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3 years ago

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Monochromatic (single-wavelength) light falling on two slits 0.016 mm apart produces the fifth-order bright spot at an 8.8◦angle

. What is the wavelength of the light used?

2 answers:

Elena L [17]3 years ago

8 0

Given Information:

distance = d = 0.016 mm = 0.0016 cm = 0.000016 m

Angle = θ = 8.8°

bright spot = m = 5

Required Information:

Wavelength = λ = ?

Answer:

Wavelength = 4.89x10⁻⁷ m

Explanation:

Monochromatic light passes through a double slit. The corresponding diffraction is given by the equation:

dsinθ = mλ

where d is the distance between two slits, m is the order of the diffraction, θ is the angle and λ is the wavelength of the light wave.

λ = dsinθ/m

λ = 0.016*sin(8.8°)/5

λ = 4.89x10⁻⁷ m

Therefore, the wavelength of the monochromatic light falling on two slits is 4.89x10⁻⁷ m

dimaraw [331]3 years ago

3 0

Answer:

the wavelength of the light used is 4.8 × 10⁻⁷m

Explanation:

We have given distance between the two slits

$d=1.6\times 10^{-2}mm=1.6\times 10^{-5}m$

Angle $\Theta =8.8^{\circ}$

We know that for nth order fringe

$dsin\Theta =n\lambda$

In question it is given that 5th order so n =5

$So 1.6\times 10^{-5}sin\ 8.8 =5\lambda\\\lambda = 4.8\times 10^{-7}m$

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a race car accelerates uniformly from 18.5m/s to 46.1m/s in 2.4 seconds. determine the acceleration of the car and the distance

8_murik_8 [283]

The car's (average) acceleration would be

$a=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.4\,\mathrm s}=11.5\,\dfrac{\mathrm m}{\mathrm s^2}$

The car's position over time would be given by

$x=v_0t+\dfrac12at^2$

so that after 2.4 seconds, the car will have traveled a distance of

$x=\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)(2.4\,\mathrm s)+\dfrac12\left(11.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm s)^2$

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3 years ago

Read 2 more answers

A convex mirror of focal length 34 cm forms an image of a soda bottle at a distance of 18 cm behind the mirror. The height of th

Mademuasel [1]

<h2>Answers:</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

<h2 /><h2>a) Position of the soda bottle (the object) </h2><h2 />

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f$ is the focal distance

$u$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $v$ , let's find $u$ from (1):

$u=\frac{v.f}{v-f}$ (2)

Taking into account the explanation at the beginign of this asnswer:

$f=-34cm$ and $v=-18cm$

The <u>negative signs</u> indicate the <u>focal distance</u> and the <u>distance between the image and the mirror are virtual </u>

Then:

$u=\frac{(-18cm)(-34cm)}{-18cm-(-34cm)}$

$u=38.25cm$ (3) >>>>Position of the soda bottle

<h2>b) magnification of the image </h2><h2 />

The magnification $m$ of the image is given by:

$m=-\frac{v}{u}$ (4)

$m=-\frac{(-18cm)}{38.25cm}$

$m=0.47$ (5)>>>the image is 0.47 smaller than the object

<h2>The fact that this value is positive means the image is upright </h2>

<h2>c) Describe the image </h2><h2 />

According to the explanations and results obtained in the prior answers, the correct option is 9:

<h2>The image is <u>virtual, upright, smaller</u> than the object </h2><h2 /><h2>d) Height of the soda bottle (object) </h2>

Another way to find the magnification is by the following formula:

$m=\frac{h_{i}}{h_{o}}$ (6)

Where:

$h_{i}$ is the image height

$h_{o}$ is the object height

We already know the values of $m$ and $h_{i}$ , let's find $h_{o}$ :

$h_{o}=\frac{h_{i}}{m}$ (7)

$h_{o}=\frac{6.8cm}{0.47}$

$h_{o}=14.46cm$ (8) >>>height of thesoda bottle

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3 years ago

A rope is pulling a sled is sloped at 25 degrees upward at the point of contact. The tension in the rope is 40.0N. What is the h

pshichka [43]

T = 40.0 N

angle = 25°

Trigonometric ratios:

sin(25°) = vertical component of the force / force

cos(25°) = horizontal component of the force / force

tan (25°) = vertical compoent of the force / horizontal component of the force.

From cos(25°) you can find the horizontal component of the force:

horizontal component of the force = force * cos(25°)

The force is the tension, 40.0 N.

horizontal component of the force = 40.0 N * cos (25°) = 36.25 N

Answer: 36.25 N

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4 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

$\Sigma F_{y} = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_{k}\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

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Work package. Hope this helps!

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