Why is it easier to slide a heavy box over a floor that it is to start it sliding in the first place?

Answer:

thickness t = 528.433 nm

Explanation:

given data

wavelength λ1 = 477.1 nm

wavelength λ2 = 668.0 nm

n = 1.58

solution

we know for constructive interference condition will be

2 × t × μ = (m1+0.5) × λ1     ....................1

2 × t × μ = (m2+0.5) × λ2     ....................2

so we can say from equation 1 and 2

(m1+0.5) × λ1 = (m2+0.5) × λ2

so

[tex]\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}[/tex]     ..............3

put here value and we get  

[tex]\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}[/tex]  

[tex]\frac{m1+0.5}{m2+0.5}[/tex]   = 1.4

[tex]\frac{m1+0.5}{m2+0.5} = \frac{7}{5}[/tex]   ...................4

so we here from equation 4

m1+0.5  = 7

m1 = 3    .................5

m2+0.5 = 4

m2 = 2    .................6

so now put value in equation  1

2 × t × μ = (m1+0.5) × λ1

2 × t × 1.58 = (3+0.5) ×  477.1

solve it we get

thickness t = 528.433 nm

Using the formula for thin film interference, the minimum thickness of the plastic film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

The question is asking for the minimum thickness of the film for which maximum light is reflected for the provided wavelengths. This is a classic example of thin film interference.

For constructive interference (maximum reflection), the thickness of the film (t) is given by the formula: t = mλ/2n. Here 'm' is the order of the bright fringe, 'λ' is wavelength, and 'n' is the index of refraction.

Considering the first-order maximum, we find the thickness for each wavelength:
t1 = (1)(477.1 x 10^-9 m)/2(1.58) = 1.51 x 10^-7 m (or 151 nm for 477.1 nm light)
t2 = (1)(668.0 x 10^-9 m)/2(1.58) = 2.12 x 10^-7 m (or 212 nm for 668.0 nm light)

Therefore, the minimum thickness of the film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.

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Answer:764.4N

Explanation:

Mass=78kg

Acceleration=9.8m/s^2

force=mass x acceleration

Force=78 x 9.8

Force=764.4

Force=764.4N

Complete Question

Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?

Answer:

The time it would take at the given speed is  [tex]x = 43.00 \ minutes[/tex]

Explanation:

From the question we are told that

      The time taken to exist the highway is  [tex]t = 55 min[/tex]

      The distance to the exist is  [tex]d = 43\ mi[/tex]

       Alberta speed is [tex]v = 60 mph[/tex]

The time it would take travelling at the given speed is mathematically represented as

        [tex]t_z = \frac{d}{v}[/tex]

  substituting values

        [tex]t_z = \frac{43}{60}[/tex]

        [tex]t_z = 0.71667\ hrs[/tex]

Converting to minutes

         1  hour =  60 minutes

So      0.71667 hours = x minutes

   Therefore

               [tex]x = 0.71667 * 60[/tex]

                [tex]x = 43.00 \ minutes[/tex]

At a speed of 60 mph, it would take Alberta approximately 43 minutes to travel the 43 miles to her grandmother's house, which is within the 55 minutes time frame she has to avoid being late.

To determine how long it will take Alberta to reach her grandmother's house if she travels at a constant speed of 60 mph, we need to use the formula for time which is time = distance ÷ speed. Alberta's exit is 43 miles away and the speed limit is 60 mph.

First, we calculate the time it would take her to travel 43 miles at 60 mph:

Time = Distance ÷ Speed
= 43 miles ÷ 60 mph
= 0.7167 hours

Since time in hours is not always intuitive, let's convert it to minutes by multiplying by 60 (since there are 60 minutes in one hour):

Time in minutes = 0.7167 hours × 60 minutes/hour
= 43 minutes

Thus, it will take Alberta approximately 43 minutes to reach her exit at the posted speed of 60 mph.

Answer:

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Explanation:

given data

radius R = 26 mm

plate separation d = 4.0 mm

potential difference Vm = 220 V

frequency f = 76 Hz

V = (220 V) sin[2π(76 Hz)t]

solution

we know that E will be

E = V ÷ d     ............1

put here value

E = [tex]\frac{220 \times sin(2\pi 76\times t)}{d}[/tex]  

and here we take as given r = R

so A = π R²    .................2

and

ФE  = E × A

ФE = [tex]\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}[/tex]  .....................3

so use use here now  Ampere's Law that is

∫ B ds = [tex]\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}[/tex]      .....................4

and

here [tex]I_{encl}[/tex]  is = 0  and r = R

so

[tex]2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}[/tex]      .....................5

and put here value we get

B =  [tex]\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}[/tex]        .....................6

put here value  for B maximum cos(2πft) = 1

and we get B (max)

B(max) = [tex]\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}[/tex]     ....................7

put here all value

B(max) = [tex]\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}[/tex]      

solve it we get

B(max) =  3.7971 × [tex]10^{-12}[/tex]  T

Answer:

1.76 eV

Explanation:

Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal

K.E' = (hc/λ)-∅.................. Equation 1

Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.

make ∅ the subject of the equation

∅ = (hc/λ)-K.E'.................. Equation 2

Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J

Substitute into equation 2

∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹

∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)

∅ = 3.21×10⁻¹⁹ J.

The maximum kinetic energy of the photo electrons when the wave length is 330 nm is

K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)

K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)

K.E' = 2.82×10⁻¹⁹ J

K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹

K.E' = 1.76 eV

Answer:they repel

Explanation:

Like charges repels, unlike charges attracts

The correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Explanation

The moon is the most popular and well-known natural satellite on planet earth, it is a satellite widely studied by humans, they have even walked on the moon. However, from Earth, at night when you see the moon you always see the same face of the moon, which has caused people to wonder why this phenomenon. The answer to this phenomenon is that we always see the same face of the moon because it takes the same time to rotate once on itself as it does to go around the Earth (about 27 days). The result is that the same part of the moon always points towards the earth. According to the above, the correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.

Answer:

20m

Explanation:

Pressure = pgh

p = density of water 1000

kg/m^3

g = acceleration due to gravity 9.81 m/s^2

h is the depth of water

Pressure = 201 kPa = 201 x 10^3 Pa

201 x 10^3 = 1000 x 9.81 x h

201 x 10^3 = 9810h

h = 20.49 m

Approximately 20 m

Answer:

The final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Explanation:

It is given that,

Energy, Q = 10 cal

Mass of ice, m = 2 g

Initial temperature, [tex]T_1=-30^{\circ}\ C[/tex]

We need to find the final temperature of the ice. We know that the specific heat of ice is [tex]0.5\ cal\ g^{-1} ^{\circ} C^{-1}[/tex]

The heat added in terms of specific heat is given by :

[tex]Q=mc(T_1-T_2)[/tex]

[tex]T_2[/tex] final temperature of the ice

c is specific heat of ice

[tex]T_2=T_1-\dfrac{Q}{mc}\\\\T_2=(-30)-\dfrac{10}{2\times 0.5}\\\\T_2=-40^{\circ}[/tex]

So, the final temperature of the ice is [tex]-40^{\circ}\ C[/tex].

Answer: the correct answer is -20

Explanation: