PHYSICS COLLEGE

You perform a double‑slit experiment in order to measure the wavelength of the new laser that you received for your birthday. You set your slit spacing at 1.09 mm and place your screen 8.61 m from the slits. Then, you illuminate the slits with your new toy and find on the screen that the tenth bright fringe is 4.53 cm away from the central bright fringe (counted as the zeroth bright fringe). What is your laser's wavelength ???? expressed in nanometers?

Answers

Answer 1

Answer:

Answer:

λ = 5.734 x 10⁻⁷ m = 573.4 nm

Explanation:

The formula of the Young's Double Slit experiment is given as follows:

$\Delta x = (\lambda L)/(d)\n\n\lambda = (\Delta x d)/(L)$

where,

λ = wavelength = ?

L = distance between screen and slits = 8.61 m

d = slit spacing = 1.09 mm = 0.00109 m

Δx = distance between consecutive bright fringes = $(4.53\ cm)/(10)$ = 0.00453 m

Therefore,

$\lambda = ((0.00453\ m)(0.00109\ m))/(8.61\ m)$

λ = 5.734 x 10⁻⁷ m = 573.4 nm

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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.

Convert 7 (gcm^2)/(min^2) into a value in standard S.I. units. Be sure to use scientific notation if necessary. You do not need to answer units.

Answers

The required value is required in SI units.

The required answer is $1.94*10^(-10)\ \text{kg m}^2/\text{s}^2$

SI units

The SI unit of mass, length and time is kg, m and s respectively.

In order to convert one unit into another it has to be multiplied or divided by the conversion factors.

A definite magnitude which has some quantity which is defined by convention or law is called a unit.

The conversion factors are

$1\ \text{g}=10^(-3)\ \text{kg}$

$1\ \text{cm}=10^(-2)\ \text{m}$

$1\ \text{cm}^2=10^(-4)\ \text{m}^2$

1 min = 60 s

$1\ \text{min}^2=60*60\ \text{s}^2$

So,

$7\ \text{g cm}^2/\text{min}^2=7* (10^(-3)* 10^(-4))/(60* 60)\n =1.94*10^(-10)\ \text{kg m}^2/\text{s}^2$

Learn more about SI units:

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Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?

Answers

The position vector of the bullet has components

$x=v_0t$

$y=1.4\,\mathrm m-\frac g2t^2$

The bullet hits the ground when $y=0$ , which corresponds to time $t$ :

$1.4\,\mathrm m-\frac g2t^2=0\implies t=0.53\,\mathrm s$

The bullet travels 168 m horizontally, which would require a muzzle velocity $v_0$ such that

$168\,\mathrm m=v_0(0.53\,\mathrm s)$

$\implies v_0\approx320\,(\mathrm m)/(\mathrm s)$

Final answer:

In the given physics problem, the bullet travels horizontally 168 meters before hitting the ground from a height of 1.4 meters. By calculating the time it takes for the bullet to fall to the ground due to gravity and then applying that time to the horizontal distance traveled, we find that the speed of the bullet when it exited the rifle was approximately 313.43 m/s.

Explanation:

The scenario defined is a classic Physics problem where an object is fired horizontally and falls to the ground due to gravity. We can calculate the horizontal speed of the bullet using the equations of motion associated with the vertical, free-fall motion of the bullet.

Gravity causes the bullet to fall to the ground. As we know that the height from the ground is 1.4 meters, we can calculate the time taken for the bullet to hit the ground using the equation: time = sqrt(2 * height / g), where g is the gravitational constant (approx. 9.8 m/s^2).

Substituting the given value, we get time = sqrt(2 * 1.4 / 9.8), which is around 0.536 seconds. The bullet travels 168 meters in this time horizontally, therefore its horizontal speed will be distance / time, which is 168 meters / 0.536 seconds = 313.43 m/s. So, Madelin's bullet had a speed of around 313.43 m/s when it exited the rifle.

Learn more about Projectile Motion here:

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An iceskater is turning at a PERIOD of (1/3) second with his arms outstretched. a) What is his ANGULAR VELOCITY w? b) If he pulls his arms towards his body to reduce his MOMENT OF INTERTIA by 1/2, what is his ANGULAR VELOCITY w? c) How much does his ROTATIONAL KINETIC ENERGY change? That is, if the initial Kinetic Energy is (KE)initial, what is the final KE? d) Where did that ENERGY come from, or go to?

Answers

Answer:

Explanation:

a )

Time period T = 1/3 s

angular velocity = 2π / T

= 2 x 3.14 x 3

ω = 18.84 radian / s

b )

Applying conservation of angular momentum

I₁ ω₁ = I₂ ω₂

I₁ / I₂ = ω₂ / ω₁

2 = ω₂ / ω

ω₂ = 2 ω

c )

(KE)initial = 1/2 I₁ ω²

(KE)final = 1/2 I₂ ω₂²

= 1/2 (I₁ / 2) (2ω)²

= I₁ ω²

c )

Change in rotational kinetic energy

= I₁ ω² - 1/2 I₁ ω²

= + 1/2 I₁ ω²

d )

This energy comes from the work done by centripetal force which is increased to increase the speed of rotation.

when a ball is dropped off a cliff in free fall, it has an acceleration of 9.8 m/s^2. what is its acceleration as it gets closer to the ground

Answers

acceleration due to gravity is contract for the purposes of this question, so the acceleration would remain at 9.8 m/s^2

If the ball, the cliff, and the ground are all on the Earth, and everything is bathed in an ocean of air, then the ball's acceleration will decrease as it falls, because of the friction of air resistance. If it has far enough to fall, it's possible that its acceleration may even become zero, and the ball settle on a constant speed (called "terminal velocity") before it hits the ground.

But until we get to College-level Physics and Engineering, we ALWAYS ignore that stuff, and assume NO AIR RESISTANCE. The ball is in FREE FALL, and the ONLY force acting on it is the force of gravity. We also assume that the distance of the fall is small enough so that the value of gravity is constant over the entire fall.

Under those assumptions, there's nothing present to change the acceleration of the falling ball. It's 9.81 m/s² when it rolls off the edge of the cliff, and it's 9.81 m/s² when it hits the ground.

A long copper cylindrical shell of inner radius 5 cm and outer radius 8 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm,
(b) 1.5 cm,
(c) 2.5 cm,
(d) 3.5 cm,
(e) 7 cm.

Answers

Answer:

a. 0

b. 8.4N/C

c. 5.04N/C

d. 3.6 N/C

e. 1.8N/C

Explanation:

The following data are given

inner cylindrical radius,r=5cm

outer cylindrical radius R=8cm

Charge density,p=7pc/m

radius of rod= 1cm

a. at distance 0.5cm from the center of the rod, this point falls on the rod itself and since the charge spread out on the surface of the rod, there wont be any electric field inside the rod itself

Hence E=0 at 0.5cm

b. at 1.5cm i.e 0.015m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.015)\nE=8.4N/C$

The direction of the field depends on the charge on the rod

c. at 2.5cm i.e 0.025m

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.025)\nE=5.04N/C$

The direction of the field depends on the charge on the rod

d. at 3.5cm i.e 0.035m this point is still within the rod and the inner cylinder

the electric field is expressed as

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.035)\nE=3.6N/C$

The direction of the field depends on the charge on the rod

e. at 7cm which is a point outside the rod and the cylinder, the electric field is

$E=(2*charge density)/(4\pi E_(0)r)\nE=(2*7*10^(-12) *9*10^(9))/(0.07)\nE=1.8N/C$

The direction of the field depends on the charge on the rod

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