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4 years ago

Distinguish between a converging lens and a diverging lens.

1 answer:

5 0

Answer:A converging lens is thickest in the middle and causes parallel light rays to converge through the focal point on the opposite side of the lens. A diverging lens is thinner in the middle and causes parallel light rays to diverge away from the focal point on the same side of the lens.

Explanation:

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Determine the total impedance of an LRC circuit connected to a 10.0- kHz, 725-V (rms) source if L = 36.00 mL, R = 10.00 kÎ©, and

SOVA2 [1]

Answer:

10042.6 ohm

Explanation:

f = 10 kHz = 10000 Hz, L = 36 mH = 0.036 H, R = 10 kilo Ohm = 10000 ohm

C = 5 nF = 5 x 10^-9 F

XL = 2 x π x f x L

XL = 2 x 3.14 x 10000 x 0.036 = 2260.8 ohm

Xc = 1 / ( 2 x π x f x C) = 1 / ( 2 x 3.14 x 10000 x 5 x 10^-9)

Xc = 3184.7 ohm

Total impedance is Z.

Z^2 = R^2 + (XL - Xc)^2

Z^2 = 10000^2 + ( 2260.8 - 3184.7 )^2

Z = 10042.6 ohm

4 0

3 years ago

91.0x26x504 ...............

Hoochie [10]

Answer:

1192464

Explanation:

91.0 times 26 = 2366

2366 times 504 = 1192464

7 0

3 years ago

Read 2 more answers

A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holdi

Elis [28]

Answer: $115.52\ N$

Explanation:

Given

Length of plank is 1.6 m

Force $F_1$ is applied on the left side of plank

Force $F_2$ is applied 43 cm from the left end O.

Mass of the plank is $m=13.7\ kg$

for equilibrium

Net torque must be zero. Taking torque about left side of the plank

$\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N$

Net vertical force must be zero on the plank

$\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N$

8 0

3 years ago

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$