PHYSICS COLLEGE

Chuck wants to investigate how gas moleculesmove in a container. Which model would be most
helpful to represent this motion?
A. stacking blocks to build a tower
B. freezing water in an ice cube tray
C. bouncing elastic balls off of each other and
the walls of a room
D. placing a closed, water-filled plastic bag in
the sun and watching condensation form

Answers

Answer 1
Answer: The answers C the molecules in gas move rapidly and all around they are spread out and bounce off each-other

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A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop isa. counterclockwiseb. clockwisec. there is no current in the loop because magnetic field is uniformd. there is no current in the loop because magnetic field does not change
Two charged particles are separated by 10 cm. suppose the charge on each particle is doubled. By what factor does the electric force between the particles change?
. 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 byB = 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/Rwhere 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
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.

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelength of d/15. The lasers produce separate interference patterns on a screen a distance 4.90 m away from the slits.a. What is the distance Δ ymax-max between the first maxima (on the same side of the central maximum) of the two patterns?
b. What is the distance Δymax-min between the second maximum of laser 1 and the third minimum of laser 2, on the same side of the central maximum?

Answers

Answer:

a)Δy = 81.7mm

b)Δy = 32.7cm

Explanation:

To calculate the distance between any point of the interference pattern, simply use the trigonometric ratio of the tangent:

Tan \theta = (y)/(D)

where  D  is the separation between the slits and the screen where the interference pattern is observed.

a) In this case:

Δy  =  |y1max  (λ1) −  y1max  (λ2)|

Δy = |(D\lambda _1)/(d) - (D\lambda _2)/(d) |

Δy = D |(d/20)/(d) - (d/15)/(d) |

Δy = D |(1)/(20) - (1)/(15) |

Δy = 4.90 |(1)/(20)- (1)/(15) |

Δy = 81.7mm

The separation between these maxima is 81.7 mm

b)

Δy  =  |y₂max  (λ1) −  y₂max  (λ2)|

Δy = D|(2(d/20))/(d) - (5(d/15))/(2d) |

Δy = 4.90|(1)/(10) - (1)/(6) |

Δy = 32.7cm

The separation between the maximum interference of the 2nd order (2nd maximum) of the pattern produced by the laser 1 and the minimum of the 2nd order (3rd minimum) of the pattern produced by the laser 2 is 32.7 cm.

Final answer:

We can solve the problem using the concepts of waveinterference and the formulas for maxima and minima positions (i.e., y = L*m*λ/d and y = L*(m+1/2)*λ/d respectively). The difference between the first maxima of the two patterns is 4.9/60 m and the difference between the second maximum of laser 1 and the third minimum of laser 2 is also 4.9/60 m.

Explanation:

The problem described deals with wave interference and can be addressed using the formulas for path difference and phasedifference.

To answer part a, we need to find the difference between the positions of the first maxima for the two lasers. The position of any maxima in an interference pattern can be found using the formula: y = L * m * λ / d, where L is the distance from the slits to the screen, m is the order of the maxima, λ is the wavelength, and d is the slit separation.

So for the first laser (λ1=d/20) the position of the first maxima would be y1 = 4.9m * 1 * (d/20) / d =4.9/20 m.

And for the second laser (λ2 = d/15) the position of the first maxima would be y2= 4.9m * 1 * (d/15) / d =4.9/15 m.

Then, the distance Δ ymax-max between the first maxima of the two patterns is y2-y1= 4.9/15 m - 4.9/20 m = 4.9/60 m.

Answering part b involves finding the positions of the second maximum of laser 1 and the third minimum of laser 2. The position of any minimum in an interference pattern can be calculated using the formula: y = L * (m+1/2) * λ / d. For the second maximum of laser 1, we have y1max2 = 4.9 m * 2 * (d/20) / d = 4.9/10 m. For the third minimum of laser 2, we have y2min3 = 4.9m * (3.5) * (d/15)/d = 4.9*7/30 m. The difference Δymax-min is y2min3-y1max2= 4.9*7/30 m - 4.9/10 m = 4.9/60 m.

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A bridge is made with segments of concrete 91 m long (at the original temperature). If the linear expansion coefficient for concrete is 1.2 × 10−5 ( ◦C)−1 , how much spacing is needed to allow for expansion for an increase in temperature of 56◦F? Answer in units of cm.

Answers

Answer:

change in length is 3.397 cm

Explanation:

Given data

long = 91 m = 9100 cm

coefficient for concrete (a) =  1.2 × 10−5 ( ◦C)−1

temperature = 56 F = (56× 5/9) ◦C

to find out

how much spacing is needed to allow

solution

we know allow space is given by this formula

change in length = coefficient for concrete × given length × temperature     .............1

put all value in equation 1

change in length = 1.2 × 10−5  × 9100 × (56× 5/9)

change in length = 3.397 cm

so change in length is 3.397 cm

A ​46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this​ causeway?

Answers

The required force parallel to the incline to hold the monolith on this​ causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), F_t = F Sin a

By substituting the values,

                                       = 46 × 0.064

                                       = 2.9 tons

Thus the response above is appropriate answer.

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Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

In an electric circuit, resistance and current are ____A. directly proportional
B. inversely proportional
C. have no effect on each other

Answers

In an electric circuit, resistance and current are ____

A. directly proportional

B. inversely proportional

C. have no effect on each other

Explanation:

A

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 290 Hz. What is the speed of the train before and after slowing down?

Answers

To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

f = f_0 ((v_0)/(v_0-v))

Here,

f_0 = Frequency of Source

v_s = Speed of sound

f = Frequency heard before slowing down

f' = Frequency heard after slowing down

v  = Speed of the train before slowing down

So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,

The first equation is,

f = f_0 ((v_0)/(v_0-v))

300 = f_0 ((343)/(343-v))

(300*343) - 300v = 343f_0

Now the second expression will be,

f' = f_0 ((v_0)/(v_0-v/2))

290 = (343)((v_0)/(343-v/2))

290*343-145v = 343f_0

Dividing the two expression we have,

((300*343) - 300v)/(290*343-145v) = 1

Solving for v, we have,

v = 22.12m/s

Therefore the speed of the train before and after slowing down is 22.12m/s

Final answer:

The speed of the train can be determined using the Doppler effect formula.

Explanation:

The question involves the Doppler effect, which is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave. In this case, the train whistle's frequency changes from 300 Hz to 290 Hz as the train approaches the station.

To find the speed of the train before and after slowing down, we can use the formula for the Doppler effect:

f' = f((v + v_o)/(v - v_s))

Where:

  • f' is the observed frequency
  • f is the source frequency
  • v is the speed of sound
  • v_o is the speed of the observer (here it is the train)
  • v_s is the speed of the source (here it is the speed of sound)

By substituting the given values for observed frequency (290 Hz), source frequency (300 Hz), and the speed of sound (343 m/s), we can solve for the speed of the train before and after slowing down.

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A Michelson interferometer operating at a 400 nm wavelength has a 3.95-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atmatm pressure is 1.00028.Required:
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

55.3

Explanation:

The computation of the number of bright-dark-bright fringe shifts observed is shown below:

\triangle m = (2d)/(\lambda) (n - 1)

where

d = 3.95 * 10^(-2)m

\lambda = 400 * 10^(-9)m

n = 1.00028

Now placing these values to the above formula

So, the  number of bright-dark-bright fringe shifts observed is

= (2 *3.95 * 10^(-2)m)/(400 * 10^(-9)m) (1.00028 - 1)

= 55.3

We simply applied the above formula so that the number of bright dark bright fringe shifts could come
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