PHYSICS COLLEGE

Two narrow slits separated by 0.30 mm are illuminated with light of wavelength 496 nm. (a) How far are the first three bright fringes from the center of the pattern if observed on the screen 130 cm distant? (b) How far are the first three dark fringes from the center of the pattern?

Answers

Answer 1

Answer:

Answer:

Explanation:

d = separation of the slits = 0.30 mm = 0.30 x 10⁻³ m

λ = wavelength of the light = 496 nm = 496 x 10⁻⁹ m

n = order of the bright fringe

D = screen distance = 130 cm = 1.30 m

$x_(n)$ = Position of nth bright fringe

Position of nth bright fringe is given as

$x_(n) =( n D \lambda )/(d)$

For n = 1

$x_(1) =( (1) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(1) = 2.15* 10^(-3)m$

For n = 2

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 4.30* 10^(-3)m$

For n = 3

$x_(2) =( (2) (1.30)(496* 10^(-9)))/(0.30* 10^(-3))$

$x_(2) = 6.45* 10^(-3)m$

Position of nth dark fringe is given as

$y_(n) =( (2n+1) D \lambda )/(2d)$

For n = 1

$y_(1) =( (2(1)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(1) = 3.22* 10^(-3)m$

For n = 2

$y_(2) =( (2(2)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$y_(2) = 5.4* 10^(-3)m$

For n = 3

$y_(3) =( (2(3)+1) (1.30)(496* 10^(-9)))/(2(0.30* 10^(-3)))$

$x_(3) = 7.5* 10^(-3)m$

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Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling without slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?

Answers

Final answer:

Given the same initial linear speed, a solid ball, solid disk, and hoop will expend energy on both rotation and translation. The solid ball, having the lowest moment of inertia, uses the most energy for translation and, therefore, will travel the highest up an incline.

Explanation:

In the context of this problem related to physics, the deciding factor is the distribution of mass, which influences each object's moment of inertia. Objects set to roll tend to use energy in two ways: translation (moving along the incline) and rotation (spinning about their center). Moment of inertia essentially measures how much of the object's energy goes towards rotation.

For a solid ball, solid disk, and hoop with the same mass and radius, the hoop has the highest moment of inertia with all of its mass at the maximum distance from the center. Followed by the solid disk, with its mass spread evenly from the center to its edge. Lastly, the solid ball has the lowest moment of inertia as its mass is concentrated towards the center.

This means that, given the same initial linear speed, the hoop will expend most of its energy on rotation rather than moving up the incline (translation). The solid disk will have a more balanced distribution between translation and rotation, and finally, the solid ball will use the least amount of energy on rotation and the most on translation. As such, the solid ball will go the farthest up the incline.

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What happens to a black body radiator as it increases in temperature? A. it gives off a range of electromagnetic radiation of shorter wavelengths.
B. It gives off only one wavelength of electromagnetic radiation
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D. It becomes hotter but gives off less electromagnetic radiation

Answers

The black body radiator as it increases in temperature gives off a range of electromagnetic radiation of shorter wavelengths so, the option A is correct.

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The effects of such an energy transfer to living matter, including the typical effects on numerous biological processes, are given a great deal of attention (e.g., photosynthesis in plants and vision in animals).

Thus, the black body radiator gives off a range of electromagnetic radiation of shorter wavelengths.

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Answer: A

Explanation:

Answer is a hope this helps guys!

A 0.010 kg ball is shot from theplunger of a pinball machine.Because of a centripetal force of0.025 N, the ball follows a
circulararc whose radius is 0.29 m. What isthe speed of the
ball?

Answers

Answer:

v = 0.85 m/s

Explanation:

Given that,

Mass of the ball, m = 0.01 kg

Centripetal force on the ball, F = 0.025 N

Radius of the circular path, r = 0.29 m

Let v is the speed of the ball. The centripetal force of the ball is given by :

$F=(mv^2)/(r)$

$v=\sqrt{(Fr)/(m)}$

$v=\sqrt{(0.025* 0.29)/(0.01)}$

v = 0.85 m/s

So, the speed of the ball is 0.85 m/s. Hence, this is the required solution.

The starter motor of a car engine draws a current of 170 A from the battery. The copper wire to the motor is 4.60 mm in diameter and 1.2 m long. The starter motor runs for 0.930 s until the car engine starts How much charge passes through the starter motor?

Answers

Answer:

The charge that passes through the starter motor is $\Delta Q=158.1 C$ .

Explanation:

Known Data

Avogadro's Number $N_(A)=6.02x10^(23)$
Current, $I=170A=170(C)/(s)$
Charge in an electron, $q=1.60x10^(-19)C$
Time, $\Delta t=0.930s$
Diameter, $d=4.60mm=0.0046m$
Transversal Area, $A=((d)/(2))^(2) \pi=((0.0046m)/(2))^(2) \pi=1.66x10^(-5) m^(2)$
Volume, $V=Length*A=(1.2m)(1.66x10^(-5) m^(2))=1.99x10^(-5) m^(3)$

First Step: Find the number of the electrons per unit of volume in the wire

We use the formula $n=(N_(A))/(V)= (6.02x10^(23) electrons)/(1.99x10^(-5) m^(3)) =3.02x10^(28)el/ m^(3)$ .

Second Step: Find the drag velocity

We can use the following formula $v_(d)=(I)/(nqA)=(170C/s)/((3.02x10^(28)m^(-3))(1.60x10^(-19)C)(1.66x10^(-5) m^(2))) =2.11x10^(-3) m/s$

Finally, we use the formula $\Delta Q=(nAv_(d)\Delta t)q=(3.02x10^(28) m^(-3))(1.66x10^(-5) m^(2))(2.11x10^(-3) m/s)(0.930s)(1.60x10^(-19)C)=158.1 C$ .

Which sling can the crane use to lift the 1000kg pipe?A.
800kg rated sling
B. 1000kg rated sling
C. 2000kg rated sling
D. Band C

Answers

Answer:

C. 2000kg rated sling

Explanation:

ensures better safety and can carry twice more mass than current mass.

A can of soup has a mass of 0.35 kg. The can is moved from a shelf that is 1.2 m off the ground to a shelf that is 0.40m off the ground. How does the gravitational potential energy of the can change?

Answers

Answer:

2.744 difference

Explanation:

Use Pe=mgh

So when the soup is at a height of 1.2m, its Pe is (.35kg)(9.8m/ $s^(2)$ )(1.2m)=4.116

when the soup is at a height of .40m, its Pe is (.35kg)(9.8m/ $s^(2)$ )(.40m)=1.372

So youre looking at a 2.744 difference in pe

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