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Katyanochek1 [597]
3 years ago
10

The 2nd maximum of a double slit diffraction pattern makes an angle 20 degrees when a wavelength of 600 nm is used. What is the

angle for the 5th maximum when the slit separation d is increased to 1.5d keeping the wavelength the same. g
Physics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

θ = 34.77°

Explanation:

From diffraction equation:

m\lambda = dSin\theta

where,

m = order of diffraction

λ = wavelength of light used

d = slit separation

θ = angle

Therefore, for initial case:

m = 2

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = ?

θ = angle 20°

Therefore,

(2)(6\ x\ 10^{-7}\ m)=d(Sin\ 20^o)\\\\d = \frac{12\ x 10^{-7}\ m}{0.342}\\\\d = 3.5\ x\ 10^{-6}\ m

Now, for the second case:

m = 5

λ = 600 nm = 6 x 10⁻⁷ m

d = slit seperation = (1.5)(3.5 x 10⁻⁶ m) = 5.26 x 10⁻⁶ m

θ = angle = ?

Therefore,

(5)(6\ x\ 10^{-6}\ m) = (5.26\ x\ 10^{-6}\ m)Sin\theta\\\\Sin\theta = \frac{(5)(6\ x\ 10^{-7}\ m)}{(5.26\ x\ 10^{-6}\ m)}\\\\\theta = Sin^{-1}(0.5703)

<u>θ = 34.77°</u>

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A car is traveling with a velocity of 24.4 m/s. It accelerates at a constant rate of 3.2m/s2. If the acceleration lasts for 5.6
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3 years ago
An applied pulling force of 100 N is used to accelerate an object to the right along a rough surface that offers 40 N of frictio
blondinia [14]

Answer:

9.8 m/s^2

Explanation:

First of all, we analyze the vertical direction.

Along this direction, the object is in equilibrium, so its acceleration is zero, and therefore  the net force on it is zero. There are only two forces acting on the object vertically: the weight, W (downward), and the normal force, N (upward), and since the net force must be zero, we can write

W=N

And since N = 60 N, the weight of the object is

W = 60 N

From this, we can also find the mass of the object, using the equation:

W=mg \rightarrow m = \frac{W}{g}=\frac{60}{9.8}=6.1 kg

where g=9.8 m/s^2 is the acceleration of gravity.

Now we analyze the forces along the horizontal direction. We have:

A pulling force of F=100 N forward

A frictional resistance of F_f = 40 N backward

So the equation of motion in this direction is

F-F_f = ma

And solving for a, we find the acceleration of the object:

a=\frac{F-F_f}{m}=\frac{100-40}{6.1}=9.8 m/s^2

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4 0
2 years ago
An electron that has a velocity with x component 2.6 × 106 m/s and y component 4.0 × 106 m/s moves through a uniform magnetic fi
Natasha2012 [34]

Explanation:

It is given that,

The horizontal and vertical component of velocity of an electron is:

v=(2.6i+4j)\times 10^6\ m/s

The magnetic field acting there is given by :

B=(0.037i-0.17j)\ T

(a) The magnitude of the magnetic force on the electron is given by :

F=q(v\times B)

q = e

F=e(v\times B)

F=1.6\times 10^{-19}\times ((2.6i+4j)\times 10^6\times (0.037i-0.17j))

F=-1.6\times 10^{-19}\times (-0.442\cdot10^{6}-0.1702\cdot10^{6})\ kN

F=9.79\times 10^{-14}\ kN

(b) We know that the charge on proton is :

q=+1.6\times 10^{-19}\ C

The magnetic force as same as for electron but the direction is opposite i.e.

F=-9.79\times 10^{-14}\ kN

Hence, this is the required solution.

4 0
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