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2 years ago

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Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes

are separated by 4.0 mm. What is the separation of the two slits

1 answer:

scoundrel [369]2 years ago

8 0

Answer:

The appropriate solution is "0.597 mm".

Explanation:

Given that:

Wavelength,

$\lambda = 519 \ nm$

or,

$=519\times 10^{-9}$

Distance,

$D=4.6 \ m$

Separated by,

$\beta = 4.0 \ mm$

or,

$=4.0\times 10^{-3}$

As we know,

⇒ $\beta=\frac{\lambda D}{d}$

or,

The separation of two slits will be:

⇒ $d=\frac{\lambda D}{\beta}$

By putting the values, we get

$=\frac{(519\times 10^{-9})(4.6)}{4.0\times 10^{-3}}$

$=5.97\times 10^{-4}$

or,

$=0.597 \ mm$

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If a participant were holding two different weights in their hands and the jnd for a 10-gram weight was 1 gram, what should the

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The jnd for a 100-gram weight, according to Weber's law will be 10 gram.

<h3>What is Weber's law?</h3>

It should be noted that Weber's law asserts that the nature of any given stimulus will always affect how change is perceived. In other words, the size, weight, importance, etc. of the prior situation and the significance of the change both influence whether a change will be observed.

In this case, it was given that the jnd for a 10-gram weight was 1 gram, therefore, the jnd for 100 gram will be;

= 100 / 10

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1 year ago

A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate

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Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

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The plate and the rod must have $Q_{total}/2\\$ . So the charge transferred from the plate to the rod is:

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2 years ago

Two forces one of 12N and another of 5N act on a body in such away that they makes an angle of 90 digre with each other,what is

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3 years ago

Can anyone solve these for my by using unit vectors? Can you also please show your work

Oxana [17]

4. The Coyote has an initial position vector of $\vec r_0=(15.5\,\mathrm m)\,\vec\jmath$ .

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where $\vec a$ is the Coyote's acceleration vector at time $t$ . He experiences acceleration only in the downward direction because of gravity, and in particular $\vec a=-g\,\vec\jmath$ where $g=9.80\,\frac{\mathrm m}{\mathrm s^2}$ . Splitting up the position vector into components, we have $\vec r=r_x\,\vec\imath+r_y\,\vec\jmath$ with

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5. The shell has initial position vector $\vec r_0=(1.52\,\mathrm m)\,\vec\jmath$ , and we're told that after some time the bullet (now separated from the shell) has a position of $\vec r=(3500\,\mathrm m)\,\vec\imath$ .

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We find the shell hits the ground at

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5b. The horizontal component of the bullet's position vector is

$r_x=v_0t$

where $v_0$ is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for $v_0$ :

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3 years ago

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Answer:

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If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

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In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

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Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

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v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ = $\frac{-17.2}{16.6} =-1.036$

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0

3 years ago