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1 year ago

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If I0 is the intensity of the unpolarized light incident on the first polarizer, and I1 and I2 denote the intensity of the light

after the first and second polarizers, respectively, which of the following quantities are known?
a) I1
b) theta1
c) I0
d) I2
e) theta2

1 answer:

e-lub [12.9K]1 year ago

7 0

E or C 10 hope this helps

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Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)

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Answer with Explanation:

We are given that

Restoring force, $(FS_p)s=-k\Delta s-q(\Delta s)^3$

$k=350N/m$

$q=750 N/m^3$

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$\Delta s=15 cm=0.15 m$

Using 1 m=100 cm

Work done= $\int_{0}^{0.15}-Fd(\Delta s)$

W= $-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)$

$W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}$

$W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750$

$W=4.033 J$

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Percentage increase in work= $\frac{4.033-3.938}{3.928}\times 100=2.4$ %

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2 years ago

Two flat conductors are placed with their inner faces separated by 6.0 mm. If the surface charge density on one of the inner fac

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Explanation:

Relation between electric field and charge density is as follows.

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So, $E_{\text{outside}}$ = 0

$E_{inside} = \frac{+\sigma}{2 \epsilon} - \frac{-\sigma}{2 \epsilon}$

or, $E_{inside} = \frac{\sigma}{\epsilon}$

Now, formula to calculate the potential difference of two conductors is as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

It is given that,

d = 6.0 mm = $6 \times 10^{-3} m$

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Hence, we will calculate the magnitude of the electric potential differences between the two conductors as follows.

$V_{1} - V_{2} = \frac{\sigma \times d}{\epsilon}$

= $\frac{40 \times 10^{-12} \times 6 \times 10^{-3}}{8.85 \times 10^{-12}}$

= 0.0271 volts

thus, we can conclude that value of the magnitude of the electric potential differences between the two conductors is 0.0271 volts.

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2 years ago

Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

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Answer:

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E = 2.07x10⁵ MPa

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The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$

b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is

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550 nm is in the visible spectrum. 400 nm is in the ultra-violet spectrum. The shorter the wavelength, the bigger the energy of the light photon emitted. Therefore a star that emits 400 nm photons, is hotter than the Sun. A star that is hotter is either bigger than the Sun (and has about the same age), either younger than the Sun (and has about the same dimension).
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leva [86]

Answer:

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Explanation:

t = Time taken

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s = Displacement = 99.4

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If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

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1 year ago