Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variabl
1 answer:
Answer:
D) D = , E) (C, D) = (
Explanation:
Part D) two expressions are indicated
3C + 4D = 5
2C +5 D = 2
let's simplify each expression
3C + 4D = 5
4D = 5 - 3C
we divide by 4
D =
The other expression
2C +5 D = 2
2C = 2 - 5D
C =
we can see that the correct result is 1
Part E.
It is asked to solve the problem by the substitution method, we already have
D =
we substitute in the other equation
2C +5 D = 2
2C +5 (5/4 - ¾ C) = 2
we solve
C (2 - 15/4) + 25/4 = 2
-7 / 4 C = 2 - 25/4
-7 / 4 C = -17/4
7C = 17
C =
now we calculate D
D =
D = 5/4 - 51/28
D =
D = - 16/28
D =
the result is (C, D) = ( )
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Answer:
The equations are:
C. 2t=(m+12)λn(m=0,1,2,...) for constructive interference. (Here is a mistake in this equation i.e. m + 12 (wrong) and m + 1/2 (correct)). So, the correct equation for constructive interference is:
2t = (m + 1/2) λ n (m=0,1,2,...)
D. 2t=mλn(m=0,1,2,...) for destructive interference.
Explanation:
Constructive Interference occurs when two waves’ crests or troughs add together therefore, the resulting wave amplitude is equal to the sum of the each wave amplitude (in this case the amplitude of each wave is of same charge).
Destructive Interference occurs when a waves’ crest and trough cancel each other therefore, the resulting wave amplitude is zero (in this case the amplitude of each wave is oppositely charged).
When the refractive index of the medium in which the light travels is lower than the refractive index of the thin film medium then a 180° phase shift occurs.
When the light wave strikes the top surface of thin film, the wave reflects from the top surface and the reflecting light wave shifts by half of wavelength i.e. λ/2
When the light wave strikes the bottom surface of thin film then the light wave does not shift however, the light cover extra distance (2t), reflecting from the bottom surface to travel back to air.
Since, oil has refractive index (n = 1.5) which is higher than the refractive index of air (n = 1) so, when light wave strikes on the top surface of thin film of oil, the reflecting wave will shift by λ/2 (because nair < noil) and when light wave strikes on the bottom surface of thin film of oil, it will cover an extra distance to travel back to air and no wavelength shift will occur (because noil > nair).
In light of the above situation, the constructive interference of light reflected from the thin oil film is:
2t = (m + 1/2) λ n
The destructive interference of light reflected from the thin film is:
2t = m λ n
Terminal velocity is the maximum velocity attainable by an object as it falls through a fluid and is given by,
Where,
m = mass of the falling object.
g = the acceleration due to gravity.
\rho = the density of the fluid the object is falling through.
A = the projected area of the object.
C = the drag coefficient.
A) Because a drop is difficult to approximate to a certain form, it usually tends to be considered a spectrum, the terminal velocity for a sphere is given by
Whatever our appreciations, all the variables are constant, except for the Diameter, we can realize that the terminal velocity is proportional to the radius of the object, the greater the radius - the larger the drop - the greater the terminal velocity.
B) Since there is a "constant" terminal velocity at the end of the path, at which point the forces are balanced, the acceleration will be 0. For both objects.