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3 years ago

14

Aspect ratio is the ratio of images width to its height. a photograph is 17.5 cm wide by 14 cm high what is the aspect ratio in

its lowest term

1 answer:

Svet_ta [14]3 years ago

5 0

Try 5:4 and see if it is correct

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Problem #1- In the first scene in which Miss Clark appears, she is supposed to be wearing a BLACK dress.

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Answer:

Blue Lighting

Explanation:

In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.

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A transmission diffraction grating with 420 lines/mm is used to study the light intensity of di event orders (n). A screen is lo

Goshia [24]

Answer:

Explanation:

Diffraction grating is used to form interference pattern of dark and bright band.

Distance between adjacent slits (a ) = 1 / 420 mm

= 2.38 x 10⁻³ mm

2.38 x 10⁻⁶ m

wave length of red light

= 680 x 10⁻⁹ m

For bright red band

position x on the screen

= n λD / a , n = 0,1,2,3 etc

D = distance of screen

putting n = 1 , 2 and 3 , we can get three locations of bright red band.

x₁ = λD / a

= 680 x 10⁻⁹ x 2.8 / 2.38 x 10⁻⁶

= .8 m

= 80 cm

Position of second bright band

= 2 λD / a

= 2 x 80

= 160 cm

Position of third bright band

= 3 λD / a

= 3 x 80

= 240 cm

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3 years ago

Which of the given will facilitate a normal Diels–Alder reaction?

babymother [125]

Answer:

D

Explanation:

- The rate of the Diels-Alder is orders of magnitude faster if there is an electron-withdrawing group on the dienophile. For example, replacing a hydrogen on ethene with the electron-withdrawing group CN results in about a 10^5 increase in the reaction rate.

- Other common electron withdrawing functional groups that will accelerate the Diels Alder reaction of dienophiles include aldehydes, ketones, and esters.

- In short, any functional group conjugated with the pi bond which can act as a pi acceptor will accelerate a Diels-Alder reaction with a typical diene.

- See attachment for graphical explanation.

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A particle with a charge of − 5.10 nC is moving in a uniform magnetic field of B⃗ =−( 1.20 T )k^. The magnetic force on the part

marta [7]

Answer:

Explanation:

Given that,

Charge q=-5.10nC

Magnetic field B= -1.2T k

And the magnetic force

F =−( 3.30×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F= q(v×B)

−( 3.30×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.2k

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2x i×k - 1.2y j×k - 1.2z k×k)

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( 1.2xj - 1.2y i )

−( 3.30×10−7N )i+( 7.60×10−7N )j=

q( -1.2y i + 1.2x j)

So comparing comparing coefficients

let compare x axis component

-( 3.30×10−7N )i=-1.2qy i

−3.30×10−7N = -1.2qy

y= -3.3×10^-7/-1.2q

y= -3.3×10^-7/-1.2×-5.10×10^-9)

y=-53.92m/s

Let compare y-axisaxis

7.6×10−7N j = 1.2qx j

7.6×10−7N = 1.2qx

x= 7.6×10^-7/-1.2q

x= 7.6×10^-7/1.2×-5.10×10^-9)

x=-124.18m/s

a. Then, the velocity of the x component is x= -124.18m/s

b. Also, the velocity component of the y axis is =-53.92m/s

c. We will compute

V•F

V=-124.18i -53.92j

F=−( 3.30×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-124.18i-53.92j)•−(3.30×10−7N)i+(7.60×10−7 N )j =

4.1×10^-5 - 4.1×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cos

Cosx=0

x= arccos(0)

x=90°

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