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3 years ago

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Monochromatic light falls on two very narrow slits 0.048 mm apart. Successive fringes on a screen 6.50 m away are 8.5 cm apart n

ear the center of the pattern. Determine the wavelength and frequency of the light.

1 answer:

vfiekz [6]3 years ago

3 0

Answer:

The wavelength is $\lambda = 6.28 *10^{-7}=628 nm$

The frequency is $f = 4.78 Hz$

Explanation:

From the question we are told that

The slit distance is $d = 0.048 \ mm = 4.8 *0^{-5}\ m$

The distance from the screen is $D = 6.50 \ m$

The distance between fringes is $Y = 8.5 \ cm = 0.085 \ m$

Generally the distance between the fringes for a two slit interference is mathematically represented as

$Y = \frac{\lambda * D}{d}$

=> $\lambda = \frac{Y * d }{D}$

substituting values

$\lambda = \frac{0.085 * 4.8*10^{-5} }{6.50 }$

$\lambda = 6.27 *10^{-7}=628 nm$

Generally the frequency of the light is mathematically represented as

$f = \frac{c}{\lambda }$

where c is the speed of light with values

$c = 3.0 *10^{8} \ m/s$

substituting values

$f = \frac{3.0*10^8}{6.28 *10^{-7}}$

$f = 4.78 Hz$

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Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.

Sergeu [11.5K]

The work done to pull the sled up to the hill is given by
$W=Fd$
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is $W=1900 J$ and the distance through which the force is applied is $d=400 m$ , so we can calculate the average force by re-arranging the previous equation and by using these data:
$F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N$

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3 years ago

irga5000 [103]

Answer:

Explanation:

solution is found below

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3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

Consider the points below. P(1, 0, 1), Q(−2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the poi

kozerog [31]

Answer:

a) (0, -33, 12)

b) area of the triangle : 17.55 units of area

Explanation:

<h2>a) </h2>

We know that the cross product of linearly independent vectors $\vec{A}$ and $\vec{B}$ gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.

Luckily for us, we know that vectors $\vec{A} = \vec{P}-\vec{Q}$ and $\vec{B} = \vec{R} - \vec{Q}$ are living in the plane through the points P, Q, and R, and are linearly independent.

We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).

If they weren't linearly independent, we will obtain vector zero as the result of the cross product.

So, for our problem:

$\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)$

$\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)$

$\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}$

$\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}$

$\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}$

$\vec{A} \times \vec{B} =(0, -33, 12)$

<h2>B)</h2>

We know that $\vec{A}$ and $\vec{B}$ are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:

$|\vec{A} \times \vec{B} | = 2 * area_{triangle}$

so:

$\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}$

$\sqrt{1233} = 2 * area_{triangle}$

$35.114= 2 * area_{triangle}$

$17.55 \ units \ of \ area = area_{triangle}$

5 0

3 years ago

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

3 years ago