A proton with a speed of 2 x 105 m/s falls through a potential difference V and thereby increases its speed to 6 x 105 m/s. Through what potential difference did the proton fall

Answer:

a) L = 1.17 m

b) width of central maxima = 1.28 mm

Explanation:

Given:-

- The wavelength of light, λ = 588.0 nm

- The slit of width, a = 0.74 mm

Find:-

(a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.93 mm from the central maximum

(b) Calculate the width of the central maximum.

Solution:-

- The results of Young's single slit experiment are given in form of a relation as the angle of separation between fringes ( θ ) as function of fringe order ( m ) and wavelength ( λ ).

- Destructive interference produces the dark fringes ( minimum ).  Dark fringes in the diffraction pattern of a single slit are found at angles θ for which:

                        a*sin ( θ ) = m*λ

Where,

            m : The order number for the minimum ( dark fringe ).

- We are to investigate for the first (m = 1 )) dark fringe (minima) which is y = 0.93 mm from central order ( m = 0 ) for which the screen must be placed at a distance L:

                        a*y/L = m*λ

                        L = a*y / m*λ      

                        L = (0.74*0.93) / (1*588*10^-9)    

                        L = 1.17 m

- The distance L of screen should be 1.17 m away from slit.

- The central maximum - central bright fringe. The maxima lie between the minima and the width of the central maximum is simply the distance between the 1st order minima from the centre of the screen on both sides of the centre.

                                tanθ ≈ θ ≈ y/L = w*λ

                                y = w*λ*L

The width of the central maximum is simply twice this value:

- Width of central maximum = 2λLw = 2*588*10^-6*1170*0.93  = 1.28 mm                                

Answer:

a) 1.17 m

b) 0.929mm

Explanation:

(a) to find the distance to the screen you use

[tex]y=\frac{m\lambda D}{d}[/tex]

m: order of the fringe

lambda: wavelength of the light = 588*10^{-9} m

D: distance to the screen

d: distance between slits = 0.74*10^-3 m

by doing D the subject of the formula and replacing the values of the other parameters you obtain:

[tex]D=\frac{dy}{m\lambda}=\frac{(0.74*10^{-3}m)(0.93*10^{-3})}{(1)(588*10^{-9}m)}=1.17m[/tex]

the distance to the screen is 1.17m

(b) to find the width of the central maximum you calculate the position of the first dark fringe:

[tex]y=\frac{(1)(588*10^{-9}m/2)(1.17m)}{0.74*10^{-3}m}=4.64*10^{-4}m=0.464mm[/tex]

[tex]y=\frac{m(\lambda/2)D}{d}[/tex]

2y = 2(0.464mm)=0.929mm is the width of the central maximum

Answer: 0.88 m/s

Explanation:

given

Force constant of the spring, k = 965 N/m

Compression of the spring, x = 3.7 cm = 0.037m

Mass of the block, m = 1.7 kg

To solve this, we would be using law of conservation of energy, where initial energy is all kinetic and elastic energy is elastic but potential.

E(f) = E(i)

1/2kx² = 1/2mv²

kx² = mv²

965 * 0.037² = 1.7 * v²

1.321 = 1.7 v²

v² = 1.321 / 1.7

v² = 0.777

v = 0.88 m/s

Thus, the initial speed of the block is v = 0.88 m/s

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

Answer:

  h ’= 3.75 inch

Explanation:

For this exercise we use the definition of magnification which is the height of the image over the height of the object

                 m = h ’/ h

                  h’= m h

Let's calculate

               h ’= 5  3/4

              h ’= 3.75 inch

Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

[tex]\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}[/tex]

[tex]\dot n_{o} = 100\,\frac{rev}{s}[/tex]

Deceleration rate needed to stop the circular saw is:

[tex]\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}[/tex]

[tex]\ddot n = - 50\,\frac{rev}{s^{2}}[/tex]

The number of turns associated with such deceleration rate is:

[tex]\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}[/tex]

[tex]\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}[/tex]

[tex]\Delta n = 100\,rev[/tex]

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.

Answer:

The acceleration is   [tex]a = 3.45*10^{3} m/s^2[/tex]

Explanation:

 From the question we are told that

         The radius is  [tex]d = 6.5 cm = \frac{6.5}{100} = 0.065 m[/tex]

           The magnitude of the magnetic field is  [tex]B = 5.5 T[/tex]

           The rate at which it decreases is  [tex]\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s[/tex]

             The distance from the center of field is  [tex]r = 1.5 cm = \frac{1.5}{100} = 0.015m[/tex]

  According to Faraday's law

          [tex]\epsilon = - \frac{d \o}{dt}[/tex]

and   [tex]\epsilon = \int\limits {E} \, dl[/tex]

 Where  the magnetic flux [tex]\o = B* A[/tex]

             E is the electric field  

             dl is a unit length

 So

         [tex]\int\limits {E} \, dl = - \frac{d}{dt} (B*A)[/tex]

         [tex]{E} l = - \frac{d}{dt} (B*A)[/tex]

Now [tex]l[/tex] is the circumference of the circular loop formed by the magnetic field and it mathematically represented as  [tex]l = 2\pi r[/tex]

A is the area  of the circular loop formed by the magnetic field and it mathematically represented as  [tex]A= \pi r^2[/tex]

So

    [tex]{E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}[/tex]

    [tex]E = \frac{r}{2} [ - \frac{db}{dt} ][/tex]  

Substituting values  

    [tex]E = \frac{0.015}{2} (24*10^{-4})[/tex]

         [tex]E = 3.6*10^{-5} V/m[/tex]

The negative signify the negative which is counterclockwise

  The force acting on the proton is mathematically represented as

                       [tex]F_p = ma[/tex]

        Also       [tex]F_p = q E[/tex]

So

           [tex]ma = qE[/tex]

 Where m is the mass of the the proton which has a value of  [tex]m = 1.67 *10^{-27} kg[/tex]

 [tex]q = 1.602 *10^{-19} C[/tex]

     So

            [tex]a =\frac{1.60 *10^{-19} *(3.6 *10^{-5}) }{1.67 *10^{-27}}[/tex]

               [tex]a = 3.45*10^{3} m/s^2[/tex]

Answer: 3 m/s²

Explanation:

m = 3 kg

F = 9 N

a = ?

F = m × a

9 = 3 × a

[tex]\frac{9}{3}[/tex] = a

a = 3 m/s²

Answer:

Explanation:

Solution is attached below

Answer: A cold front occurs when cold, denser air replaces the rising, less dense air mass. The reason this front brings in the rain is that as the rising warm air cools (as it rises to the cooler upper atmosphere) the moisture in it condenses into clouds that precipitate down as rain or snow.

Answer:

cold front

Explanation:

The number of revolutions is 301

Explanation:

Angular acceleration, α = 3.5 rad/s

Time, t = 9s

Number of revolutions, n = ?

We know,

[tex]\alpha = \frac{w}{t}[/tex]

where,

ω = angular velocity

t = time

It can also be written as:

[tex]\alpha = \frac{\pi n}{t X 30}[/tex]

On substituting the value, we get:

[tex]3.5 = \frac{\pi X n}{9 X 30} \\\\n = 300.8[/tex]

Therefore, the number of revolutions is 301

The  answer is 22 complete revolutions.

The number of complete revolutions the disk makes in 9 seconds is given by the formula for angular displacement due to constant angular acceleration:

[tex]\[ \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2 \][/tex]

where:

-[tex]\( \theta \)[/tex] is the final angular displacement in radians,

-[tex]\( \theta_0 \)[/tex] is the initial angular displacement, which is zero since the disk starts from rest,

-[tex]\( \omega_0 \)[/tex] is the initial angular velocity, which is also zero since the disk starts from rest,

- [tex]\( \alpha \)[/tex] is the angular acceleration, which is 3.5 [tex]rad/s\( ^2 \)[/tex],

- [tex]\( t \)[/tex] is the time in seconds, which is 9 seconds.

Since the disk starts from rest, [tex]\( \theta_0 = 0 \)[/tex] and [tex]\( \omega_0 = 0 \)[/tex], the equation simplifies to:

[tex]\[ \theta = \frac{1}{2} \alpha t^2 \][/tex]

Substituting the given values:

[tex]\[ \theta = \frac{1}{2} \times 3.5 \, \text{rad/s}^2 \times (9 \, \text{s})^2 \][/tex]

[tex]\[ \theta = \frac{1}{2} \times 3.5 \times 81 \][/tex]

[tex]\[ \theta = 1.75 \times 81 \][/tex]

\[ \theta = 141.75 \, \text{radians} \]

To find the number of complete revolutions, we need to convert the angular displacement from radians to revolutions. Since there are [tex]\( 2\pi \)[/tex] radians in one revolution, we divide the angular displacement by [tex]\( 2\pi \)[/tex]:

[tex]\[ \text{Number of revolutions} = \frac{\theta}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{2\pi} \][/tex]

[tex]\[ \text{Number of revolutions} = \frac{141.75}{6.2831853071796} \][/tex]

[tex]\[ \text{Number of revolutions} \approx 22.57 \][/tex]

Answer: you'll see cyan color on the screen

Explanation:

Saturating the red cone causes them to stop functioning, hence you can't perceive the red part of white light. White light is made up of three main colors which are blue, red and green. When one can no longer perceive the red part of light, one is left with the grean and blue part. The green and blue part of light will superimpose to give a cyan color.

Answer:

Cyan color

Explanation:

When an individual stares at a red object and immediately look at a white area afterward, there will be an image formation almost immediately that is the same size and shape as the previous one. The only difference is there will be a color change to blue-green, or cyan.

This is due to the eyes generally using the red, blue and green cone cells to perceive white light. Since the red cone cells are fatigued due to long view of it and cone cells being saturated , the blue and green color which are the remaining color cone cells use to perceive white light is viewed. This gives rise to the blue green or cyan color.

Answer:

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s

Explanation:

as per the system, it conserves the linear momentum,

so along x axis :

Mp V1 (i) = Mp V1 (f) cos θ1 + Mp V2 (f) cos θ2

along y axis :

0 = -Mp V1 (f) sin θ1 + Mp V2 (f) sin θ2

let us assume before collision it was moving on positive x axis, hence target angle will be θ2 = 45° from x axis

V2(f) = V1 (i) sin θ1 / ( cosθ2 sin θ1 + cos θ1 sin θ2)

      =  1100 * sin 45 / ( cos 45 sin 45 + cos 45 sin 45 )

     =  1100 * 0.7071 /( 0.7071 * 0.7071 +0 .7071 * 0.7071 )

     = 777.82 /( 0.5 + 0.5)

    = 777.82 m/s

(b) 

 the speed of projectile , V1 (f) = sinθ2 * V2(f) / sinθ1

                                                              = sin 45 * 777.82 / sin 45

                                                              = 777.82 m/s

So, the target proton's speed is  777.82 m/s

And, the projectile proton's speed is  777.82 m/s

Answer:

Explanation:

solution solved below

Answer:

Answer: The thickness of film is equal to 1/2 of wavelength in the film

Explanation:

Consider a thin film surrounded with two other transparent materials. Index of refraction of the top material is 1.2; index of refraction of a film 1.3 and of the bottom material 1.4 For normal incidence this film will look bright for light which wavelength satisfies the following condition a. Thickness of a film is equal to of wavelength of light in bottom material b. Thickness of a film is equal ½ of wavelength in the film c. Thickness of a film is equal ¼ of wavelength in the film. d. Thickness of a film is equal to ½ of wavelength of light in upper material e. Thickness of a film is equal ¼ of wavelength of light in upper material f. Thickness of a film is equal 4 of wavelength in bottom material

For constructive interference

     Zt = ∧/n

    t=∧/n

        2

thickness = 1/2 x wavelength in the film

The thickness of film is equal to 1/2 of wavelength in the film

The condition for a thin film to look bright is when its thickness is ½ of the wavelength in the film itself, due to constructive interference resulting from the phase shift upon reflection.

The question deals with thin film interference, a phenomenon where light waves reflected off the top and bottom surfaces of a thin film interfere with each other. For a thin film to appear bright for light of a certain wavelength, the condition for constructive interference must be satisfied. This often involves the film's thickness being a multiple or a fraction of the wavelength of light in the film itself or in the surrounding media.

In the context of thin films surrounded by materials with different indices of refraction, the correct condition for the film to look bright is when its thickness is equal to ½ of the wavelength in the film itself. This accounts for the phase shift that occurs upon reflection, leading to constructive interference when the optical path difference is a multiple of the wavelength.

It's important to note that the index of refraction of the material influences the effective wavelength of light within it. Therefore, the wavelength in the material can be determined by dividing the wavelength in vacuum by the material's index of refraction.

Answer: A. The total number of protons and the total number of neutrons both remain the same.

Explanation:

Nuclear fission is the process in which a large nucleus splits into two smaller nuclei with the release of energy. In other words, fission is the process in which a nucleus is divided into two or more fragments, and neutrons and energy are released.

In fission processes, the total number of protons and neutrons both remain the same because nucleons are neither created nor destroyed, they just rearrange into new nuclei.

In fission processes, it is true that the overall number of protons and the total number of neutrons remains the same after the event. This is because nucleons, which include both protons and neutrons, are neither created nor destroyed during fission; they simply rearrange themselves into new nuclei. The mass number, which is the sum of protons and neutrons, is preserved, and though there can be a conversion between protons and neutrons (e.g., via beta decay), the total count of nucleons stays constant. However, the total mass of the product nuclei is less than the mass of the reactants due to the release of nuclear energy as a result of the conversion of mass to energy according to Einstein's equation E=mc2.

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Answer:

216.59 s.

Explanation:

Using,

F = ma................. Equation 1

Where F = force exerted by the tugboat, m = mass of the ocean liner, a = acceleration of the ocean liner

make a the subject of the equation

a = F/m.................. Equation 2

Given: F = 195 kN = 195000 N, m = 38000 Mg = 38000000 kg.

Substitute into equation 2

a = 195000/38000000

a = 5.13×10⁻³ m/s²

Also using,

Assuming the liner is decelerating

a = (u-v)/t............ Equation 3

Where v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (u-v)/a............. Equation 4

Given: u = 4 km/h = 4(1000/3600) = 1.111 m/s, v = 0 m/s, a = 0.00513 m/s²

Substitute into equation 4

t = (1.111-0)/0.00513

t = 216.59 s.

Answer:a low pressure system is moving into the area

Explanation:

This is the only reasonable answer

Answer: a low pressure system is moving into the area

Explanation: It says it on study island

Answer:

Explanation:

Given that,

Surface area A= 17m²

The speed at the top v" = 66m/s

Speed beneath is v' =40 m/s

The density of air p =1.29kg/m³

Weight of plane?

Assuming that,

the height difference between the top and bottom of the wind is negligible and we can ignore any change in gravitational potential energy of the fluid.

Using Bernoulli equation

P'+ ½pv'²+ pgh' = P'' + ½pv''² + pgh''

Where

P' is pressure at the bottom in N/m²

P" is pressure at the top in N/m²

v' is velocity at the bottom in m/s

v" is velocity at the top in m/s

Then, Bernoulli equation becomes

P'+ ½pv'² = P'' + ½pv''²

Rearranging

P' — P'' = ½pv"² —½pv'²

P'—P" = ½p ( v"² —v'²)

P'—P" = ½ × 1.29 × (66²-40²)

P'—P" = 1777.62 N/m²

Lift force can be found from

Pressure = force/Area

Force = ∆P ×A

Force = (P' —P")×A

Since we already have (P'—P")

Then, F=W = (P' —P")×A

W = 1777.62 × 17

W = 30,219.54 N

The weight of the plane is 30.22 KN

Answer:

Weight of plane ; W = 30219.54 N

Explanation:

For us to determine the lift force of the system, let's multiply the pressure difference with the effective wing surface area given that the area is obtained by Bernoulli equation. Thus,

P_b + (1/2)ρ(v_b)² + ρg(y_b) = P_t + (1/2)ρ(v_t)² + ρg(y_t)

Now, since the flight is level, the height is constant.

Thus, (y_b) = (y_t)

So, we now have;

P_b + (1/2)ρ(v_b)² = P_t + (1/2)ρ(v_t)²

Rearranging, we have ;

P_b - P_t = (1/2)ρ(v_t)² - (1/2)ρ(v_b)²

P_b - P_t = (1/2)ρ[(v_t)² - (v_b)²]

Now, weight is given by the formula;

W = (P_b - P_t) •A

Thus,

W = (1/2)ρ[(v_t)² - (v_b)²] •A

From the question,

Density; ρ = 1.29 kg/m³

Velocity over top of wings; v_t = 66 m/s

Velocity beneath the wings; v_b = 40 m/s

Surface Area; A = 17 m²

Thus;

W = (1/2)1.29[(66)² - (40)²] •17

W = (1/2)•1.29•17[2756]

W = 30219.54 N

Explanation:

Given that,

Mass of the truck, m = 2268 kg

Speed of the truck, v = 30 m/s

The friction force has an average magnitude of 700 N, f = -700 N

(a) The initial kinetic energy of the truck is given by :

[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2268\times (30)^2\\\\K=1.02\times 10^6\ J[/tex]

(b) Finally, the stops due to an effective friction force that the road, final speed is 0. Let d is the stopping distance of the truck. using third equation of motion to find it as :

[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}[/tex]

Since, f = ma

[tex]d=\dfrac{-u^2m}{2f}\\\\d=\dfrac{-(30)^2\times 2268}{-2\times 700}\\\\d=1458\ m[/tex]

So, the stopping distance of the truck is 1458 meters.

Answer:

8.049 MW

Explanation:

The expression for gravitational potential energy is given as

Ep = mgh............. Equation 1

Ep = gravitational potential energy, m = mass of water, h = height, g = acceleration due to gravity.

Given: m = 58.4×10³ kg, h = 20.1 m, g = 9.81 m/s²

Substitute into equation 1

Ep =  58.4×10³(20.1)(9.81)

Ep = 1.6098×10⁷ J.

If one half the gravitational potential energy of the water were converted to electrical energy

Electrical energy = Ep/2

Electrical energy = (1.6098×10⁷)/2

Electrical energy = 8.049×10⁶ J

In one seconds,

The power generated = 8.049×10⁶ W

Power generated = 8.049 MW

Answer:

a) P ( X ≤ 49 ) = 0.9149

b) P ( X ≥ 48 ) = 0.2119

c) 0.9876

Explanation:

Solution:-

- Lets define a random variable X: the maximum speed of a moped.

- The random variable follows normal distribution with the following parameters:

                    X ~ Norm ( u , σ^2 )

Where,

              u = Mean = 46.6 km/h

              σ = Standard deviation 1.75 km/h

Hence,

                    X ~ Norm ( 46.6 , 1.75^2 ).

a) The probability that the maximum speed of mopeds is atmost 49 km/h?

- To evaluate the probability of P ( X ≤ 49 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{49-46.6}{1.75} \\\\Z-score = 1.37142[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z < 1.37142 ) = 0.9149

Hence,

                     P ( X ≤ 49 ) = 0.9149  

b) The probability that the maximum speed of mopeds is at-least 48 km/h?

- To evaluate the probability of P ( X ≥ 48 ). We will find the standard Z-score value using the following formula:

                       [tex]Z-score = \frac{x-u}{s} \\\\ Z-score = \frac{48-46.6}{1.75} \\\\Z-score = 0.8[/tex]

- Now use the standard normal tables to determine the required probability of:

                     P ( Z ≥ 0.8 ) = 0.2119

Hence,

                     P ( X ≥ 48 ) = 0.2119                          

c) The probability that the maximum speed of mopeds differs from mean by at-most 2.5 standard deviation.

- The required probability is the standard error from the mean value "u" of 2.5 standard deviation.

- We don't need to evaluate the test statistics as we are already given the standard error about mean.

- Using, the standard normal Z-score: The required probability is:

                P ( -2.5 < Z < 2.5 ) = 2*P ( Z < 2.5 ) - 1

                                              = 2*0.9938 - 1

                                              = 0.9876

Explanation:

Case 1,

A force F is pushing perpendicular on an object a distance L/2 from the rotation axis. Torque is given by :

[tex]\tau_1=Fd\sin \theta\\\\\tau_1=\dfrac{FL}{2}\sin (90)\\\\\tau_1=\dfrac{FL}{2}\ ..........(1)[/tex]

Case 2,

The same force is pushing at an angle of 30 degrees a distance L from the axis. New torque is given by :

[tex]\tau_2=Fd\sin \theta\\\\\tau_2=FL\sin (30)\\\\\tau_2=\dfrac{FL}{2}\ ..........(2)[/tex]

From equation (1) and (2), it is clear that the force in both cases is same.

Answer:

d. Planets formed from the disk of gas and dust that surrounded the Sun, and such disks are common around young stars.

Explanation:

A star passes through various stages before it becomes a full fledged star with its own planetary system. The same is with Sun as well. It was born out of a Nebula. Nebula is a cloud of dust and gases. The dust and gases start accumulating to form what we call as a protostar.

A lot of material at this stage is thrown out from the young star, this material forms a disk around the star. This disk is known as proto-planetary disk. The gases and dust of this disk then coalesce together to make planets and other objects of the star system.

We have observed such disks in the Orion nebula.

The solar nebula theory suggests that planets are formed from the disk of gas and dust that surrounds young stars. Such disks are commonly observed around stars, leading to the inference that planet formation is a universal occurrence.

The solar nebula theory, or nebular hypothesis, explains the formation of our solar system from a nebular cloud of gas and dust. According to this theory, planets are formed from the disk of gas and dust that surrounds a young star, also known as protoplanetary disk. This leads to the implication that planets are common because such disks are commonly found around young stars.

When a star condenses from a nebula, it forms a hot, spinning disk of gas and dust around it. Over time, particles within this disk begin to collide and stick together, gradually forming planetesimals and eventually, planets. Since we observe many stars surrounded by these disks, we can infer that planet formation is a common process in the universe.

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Answer: The ratio of their radii (Ratio of the radius of iron wire to that of copper wire) is 2.43

Explanation: Please see the attachments below

Final answer:

The ratio of the radius of an iron wire to that of a copper wire that have the same potential difference and current is approximately 1.9.

Explanation:

To find the ratio of the radii of an iron wire and a copper wire with the same potential difference and current, we can use Ohm's law and the formulas for resistance and resistivity.

The resistance is given by R = rho * (L/A), where rho is the resistivity, L is the length, and A is the cross-sectional area of the wire. Since the potential difference and current are the same for both wires, we can equate the resistances and cross-sectional areas, resulting in rho_iron * (L/A_iron) = rho_copper * (L/A_copper). Rearranging the equation, we get (A_iron/A_copper) = (rho_iron/rho_copper).

Since the radii are related to the areas of the wires by the equation A = pi * r^2, we can substitute A_iron = pi * (r_iron)^2 and A_copper = pi * (r_copper)^2 into the equation. This gives us (pi * (r_iron)^2)/(pi * (r_copper)^2) = (rho_iron/rho_copper). By canceling out the pi terms, we find (r_iron/r_copper)^2 = (rho_iron/rho_copper). Simplifying further, we get r_iron/r_copper = sqrt(rho_iron/rho_copper).

Substituting the given resistivities into the equation, we have r_iron/r_copper = sqrt((1.0 * 10^-7) / (1.7 * 10^-8)). Evaluating the expression, we find r_iron/r_copper ≈ 1.9. Therefore, the ratio of the radius of the iron wire to that of the copper wire should be approximately 1.9.

Answer:

1.56 - 1.67

Explanation:

Refractive index of any material is given as the ratio of the speed of light in a vacuum to the speed of light in that medium.

Mathematically, it is given as:

n = c/v

Where c is the speed of light in a vacuum and v is the speed of light in the medium.

Given that the speed of light in the optical medium varies from 1.8 * 10^8 m/s to 1.92 * 10^8 m/s, we can find the range of the refractive index.

When the speed is 1.8 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.8 * 10^8)

n = 1.67

When the speed is 1.92 * 10^8 m/s, the refractive index is:

n = (3 * 10^8) / (1.92 * 10^8)

n = 1.56

Therefore, the range of values of the refractive index of the optical medium is 1.56 - 1.67.

Final answer:

The range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.

Explanation:

The index of refraction, n, of a material can be calculated using the equation n = c/v, where c is the speed of light in a vacuum and v is the speed of light in the material.

For this specific optical medium, the speed of light varies from 1.80 × 10^8 m/s for violet light to 1.92 × 10^8 m/s for red light. To calculate the range of the index of refraction, we need to determine the ratio of the speed of light in a vacuum to the speed of light in the material for both violet and red light.

The range of the index of refraction, n, can be calculated as:

For violet light: n = c/v = (3.00 × 10^8 m/s) / (1.80 × 10^8 m/s) ≈ 1.67

For red light: n = c/v = (3.00 × 10^8 m/s) / (1.92 × 10^8 m/s) ≈ 1.56

Therefore, the range of the index of refraction for visible light in this optical medium is approximately 1.56 to 1.67.

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

[tex]\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2} =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}[/tex]

[tex]\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad[/tex]

[tex]w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}[/tex]

The velocity at the contact point is equal to:

[tex]v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}[/tex]

[tex]v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}[/tex]

Matching both expressions:

[tex]1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s[/tex]

b) The time during which the disks slip is:

[tex]t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s[/tex]

a) The angular acceleration of each disk is

[tex]\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)[/tex]

[tex]\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)[/tex]

The angular acceleration of each disk and the time during which the disks slip cannot be determined with the given information.

(a) To determine the angular acceleration of each disk, we can use the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. For disk A, the initial angular velocity is 240 rpm counterclockwise and the final angular velocity is 0 rpm. The time during which disk A stops is not given, so we cannot determine its angular acceleration at this time. For disk B, the initial angular velocity is 0 rpm and the final angular velocity is also 0 rpm. Since disk B is at rest initially and then starts moving, it undergoes an angular acceleration that brings it to rest. However, the specific time during which disk B slips is not given, so we cannot determine its angular acceleration during slipping.

(b) The time during which the disks slip is not provided in the question, so we cannot determine the exact time. However, we know that disk B makes 2 revolutions before reaching its final angular velocity, so during this period, disk B is slipping.

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Answer:

515.025J

A woman lifts a 35-kg child a distance of 1.5 m and carries her forward for 6.5 m.

How much work does the woman do in lifting the child?​

Explanation:

Given;

Mass m = 35kg

Distance of lifting l = 1.5m

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

Workdone = mgl = 35×1.5 × 9.81 = 515.025 J

The change in potential energy is equal to the work that must be done, to lift the child a certain distance. Therefore, the work done by the woman in lifting the child is:

[tex]W=\Delta U\\W=mgh_f-mgh_0\\W=mg(h_f-h_0)\\W=35kg*1.5m(1.5m-0m)\\W=515.03J[/tex]

Answer:

Explanation:

Given that,

Metal of mass

M = 0.065kg

Initial temperature of metal

θm = 210°C

The metal is drop into a beaker which contain liquid of mass

m = 0.377 kg

Initial temperature of water

θw = 26°C

The final mixture temperature is

θf = 28.14°C

Specific heat capacity of water

Cw = 4186 J/kg°C

Since the metal is hotter than the water, then the metal will lose heat, while the water will gain heat, we assume that no heat is loss by the beaker.

So,

Heat Loss = Heat gain

Now, heat loss by metal

H(loss) = M•Cm•∆θ

Where M is mass of meta

Cm is specific capacity of metal, which we are looking fro

So,

H(loss) = 0.065 × Cm × (θi - θf)

H(loss) = 0.065 × Cm × (210-28.14)

H(loss) = 11.821 •Cm

Now, Heat gain by water

H(gain) = m•Cw•∆θ

H(gain) = m•Cw•(θf - θi)

Where

m is mass of water and Cw is specific heat capacity of water

H(gain) = 0.377 ×4286 × (28.14-26)

H(gain) = 3457.86

So, H(loss) = Heat(gain)

11.821 •Cm = 3457.86

Cm = 3457.86/11.821

Cm = 292.52 J/Kg°C

The specific heat capacity of the metal ball is 292.52 J/Kg°C

Answer:

320.86J/kgC

Explanation:

To find the specific heat of the substance you take into account that the heat lost by the metal is gained by the water, that is:

[tex]Q_1=-Q_2[/tex]

Furthermore the heat is given by:

[tex]Q_1=m_1c(T_1-T)\\\\Q_2=m_2c(T_2-T)[/tex]

m1: mass of the metal

m2: mass of the water

c: specific heat

T: equilibrium temperature

T1: temperature of the metal

T2: temperature of water

By replacing all these values you can calculate c of the metal:

[tex]m_1c_1(T_1-T)=-m_2c_2(T_2-T)\\\\c_1(0.065kg)(210-28.4)\°C=-(0.377kg)(4186J/kg\°C)(26-28.4)\°C\\\\c_1=320.86\frac{J}{kg\°C}[/tex]

Hence, the specific heat of the metal is 320.86J/kgC

Answer:

[tex]W=173.48J[/tex]

Explanation:

information we know:

Total force: [tex]F=45N[/tex]

Weight: [tex]w=100N[/tex]

distance: [tex]4m[/tex]

vertical component of the force: [tex]F_{y}=12N[/tex]

-------------

In this case we need the formulas to calculate the components of the force (because to calculate the work we need the horizontal component of the force).

horizontal component: [tex]F_{x}=Fcos\theta[/tex]

vertical component: [tex]F_{y}=Fsen\theta[/tex]

but from the given information we know that [tex]F_{y}=12N[/tex]

so, equation these two [tex]F_{y}=Fsen\theta[/tex] and [tex]F_{y}=12N[/tex]

[tex]Fsen\theta =12N[/tex]

and we know the force [tex]F=45N[/tex], thus:

[tex]45sen\theta=12[/tex]

now we clear for [tex]\theta[/tex]

[tex]sen\theta =12/45\\\theta=sin^{-1}(12/45)\\\theta =15.466[/tex]

the angle to the horizontal is 15.466°, with this information we can calculate the horizontal component of the force:

[tex]F_{x}=Fcos\theta[/tex]

[tex]F_{x}=45cos(15.466)\\F_{x}=43.37N[/tex]

whith this horizontal component we calculate the work to move the crate a distance of 4 m:

[tex]W=F_{x}*D\\W=(43.37N)(4m)\\W=173.48J[/tex]

the work done is W=173.48J

To calculate the work done by the force in moving the crate, we use the formula Work = Force x Distance x cos(theta). We are given the force, distance, and the vertical component of the force, so we can find the horizontal component using trigonometry. Substituting the values into the formula gives us the work done as 174 J.

To calculate the work done by a force, we use the formula:

Work = Force x Distance x cos(theta)

In this case, the force is 45 N, the distance is 4 m, and the vertical component of the force is 12 N. We need to find the horizontal component of the force, which can be calculated using the given information.

Given that the vertical component of the force is 12 N, we can use the trigonometric relationship between the horizontal and vertical components of a force to find the horizontal component as follows:

sin(theta) = vertical component / total force

sin(theta) = 12 N / 45 N

sin(theta) = 0.267

theta = sin^(-1)(0.267)

theta = 15.1 degrees

Once we have the horizontal component of the force, we can calculate the work done as follows:

Work = Force x Distance x cos(theta)

Work = 45 N x 4 m x cos(15.1 degrees)

Work = 180 J x cos(15.1 degrees)

Work = 180 J x 0.966

Work = 174 J

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The total volume of bound charge is zero.

Explanation:

We have to the volume and surface bounded charge densities.

   ρb = - Δ . p = - Δ .k ([tex]x^{X}[/tex] +[tex]y^{Y}[/tex] +[tex]x^{Y}[/tex])

                      = - 3k

 On the top of the cube the surface charge density is

                     σb = p . z

                           = [tex]\frac{ka}{2}[/tex]

By symmetry this holds for all the other sides. The total bounded charge should be zero

        Qtot = (-3k)a³ + 6 . [tex]\frac{ka}{2}[/tex] . a² = 0

               σb = -3K σb = [tex]\frac{ka}{2}[/tex]

            Qtot = 0

Hence,  the total volume of bound charge is zero.