, in H.
Replacing in (1), by the values given of VL, IL, and L, we can solve for f, as follows:
f = VL / 2*π*IL*L = 12 V / 2*π*(3.00*10⁻³) A* (4.9*10⁻³) H = 130 Khz
Answer:
The vacuum tube was replaced with transistor.
Explanation:
- The invention of semiconductor was very useful in making solid state transistor that allowed the production of small yet faster, cheaper, and more trusted and reliable computers.
- These solid state transistor is so often used that, it nearly replaced all the use of transistor.
- This replacement took place after the invention of semiconductor in the year around 1940.
- Vacuum tubes also known as thermionic tubes are not used anymore in computers and electronics.
Before answering the question, first we have to understand a longitudinal wave.
A longitudinal wave is a type of mechanical wave in which the direction of wave propagation is parallel to the particle vibration of the medium.
In this type of wave, there will be compressions and rarefactions. Compressions are the high pressure regions where the particles of the medium are very close to each other. The rarefactions are the low pressure regions of a longitudinal wave where the particles are not so close to each other.
Hence, a longitudinal wave is a series of compressions and rarefactions.
The wavelength of a longitudinal wave is defined as the distance between two successive compressions or rarefactions.
Hence, the correct answer to the question is C) by measuring the distance between adjacent rarefactions.
Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
