Guys I need help please!!!! Which pair of graphs represent the same motion?

A laser beam of wavelength λ=632.8 nm shines at normal incidence on the reflective side of a compact disc. The tracks of tiny pi

anzhelika [568]

Answer:

The intensity of light be maximum is for angles 23.3° and 52.3°.

Explanation:

Given that,

Wave length = 632.8 nm

Distance = 1.60 μm

We need to calculate the intensity of light be maximum

Using Bragg's law

$d\sin\theta=n\lambda$

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

We need to calculate the angle for different value of n

Using Bragg's law

$\theta=\sin^{-1}(\dfrac{n\lambda}{d})$

For n₁,

Put the value into the formula

$\theta=\sin^{-1}(\dfrac{632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=23.3^{\circ}$

For n₂,

$\theta=\sin^{-1}(\dfrac{2\times632.8\times10^{-9}}{1.60\times10^{-6}})$

$\theta=52.3^{\circ}$

Hence, The intensity of light be maximum is for angles 23.3° and 52.3°.

6 0

3 years ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used to solve this question.

From the diagram, the horizontal component of the force is:

$F_X = F_{cos \ \theta}$

Replacing 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

On the other hand, the vertical component is ;

$F_Y = Fsin \ \theta$

Replacing 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

$f = \mu \ N$

Where;

$\mu$ is said to be the coefficient of the friction

N = the normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing $F_Y \ for \ F_{sin \ \theta}$ . The normal reaction can now be:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Making $\mu$ the subject of the formular in the above equation; we have the following:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Replacing the following values: i.e

$F_X \ = \ 64.65 \ N$

m = 73 Kh

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Then:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Thus, the coefficient of friction is = 0.0984

5 0

3 years ago

1)Describe the difference between an

solniwko [45]

Answer:

1) Energy transfer is the movement of energy from one location to another. ... Energy transformation is when energy changes from one form to another – like in a hydroelectric dam that transforms the kinetic energy of water into electrical energy.

2)A swinging pirate ship ride at a theme park. Kinetic energy is transferred into gravitational potential energy.

A boat being accelerated by the force of the engine. ...

Bringing water to the boil in an electric kettle.

3)Common examples of energy storage are the rechargeable battery, which stores chemical energy readily convertible to electricity to operate a mobile phone, the hydroelectric dam, which stores energy in a reservoir as gravitational potential energy, and ice storage tanks, which store ice frozen by cheaper energy at night

Explanation:

PLEASE MARK ME AS BRAINLIEST

7 0

3 years ago

A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t

vivado [14]

Answer:

The force that would be applied on the rope just to start moving the wagon is 122 N

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

μ is the coefficient of friction ( μs, in this case, static friction);

m is mass of the object and;

g is the acceleration due to gravity( a constant equal to 9.81 m/ $s^{2}$ )

from the equation we are provide with;

μs = 0.25

m = 50 kg

g = 9.81 m/ $s^{2}$

F =?

Using equation 1

F = 0.25 x 50 kg x 9.81 m/ $s^{2}$

F = 122.63 N

Therefore a force of 122 N must be applied to the rope just to start the wagon.

6 0

3 years ago

Read 2 more answers

A system consists of N particles that can occupy two energy levels: a nondegenerate ground state and a three-fold degenerate exc

cupoosta [38]

Answer:

Ng = 0.893 N, Ne = 0.107N

Explanation:

Number of particles in Ground state = Ng

Number of particles in Excited state = Ne

Ne/Ng = e^{(-ΔE)/kt}

Since excited state is 3 fold degenerate

Ne/Ng =3 x e^{(-ΔE)/kt}

ΔE = Energy difference between ground and excited states = 0.25eV

T = 960 K

Constant k = 8.617 x 10^-5 eV/K

Ne/Ng = 3 x e^{-0.25/(8.617x10^-5) x 960}

= 3 x e^(-3.188645)

= 3 x 0.0412 = 0.1237 ≅ 0.12

Ne = 0.12 Ng

but Ne + Ng = N, where is N is total number of particles, substituting Ne into equation we get,

Ng(1 + 0.12) = N

Ng = N/1.12 = 0.893N

and Ne = 0.12 x 0.893 N = 0.107 N

3 0

3 years ago