PHYSICS COLLEGE

A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm wide-that is, the two first-order diffraction minima are separated by 1.40cm What is the distance between the two second-order minima?

Answers

Answer 1

Answer:

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=(y_1)/(2)\n\Rightarrow \beta_1=(1.4)/(2)\n\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=(m_1\lambda D)/(d)\n\Rightarrow d=(m_1\lambda D)/(\beta_1)$

For second order

$\beta_2=(m_2\lambda D)/(d)\n\Rightarrow \beta_2=(m_2\lambda D)/((m_1\lambda D)/(\beta_1))\n\Rightarrow \beta_2=(m_2)/(m_1)\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\n\Rightarrow y_2=2(m_2)/(m_1)\beta_1\n\Rightarrow y_2=2(2)/(1)* 0.7\n\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

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. Using your knowledge of circular (centripetal) motion, derive an equation for the radius r of the circular path that electrons follow in terms of the magnetic field B, the electrons' velocity v, charge e, and mass m. You may assume that the electrons move at right angles to the magnetic field.2. Recall from electrostatics, that an electron obtains kinetic energy when accelerated across a potential difference V. Since we can directly measure the accelerating voltage V in this expierment, but not the electrons' velocity v, replace velocity in your previous equation with an expression containing voltage. The electron starts at rest. Now solve this equation for e/m.

You should obtain e/m = 2V/(B^2)(r^2)

3. The magnetic field on the axis of a circular current loop a distance z away is given by

B = mu I R^2 / 2(R^2 + z^2)^ (3/2)

where R is the radius of the loops and I is the current. Using this result , calculate the magnetic field at the midpoint along the axis between the centers of the two current loops that make up the Helmholtz coils, in terms of their number of turns N, current I, and raidus R.Helmholtz coils are separated by a distance equal to their raidus R. You should obtain:

|B| = (4/5)^(3/2) *mu *NI/R = 9.0 x 10^-7 NI/R

where B is magnetic field in tesla, I is in current in amps, N is number of turns in each coil, and R is the radius of the coils in meters

Answers

Answer:

Explanation:

Magnetic field creates a force perpendicular to a moving charge in its field which is equal to Bev where B is magnetic field , e is amount of charge on the moving charge and v is the velocity of charge particle .

This force provides centripetal force for creation of circular motion. If r be the radius of the circular path

Bev = mv² / r

r = mv / Be

2 ) If an electron is accelerated by an electric field created by potential difference V then electric field

= V / d where d is distance between two points having potential difference v .

force on charged particle

electric field x charge

= V /d x e

work done by field

= force x distance

= V /d x e x d

V e

This is equal to kinetic energy created

V e = 1/2 mv²

= 1/2 m (r²B²e² / m² )

V = r²B²e/ 2 m

e / m = 2 V/ r²B²

3 )

B = $(\mu* I* R^2)/(2(R^2+Z^2)^(3)/(2) )$

In Helmholtz coils , distance between coil is equal to R so Z = R/2

B = $(\mu* I* R^2)/(2(R^2+(R^2)/(4) )^(3)/(2) )$

For N turns of coil and total field due to two coils

B = $(\mu* I* N)/(R*((5)/(4))^(3)/(2) )$

= $(\mu* I* N)/(R)* ((4)/(5))^(3)/(2)$

= 9.0 x 10^-7 NI/R

Which of the following best represents stored potential energy?Air leaking from a flat tire
Stress built up in a rock fault
Heat given off by a forest fire
Water flowing through a hose

Answers

Answer:

Explanation:

stress built up on a rock fault

Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4 = 407 N , and MA = 1504 N⋅m . Express the Cartesian components of the resultant force and the couple moment in newtons and newton-meters to three significant figures separated by commas.

Answers

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

Final answer:

The Cartesian components of the resultant force and the couple moment are calculated by summing up all the forces and moments acting on the object. The resultant force is 1724 N and the couple moment is 29.764 N*m.

Explanation:

The resultant force and couple moment in the Cartesian coordinate system can be obtained by summing up all the forces and moments acting on the object. In this case, we have the forces F1, F2, F3, F4 and the couple moment MA acting on the object. The resultant force (FR) can be calculated as the sum of all the forces, i.e., FR = F1 + F2 + F3 + F4. Using the values given, FR = 510 N + 306 N + 501 N + 407 N = 1724 N. The resultant moment (MR) can be calculated as the sum of all the moments, i.e., MR = d1*F1 + d2*F2 + d3*F3 + d4*F4 - MA. Using the values given, MR = 0.880 m * 510 N + 1.11 m * 306 N + 0.560 m * 501 N + 2.08 m * 407 N - 1504 N*m = 29.764 N*m. Therefore, the Cartesian components of the resultant force and the couple moment are 1724 N and 29.764 N*m respectively.

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A moon is in orbit around a planet. The moon's orbit has a semimajor axis of 4.3 times 10 Superscript 8 Baseline m and has an orbital period of 1.516 days. Use these data to estimate the mass of the planet.

Answers

Answer:

The mass of the planet is $2.7*10^(27)\ kg$ .

Explanation:

Given that,

Semi major axis $a= 4.3*10^(8)$

Orbital period T=1.516 days

Using Kepler's third law

$T^2=(4\pi^2)/(GM)a^3$

$M=(4\pi^2)/(GT^2)a^3$

Where, T = days

G = gravitational constant

a = semi major axis

Put the value into the formula

$M=(4*(3.14)^2)/(6.67*10^(-11)(1.516*24*60*60)^2)(4.3*10^(8))^3$

$M=2.7*10^(27)\ kg$

Hence, The mass of the planet is $2.7*10^(27)\ kg$ .

Which wave diagram BEST represents a dim red sunset on the right side to the light from an intense ultraviolet bug light on the left side?

Answers

Diagram-A satisfies. High amplitude (bright) and long wavelengths are present on the left (red). & The right side has a short wavelength and low amplitude (dim) (violet).

What are light waves?

Light comes from a source as waves. Each wave has an electric and a magnetic component. Light is hence sometimes referred to as electromagneticradiation.

A large portion of the light in the universetravels with wavelengths that are too short or too long for the human eye to detect, yet our brains interpret light waves by giving distinct colours to the various wavelengths.

The infrared, microwave, and radio spectrum bands have the longest wavelengths. The ultraviolet, x-ray, and gammaradiation have the shortest wavelengths in the electromagnetic spectrum.

Diagram A is therefore satisfactory. On the left, there are long wavelengths with high amplitude (bright) (red). & The right side is dark and has a short wavelength (violet).

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The wave that best represents it is c.

A disgruntled autoworker pushes a small foreign import offacliff with a height of h. the vehicle lands a distance away
fromthe cliff. Determine how fast the vehicle was pushed off
thecliff.