Answer:
the charge carriers have an energy 2.8 10⁻¹⁹ J
Explanation:
The energy in a diode is conserved so the energy supplied must be equal to the energy emitted in the form of photons.
The energy of a photon is given by the Planck expression
E = h f
the speed of light, wavelength and frequency are related
c = λ f
we substitute
E =
a red photon has a wavelength of lam = 700 nm = 700 10⁻⁹ m
we calculate the energy
E = 6.626 10⁻³⁴ 3 10⁸/700 10⁻⁹
E = 2.8397 10⁻¹⁹J
therefore the charge carriers have an energy 2.8 10⁻¹⁹ J,
Answer:
An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.
Explanation:
Answer:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2."
Explanation:
The full question has not been provided, so I just copied this into the web and found this answer and explanation on quizlet:
"The wavelengths are the same for both. The width of slit 1 is larger than the width of slit 2.
D sin θ = m λ
if the wavelengths are the same, then if the angle is smaller, the slit width must be larger. The top photo shows a pattern that is more closely spaced. That means the angle is smaller. The slit width must be larger."
This answer/explanation should be correct, as we are looking at bright fringes and the formula being used corresponds to the parameters of the question.
Hope this helps!
Answer:
<h2>
False</h2>
Explanation:
Hope this helps! Please consider marking brainliest! Always remember, your smart and you got this! -Alycia :)
With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance). So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .