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Light from a sodium lamp (λ = 589 nm) illuminates two narrow slits. The fringe spacing on a screen 150 cm behind the slits is 4.0 mm. What is the spacing (in mm) between the two slits?

Answers

Answer 1

Answer:

The spacing between the two slits is 0.221mm.

The spacing between the two slits is given as,

$D=(\lambda L)/(y)$

Where $\lambda$ is wavelength, y is fringe spacing and L is length of screen.

Given that, $\lambda=589nm,L=150cm,y=4mm$

Substitute in above equation.

$D=(589*10^(-9)*150*10^(-2) )/(4*10^(-3) )\n \nD=2.21*10^(-4) m\n\nD=0.221mm$

Hence, the spacing between the two slits is 0.221mm.

Learn more about the sodium lamp here:

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In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Answers

Answer:

Explanation:

Let length of the pendulum be l . The expression for time period of pendulum is as follows

T = 2π $\sqrt{(l)/(g) }$

For Mars planet ,

1.5 = $2\pi\sqrt{(l)/(.38*9.8) }$

For other planet

.92 = $2\pi\sqrt{(l)/(g_1) }$

Squiring and dividing the two equations

$(1.5^2)/(.92^2) = (g_1)/(3.8*9.8)$

$g_1 = 9.9$

The second planet appears to be earth.

Why might a scientist want to use a model to study the solar system? O A. Its extreme simplicity makes it difficult to see patterns in observations. B. Its extremely slow movement makes it difficult to see the motions of different planets. C. Its extremely large size makes it difficult to see all of its parts at the same time. D. Its extremely small size makes it difficult to see planets that are far away

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A scientist wants to use a model to study the solar model because its extremely large size makes it difficult to see all of its parts at the same time. Hence, option C is correct.

What is a Solar System?

The Sun and all the smaller movable objects that orbit it make up the Solar System. The eight main planets are the largest objects in the Solar System, excluding the Sun. Mercury, Venus, Earth, and Mars are the four relatively tiny, rocky planets closest to the Sun.

The asteroid belt, which is home to millions of stony objects, lies beyond Mars. These are remains from the planets' creation 4.5 billion years ago.

Jupiter, Saturn, Uranus, and Neptune are the four gas giants that can be found on the opposite side of the asteroid belt. Despite being much larger than Earth, these planets are rather light. Their main components are hydrogen and helium.

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Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

b) The radiation with a frequency of $7.9x10^(14)Hz$ belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:

Gamma ray

X ray

Ultraviolet rays

Visible region

Lower frequencies:

Infrared

Microwave

Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of $3x10^{8]m/s$ in vacuum.

The velocity of a wave can be determined by means of the following equation:

$c = \nu \cdot \lambda$ (1)

Where c is the speed of light, $\nu$ is the frequency and $\lambda$ is the wavelength.

Then, from equation 1 the frequency can be isolated.

$\nu = (c)/(\lambda)$ (2)

Before using equation 2 to determine the range of UV-B it is necessary to express $\lambda$ in units of meters in order to match with the units from c.

$\lambda = 320nm . (1m)/(1x10^(9)nm)$ ⇒ $3.2x10^(-7)m$

$\lambda = 280nm . (1m)/(1x10^(9)nm)$ ⇒ $2.8x10^(-7)m$

$\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)$

$\nu = 9.375x10^(14)s^(-1)$

$\nu = 9.375x10^(14)Hz$

$\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)$

$\nu = 1.071x10^(15)Hz$

Hence, the UV-B has frequencies between $9.375x10^(14)Hz$ and $1.071x10^(15)Hz$

(b) In which of these three categories does radiation with a frequency of $7.9x10^(14)Hz$ belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

$\lambda = 400nm . (1m)/(1x10^(9)nm)$ ⇒ $4x10^(-7)m$

$\nu = (3x10^(8)m/s)/(4x10^(-7)m)$

$\nu = 7.5x10^(14)s^(-1)$

$\nu = 7.5x10^(14)Hz$

Hence, the UV-A has frequencies between $7.5x10^(14)Hz$ and $9.375x10^(14)Hz$ .

Therefore, the radiation with a frequency of $7.9x10^(14)Hz$ belongs to UV-A category.

Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x

Answers

Answer:

The exponent A in the equation is 3.

Explanation:

v = a^2 t^ A /x

$v = (a^2t^A)/(x) \n\nvx = a^2t^A\n\n((L)/(T))(L) = ((L)/(T^2))^2(T)^A\n\n (L^2)/(T)= ((L^2)/(T^4))(T)^A\n\n (L^2)/(T) *(T^4)/(L^2) = (T)^A\n\nT^3 = T^A\n\n(T^3)/(T^3) = (T^A)/(T^3)\n\nT^(3-3) = T^(A-3)\n\n3-3 = A-3\n\n0 = A-3\n\nA = 3$

Therefore, the exponent A in the equation is 3.

Proposed Exercise - Circular MovementConsider four pulleys connected by correals as illustrated in the figure below. One motor moves the A pulley with angular acceleration A= 20 rad/s^2/. If pulley A is initially moving with angular acceleration A= 40 rad/s^2, determine the angular speed of pulleys B and C after three seconds. Consider that the belts do not slide

Answers

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

The landing gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described determine the values of k and c that limit the amplitude of vibration of the airplane (x) to 0.1 m. assume

Answers

The land of airplane gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described

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