Newton's first law equations like velocity and stuff like that

Physics

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force. ... An object sliding across a table or floor slows down due to the net force of friction acting on the object.

Explanation:

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How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 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,

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