Answer:
The power dissipated in either one of the parallel resistors is 2 V
Explanation:
Given;
two parallel resistors, R₁ and R₂ = 2 ohms
The total resistance of the Two resistors of 2 ohms connected in parallel is;

when connected to another resistor of 1 ohm in series, the total resistance becomes;
Rt = R₁ + R₂
Rt = 1 + 1 = 2 ohms
Current in the circuit, I = voltage / total resistance
= 2 /2 = 1 A
the overall circuit has been resolved to series connection, and current flow in series circuit is constant.
Power = I²R
Thus, power dissipated in either one of the parallel 2 ohms resistors is;
Power = I²R = (1)² x 2 = 2 V
Answer:
the action of polarizing or state of being or becoming polarized: such as the action or process of affecting radiation and especially light so that the vibrations of the wave assume a definite form. Polarization, in Physics, is defined as a phenomenon caused due to the wave nature of electromagnetic radiation. There are two types of waves, transverse waves, and longitudinal waves.
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Explanation:
According to Rydberg's formula, the wavelength of the balmer series is given by:

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

Here,
is the "general" Rydberg constant,
is electron's mass and M is the mass of the atom nucleus
For hydrogen, we have,
:

Now, we calculate the wavelength for hydrogen:
![\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5Clambda%7D%3DR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5C%5C%5Clambda%3D%5BR_H%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.0967%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5646%2A10%5E%7B-7%7Dm%3D656.46nm)
For deuterium, we have
:
![R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm](https://tex.z-dn.net/?f=R_D%3D%5Cfrac%7B1.09737%2A10%5E7m%5E%7B-1%7D%7D%7B%281%2B%5Cfrac%7B9.11%2A10%5E%7B-31%7Dkg%7D%7B2%2A1.67%2A10%5E%7B-27%7Dkg%7D%29%7D%5C%5CR_D%3D1.09707%2A10%5E7m%5E%7B-1%7D%5C%5C%5C%5C%5Clambda%3D%5BR_D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D%5B1.09707%2A10%5E7m%5E%7B-1%7D%28%5Cfrac%7B1%7D%7B2%5E2%7D-%5Cfrac%7B1%7D%7B3%5E2%7D%29%5D%5E%7B-1%7D%5C%5C%5Clambda%3D6.5629%2A10%5E%7B-7%7D%3D656.29nm)
Answer:
D) The velocity of the airplane must be greater than the speed of sound
Explanation:
Sonic boom is a phenomenon that occurs when an airplane travels ay a velocity greater than the speed of sound.
When this condition is met, the sound waves produced by the airplane (mainly from the engine) is not able to travel ahead of the plane, because the plane is moving at a greater speed than the sound itself; as a result, it is only transmitted behind the plane.
In this situation, a shock wave is produced. This shock wave is associated with a very loud "bang", that can be heard clearly even at several kilometers distance (in fact, supersonic planes are normally not allowed to fly over residential areas. In the past, for example the Concorde, was only allowerd to fly at supersonic speed over the ocean).
So the correct answer is
D) The velocity of the airplane must be greater than the speed of sound