What is the si unit of potential difference

Match the following light concepts

Gelneren [198K]

Answer:

- Light is bounced back at same angle (Classical Reflection)

- Light penetrates medium at different angle due to different material densities (Refraction)

Light bounces at different angles in periodic grid (Reflected Diffraction)

Light enters medium at different angles through a grid (Transmission Diffraction)

- Light EMF field looses one axis component (Polarized filter)

Explanation:

Reflection is a phenomenon in which waves (light included) bounce back from an obstacle at the same angle of incidence

Refraction is the change in the angle of a wave as it enters the interface of two media. The change in angle is due to the difference in the densities of the two media.

Reflected diffraction occurs when an optical component with a periodic grid, splits, and diffracts light into several beams travelling in different directions. The light light bounces at an angle in the periodic grid.

Transmission diffraction is dispersion a beam of various wavelengths into a spectrum of associated lines due to the principle of diffraction. In this type of diffraction, light enters medium at different angles through a grid.

Polarized filters removes one field from the incidence electromagnetic wave like light, leaving it to vibrate in only one plane.

6 0

4 years ago

A 60-kg woman holds a 9-kg package as she stands within an elevator which briefly accelerates upward at a rate of g/4. Determine

Anuta_ua [19.1K]

Answer:

force R = 846.11 N

lifting force L = 110.36 N

if cable fail complete both R and L will be zero

Explanation:

given data

mass woman mw = 60 kg

mass package mp = 9 kg

accelerates rate a = g/4

to find out

force R and lifting force L and if cable fail than what values would R and L acquire

solution

we calculate here first reaction R force

we know elevator which accelerates upward

so now by direction of motion , balance the force that is express as

R - ( mw + mp ) × g = ( mw + mp ) × a

here put all these value and a = g/4 and use g = 9.81 m/s²

R - ( 60 + 9 ) × 9.81 = ( 60 + 9 ) × g/4

R = ( 69 ) × 9.81/4 + ( 69 ) 9.81

R = 69 ( 9.81 + 2.4525 )

force R = 846.11 N

and

lifting force is express as here

lifting force = mp ( g + a)

put here value

lifting force = 9 ( 9.81 + 9.81/4)

lifting force L = 110.36 N

and

we know if cable completely fail than body move free fall and experience no force

so both R and L will be zero

5 0

3 years ago

Verify if 83 is a term in the arithmetic sequence 11,15,19,23

EleoNora [17]

Answer:

yes, it is

Explanation:

The sequence: (+4)

23,27,31,35,39,43,47,51,55,59,63,67,71,75,79,83

Hope this helps! :)

3 0

3 years ago

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stre

velikii [3]

Complete Question

Consider a single crystal of some hypothetical metal that has the BCC crystal structure and is oriented such that a tensile stress is applied along a [121] direction. If slip occurs on a (101) plane and in a [111] direction, compute the stress at which the crystal yields if its critical resolved shear stress is 2.9 MPa.

Answer:

The stress is $\sigma = 10. 655 MPa$

Explanation:

From the question we are told that

The critical yield resolved shear stress is $\sigma = 2.9Mpa$

First we obtain the angle $\lambda$ between the slip direction [121] and [111]

$\lambda = cos^{-1} [\frac{(u_1 u_2 + v_1 v_2 + w_1 w_2}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_2^2 + v_2^2 + w_2 ^2)} } ]$

Where $u_1 ,u_2 ,v_1 , v_2 , w_1 , w_2$ are the directional indices

$\lambda = cos ^-[ \frac{(1) (-1) + (2) (1) + (1) (1)}{\sqrt{((1)^2 +(2)^2 + (1)^2)}\sqrt{((-1)^2 + (1)^2 + (1)^2 ) } } ]$

$= cos^{-1} [\frac{2}{\sqrt{6} \sqrt{3} } ]$

$= 61.87^0$

Next is to obtain the angle $\O$ between the direction [121] and [101]

$\O = cos^{-1} [\frac{(u_1 u_3 + v_1 v_3 + w_1 w_3}{\sqrt{u_1^2 + v_1 ^2+ w_1^2})\sqrt{( u_3^2 + v_3^2 + w_3 ^2)} } ]$

Substituting 1 for $u_1$ , 2 for $v_1$ , 1 for $w_1$ , 1 for $u_2$ , 0 for $v_2$ , and 1 for $w_2$

$\O = cos^{-1} [\frac{1* 1 + 2*0 + 1*1 }{\sqrt{1^2 + 2^2 + 1^2 } \sqrt{(1^2 + 0^2 + 1^2 )} } ]$

$\O = cos^{-1} [\frac{2}{\sqrt{6} \sqrt{2} } ]$

$= 54.74 ^o$

The stress is mathematically represented as

$\sigma = \frac{\tau_c}{cos \O cos \lambda }$

$= \frac{2.9}{cos 54.74^o cos 61.87^o}$

$= \frac{2.9}{0.2722}$

$\sigma = 10. 655 MPa$

5 0

3 years ago

Determine the maximum volume in gallons [gal] of olive oil that can be stored in a closed cylindrical silo with a diameter of 3

Olin [163]

Answer:

V=1601gal

Explanation:

Hello! This problem is solved as follows,

First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.

This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.

P=Poil+Patm

P=total pressure or absolute pressure=26psi=179213.28Pa

Patm= the atmospheric pressure =101325Pa

Poil=pressure due to the weight of olive oil=0.86αgh

α=density of water=1000kg/m^3

g=gravity=9.81m/s^2

h= height that olive oil reaches

solving

P=Poil+Patm

P=Patm+0.86αgh

$h=\frac{P-Patm}{0.86\alpha g } =\frac{179214.28-101325}{(0.86)(1000)(9.81)} \\h=9.23m$ [/tex]

Now we can use the equation that defines the volume of a cylinder.

V= $V=\frac{\pi }{4} D^{2} h$

D=3ft=0.9144m

h=9.23m

solving

$V=\frac{\pi }{4} (0.9144)^{2} 9.23=6.06m^3$

finally we use conversion factors to find the volume in gallons

$V=6.06m^3\frac{1000L}{1m^3} \frac{1gal}{3.785L} =1601gal$

3 0

4 years ago

What is the si unit of potential difference​

What is the si unit of potential difference