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2 years ago

Earth's lithospheric plates interact at____.

Physics

1 answer:

earnstyle [38]2 years ago

7 0

Explanation:

Tectonic plate interactions are classified into three basic types: Divergent boundaries are areas where plates move away from each other, forming either mid-oceanic ridges or rift valleys. These are also known as constructive boundaries. Convergent boundaries are areas where plates move toward each other and collide.

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The index of refraction of silicate flint glass for red light is 1.620 and for violet light is 1.660 . A beam of white light in

Sonja [21]

Answer:

10.16 degrees

Explanation:

Apply Snells Law for both wavelenghts

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

For red

(1.620)(sin 25.5) = (1)(sin r)

For red, the angle is 35.45degrees

For violet

(1.660)(sin 25.5) = (1)(sin v)

For violet, the angle is 45.6 degrees

The difference is 45.6- 35.45 = 10.16 degrees

3 0

3 years ago

an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o

ZanzabumX [31]

Answer:

The maximum amplitude ( $V_{max}$ ) will be 7.96 V.

Explanation:

We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

$S = 2 \pi f V_{max}$

where ' $f$ ' is the highest frequency signal.

Therefore, from the above equation we can write,

$&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V$

3 0

3 years ago

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$