Your high-fidelity amplifier has one output for a speaker of resistance 8 Ω. How can you arrange two 8-Ω speakers, one 4-Ω speaker, and one 12-Ω speaker so that the amplifier powers all speakers and their equivalent resistance when connected together in this way is 8 Ω? Compare the power output of your arrangement with the power output of a single 8-Ω speaker.

Answer:

A. recycling allows resources to be reused, instead of dumped in landfills.

Explanation:

Answer:

Explanation:

When 238U which is radioactive turns into 206Pb , it becomes stable and no further disintegration is done . Hence in the initial period ratio of 238U undecayed and 206Pb formed  will be very high because no of atoms of 238U in the beginning will be very high. Gradually number of 238U undecayed will go down and number of 206Pb formed will go up . In this way the ratio of 238U and 206Pb in the mixture will gradually reduce to be equal to one or even less than one .

In the given option we shall calculate their raio

1 ) ratio of 238U and 206Pb = 5

2 ) ratio of 238U and 206Pb = 4

3 )ratio of 238U and 206Pb = 1

4 ) ratio of 238U and 206Pb = 20

5 )ratio of 238U and 206Pb = 3

lowest ratio is 1 , hence this sample will be oldest.

Ranking from youngest to oldest

4 , 1 , 2 , 5 , 3 .

Answer:

The correct answer to the following question will be "41.87 m".

Explanation:

The given values are:

The speed of trooper = [tex]30.1 \ m/s[/tex]

The velocity of red car = [tex]45.4 \ m/s[/tex]

Now,

A red car goes as far as possible until the speed or velocity of the troops is the same as that of of the red car at

[tex]t=\frac{45-30}{2.7}[/tex]                                                              (∵ [tex]time=\frac{distance}{time}[/tex])

[tex]t=5.56 \ sec[/tex]

then,

The distance covered by trooper,

[tex]t1=30\times 5.56+\frac{1}{2}\times 2.7\times (5.56)^2[/tex]

   [tex]=208.33 \ m[/tex]

The distance covered by red car,

= [tex]45\times 5.56[/tex]

= [tex]250.2 \ m[/tex]

Maximum distance = [tex]250.2-208.33[/tex]

                                = [tex]41.87 \ m[/tex]                                                        

Final answer:

The trooper's car catches up with the red car after 7.65 seconds. The maximum distance ahead reached by the red car is 347.01 meters.

Explanation:

To find the maximum distance ahead reached by the red car, we need to calculate the time it takes for the trooper's car to catch up with the red car.

First, find the time it takes for the trooper's car to accelerate from 30.1 m/s to 45.4 m/s:

t = (v - v₀) / a = (45.4 - 30.1) / 2.0 = 7.65 seconds

Now, we can find the maximum distance ahead by multiplying the velocity of the red car by the time it took for the trooper's car to catch up:

Distance = velocity x time = 45.4 m/s x 7.65 s = 347.01 meters

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

Final answer:

The physics problem involves two students on skateboards applying a force to each other and moving in opposite directions. To find the velocity of the second student, the conservation of momentum is used, considering no external forces act on them.

Explanation:

The subject of this question is physics, specifically relating to the principle of conservation of momentum in a system of two students standing on skateboards. When the two students push off each other, they start moving in opposite directions due to this principle. Conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this scenario, the system comprises the two students on skateboards.

To find how fast the second student moves, one would use the formula:

Total momentum before the shove = Total momentum after the shove

0 = (mass of student 1 \(\times\) velocity of student 1) + (mass of student 2 \(\times\) velocity of student 2)

We have the mass and velocity of student 1 (77 kg, moving left at 5.7 m/s), and the mass of student 2 (59 kg), but we need to calculate the velocity of student 2 after the push to ensure momentum is conserved.

To find the double-slit characteristics of width and separation, one must use the data from a single-slit diffraction pattern to calculate the slit width and then the interference pattern from both slits to find the slit separation.

The student's question is about the characteristics of a double-slit diffraction pattern produced when a laser beam with a specific wavelength is projected through two slits. To deduce the slit width and separation, we use the information provided about the diffraction and interference patterns observed.

Firstly, we use the minima separation from the single slit diffraction to find the slit width using the diffraction formula d sin(θ) = mλ, where d is the slit width, m is an integer denoting the order of the minima, λ is the wavelength, and θ is the angle of the minima. Since the screen is 1 meter away and the minima are 6.5 mm apart, we can approximate sin(θ) as the ratio of minima separation to the distance to the screen to solve for the slit width d.

Next, we apply the formula for double-slit interference, d sin(θ) = nλ (where d is now the slit separation, n is an integer denoting the order of the maxima) to the observed interference maxima separation to calculate the slit separation.

Slit separation: [tex]\(97.23 \times 10^{-6}\)[/tex] m. Slit width: [tex]\(1194.34 \times 10^{-6}\)[/tex] m. Calculated using interference formula and provided data.

To solve this problem, we can use the double-slit interference formula:

[tex]\[ \Delta y = \frac{\lambda L}{d} \][/tex]

where:

- [tex]\( \Delta y \)[/tex] is the distance between adjacent maxima or minima on the observation screen,

- [tex]\( \lambda \)[/tex] is the wavelength of the laser beam,

- [tex]\( L \)[/tex] is the distance between the screen and the slits, and

- [tex]\( d \)[/tex] is the slit separation.

Given:

- Wavelength [tex]\( \lambda = 632.8 \)[/tex] nm = [tex]\( 632.8 \times 10^{-9} \)[/tex] m,

- Distance between the screen and the slits [tex]\( L = 1 \)[/tex] m = 1 m,

- Distance between adjacent minima [tex]\( \Delta y_{\text{min}} = 6.5 \)[/tex] mm = [tex]\( 6.5 \times 10^{-3} \)[/tex] m, and

- Distance between adjacent maxima [tex]\( \Delta y_{\text{max}} = 0.53 \)[/tex] mm = [tex]\( 0.53 \times 10^{-3} \)[/tex] m.

First, let's find the slit separation [tex]\( d \)[/tex] using the minima data:

[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} \][/tex]

Then, let's find the slit width [tex]\( w \)[/tex] using the maxima data:

[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} \][/tex]

Let's calculate these values.

First, let's calculate the slit separation [tex]\(d\)[/tex] using the minima data:

[tex]\[ d = \frac{\lambda L}{\Delta y_{\text{min}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{6.5 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ d \approx \frac{(632.8)(1)}{6.5} \times 10^{-6} \, \text{m} \][/tex]

[tex]\[ d \approx 97.23 \times 10^{-6} \, \text{m} \][/tex]

Now, let's calculate the slit width [tex]\(w\)[/tex] using the maxima data:

[tex]\[ w = \frac{\lambda L}{\Delta y_{\text{max}}} = \frac{(632.8 \times 10^{-9} \, \text{m})(1 \, \text{m})}{0.53 \times 10^{-3} \, \text{m}} \][/tex]

[tex]\[ w \approx \frac{(632.8)(1)}{0.53} \times 10^{-6} \, \text{m} \][/tex]

[tex]\[ w \approx 1194.34 \times 10^{-6} \, \text{m} \][/tex]

So, the slit separation [tex]\(d\)[/tex] is approximately [tex]\(97.23 \times 10^{-6}\)[/tex] m, and the slit width [tex]\(w\)[/tex] is approximately [tex]\(1194.34 \times 10^{-6}\)[/tex] m.

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

[tex]T_1+V_1=T_2+V_2[/tex]

[tex]0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}[/tex]

Also;

[tex]\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}[/tex]

Thus;

[tex]k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}[/tex]

where;

[tex]\delta[/tex] = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

[tex]k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}[/tex]

[tex]k = 104.82\ \ N/m[/tex]

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

[tex]T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0[/tex]

[tex]\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0[/tex]

[tex]\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0[/tex]

[tex]0.7344 \omega_3^2 = 2.128[/tex]

[tex]\omega _3 = \sqrt{\frac{2.128}{0.7344} }[/tex]

[tex]\omega _3 =1.70 \ rad/s[/tex]

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = [tex]\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}[/tex]

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= [tex]\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}[/tex]

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

1. The resultant speed for an inelastic collision is =.05 m /s

2. When The direction for the smaller object the collision when it rebounds is = -.66 m /s

Conservation of momentum is refer to a major law of physics that states that the momentum of a system is constant if no external forces are acting on the system.  when It is embodied in Newton’s First Law or The Law of Inertia.

1. We are involving the conservation of momentum law to solve the problem.

mv is = ( M +m) V, m, and M are masses of the small and large objects, v is the velocity of a small object before the collision and also V is the velocity of both the objects together after the collision.

Then .5 x .2 = (1.5 + .5)V

Therefore, V = .05 m /s

2. We shall use the formula for the velocity of the object behind the elastic collision as follows

After that, v₁ = (m₁-m₂)/(m₁+m₂)u1 + 2m₂u₂/(m₁+m₂)

m₁ and m₂ are masses of the first and second objects u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

Now, = (200-1000)/(1000+200) × 1 + 2×1000×0/(1000+200)

Thus, = - .66 m /s

When the sign is negative so it will be in opposite direction.

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Answer:

a) falso

b) verdadero

c) falso

d) falso

e) verdadero

Explanation:

A) FALSO: la distancia recorrida es independiente del punto de partida y del punto de llegada

B) VERDADERO: para que el ciclista recorra la pista de ida y vuelta es necesario que su velocidad cambie, por lo tanto la aceleración es diferente de cero.

C) FALSO

D) FALSO: en todo momento hay una distancia recorrida en un tiempo en específico, es decir, una rapidez.

E) VERDADERO: el despazamiento sí depende de la distancia entre el punto de llegada y el punto de salida.

Answer:

[tex]r_{min}=\frac{kq^2}{5m_ev^2}[/tex]

Explanation:

The total kinetic energy of both electrons will be electrostatic potential energy, when the electrons reach the minima distance due to electrostatic repulsion. Then, you have:

[tex]E_{k}=U_E\\\\\frac{1}{2}m_ev_1^2+\frac{1}{2}m_ev_2^2=k\frac{q^2}{r_{min}}[/tex]

me: mass of the electron

q: charge of the electron

k: Coulomb's constant

you take into account that v2=3v1=3v and do rmin the subject of the formula:

[tex]\frac{1}{2}m_e[v^2+9v^2]=5m_ev^2=k\frac{q^2}{r_{min}}\\\\r_{min}=\frac{kq^2}{5m_ev^2}[/tex]

The minimum separation rmin of two electrons moving towards each other due to their electric repulsion can be calculated using the equation rmin = 2q2 / (mkv2), where q is the charge of the electron, m is its mass, v is the initial speed of one electron, and k is the electrostatic constant.

When two electrons with mass m and charge q move towards each other along the x-axis, their repulsion causes them to slow down. Eventually, they reach a point where their repulsion is strong enough to push them away from each other. This point of minimum separation, denoted by rmin, can be found by equating the electrical force of repulsion to the centripetal force of motion. Since the electrical force between two point charges q1 and q2 separated by a distance r is given by F = k(q1q2)/r2, where k is the electrostatic constant, the equation for rmin is:

rmin = 2q2/ (mkv2)

where v is the initial speed of one electron towards the other.

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Answer:

the​ difference, in miles per​ hour, in the linear speeds of Brett and​ Will;

∆v = 1.38 mph

Explanation:

Given;

Angular speed w = 2.7 revolutions per minute

Converting to revolutions per hour

w = 2.7 × 60 revolutions per hour

w = 162 rev/hour

Linear speed v = angular speed × 2πr

the​ difference, in miles per​ hour, in the linear speeds of Brett and​ Will;

∆v = w × 2π(r1 - r2)

r1 = Brett radius in miles

r2 = Will radius in miles

r1 = 19ft 1in = (19×12 + 1) = 229 in

r1 = 229 × 1.57828283 × 10^-5 miles

r2 = 11 ft 11 in = (11×12 + 11) = 143 in

r2 = 143 × 1.57828283 × 10^-5 miles

Substituting the values;

∆v = 162 × 2π × (229-143)×1.57828283 × 10^-5 mph

∆v = 1.38 mph

Answer:

a) 3.92 rad/s

b) 0.373 A

d) 0.018 A

Explanation:

a) The angular frequency of rotation is given by:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{(1.6s)}=3.92\frac{rad}{s}[/tex]

b) The maximum induced current in the loop is given by:

[tex]I_{max}=\frac{emf_{max}}{R}=\frac{AB\omega}{R}[/tex]

R: resistance

A: area of the triangle loop = bh/2 = (0.34m)(0.77m)/(2) = 0.1309m^2

B: magnitude of the magnetic field

[tex]I_{max}=\frac{(0.13m^2)(1.6T)(3.92rad/s)}{2.2\Omega}=0.373A[/tex]

d) For t = 0.45s you have:

[tex]I(t)=\frac{ABcos(\omega t)}{R}\\\\I(0.45)=\frac{(0.13m^2)(1.6T)cos(3.92rad/s \ (0.45s))}{2.2\Omega}=-0.018A[/tex]

But I1 is defined to be positive if it flows in the negative y-direction.

hence, I for t=0.45 s is 0.018A

(1) The angular frequency is 3.92 rad/s

(2) the value of I(max) = 0.373A

(3) Image is required

(4) The current at 0.45s is -0.018A

Given a conducting wire formed in the shape of a right triangle with:

Base b = 34 cm

Height h = 77 cm

Resistance R = 2.2 Ω

(1) Time period of rotation is T = 1.6s

The angular frequency is given by:

ω = 2π/T

ω = 2π/1.6

ω = 3.92 rad/s

(2) The magnetic field applied is B  = 1.6T, perpendicular to the plane of the triangle.

the induced current is given by:

[tex]I_{ind}=\frac{EMF_{ind}}{R}[/tex]

where R is the resistance

[tex]I_{max}=\frac{EMF_{max}}{R}\\\\I_{max}=\frac{BA\omega}{R}\\\\I_{max}=\frac{1.6\times(1/2)\times34\times77\times3.92}{2.2}\\\\[/tex]

I(max) = 0.373A

(3) The image for the question is not attached.

(4) The instantaneous value of induced current at time t = 0.45s is given by:

[tex]I_{ind}=\frac{EMF_{ind}}{R}\\\\I(t)=\frac{BA(cos\omega t)}{R}\\\\I(t)=\frac{(1/2)\times34\times77\times1.6\timescos(3.92\imes0.45)}{2.2}[/tex]

I(t) = -0.018 A

the negative sign indicated that the current flows in the negative y-direction.

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The wavelength of the sound waves is equal to 1.714 m. Therefore, option (d) is correct.

Sound waves can transmit through liquids, and gases, as well as plasma as longitudinal waves, called compression waves. It requires a medium to travel so it can be traveled through solids in the form of both longitudinal waves and transverse waves.

Given, the distance between two speakers, d₁ = 4.52 m

The distance between two walls, d₂ = 10.2 m

The expression to calculate wavelength: pd =(n + 0.5 λ)

As there is no slit, n =0 therefore,  pd = 0.5 λ                        ...........(1)

Given, the triangle-1, the value of s₁ = 10.7 m

The value of s₂ = (4.52/2) - 2.11 = 0.15 m

The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(0.15)^2}[/tex]

s₃ = 10.7 m

Given, the triangle-2, the value of a₁ = 10.7 m

The value of a₂ = (4.52/2) + 2.11 = 4.37 m

The value of s₃ is equal to: [tex]s_3 =\sqrt{(10.7)^2+(4.37)^2}[/tex]

a₃ = 11.557 m

The constructive interference, pd =  a₃ - s₃ = 11.557 - 10.7 = 0.857 m

From the equation (1) we can determine wavelength:

pd = 0.5 λ

0.857 = 0.5 λ

λ = 1.714 m

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Answer:

The extent to which it would stretch  is [tex]\Delta L = 0.015 \ m[/tex]

Explanation:

From the question we are told that

    The initial length is  [tex]L = 1.00m[/tex]

     The area is  [tex]A = 0.500 mm^2 = \frac{0.500}{1 *10^6} = 0.500*10^6 \ m^2[/tex]

     The Young modulus of the steel is  [tex]Y = 2.0*10^{11} Pa[/tex]

     The tension   is  [tex]T =1500 N[/tex]

The Young modulus is mathematically represented as

       [tex]Y = \frac{\sigma}{e}[/tex]

Where [tex]\sigma[/tex] is the stress which is mathematically represented as

           [tex]\sigma = \frac{F}{A}[/tex]  

Substituting values

            [tex]\sigma = \frac{1500}{0.500*10^{-6}}[/tex]  

           [tex]\sigma = 3.0*10^9 N/m^2[/tex]  

And  e is the strain which is mathematically represented as

            [tex]e = \frac{\Delta L}{L }[/tex]

Where [tex]\Delta L[/tex] The extension of the steel string

Substituting these into the equation above

             [tex]Y = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]

Substituting values  

           [tex]2.0 *10^{11} = \frac{3.0*10^9}{\frac{\Delta L}{L} }[/tex]

          [tex]\Delta L = \frac{3.0*10^9 * 1}{2.0 *10^{11}}[/tex]

         [tex]\Delta L = 0.015 \ m[/tex]

Answer:[tex]f_e=16.22\ cm[/tex]

Explanation:

Given

Focal length of objective lens is [tex]f_o=89.7\ cm[/tex]

angular magnification is [tex]m_a=-5.53[/tex]

Angular magnification depends on the focal length of objective and eye piece

i.e.

[tex]m_a=-\frac{f_o}{f_e}[/tex]

[tex]f_e=-\frac{f_o}{m_a}[/tex]

[tex]f_e=-\frac{89.7}{-5.53}[/tex]

[tex]f_e=16.22\ cm[/tex]

The eyepiece should have a focal length of approximately 16.22 cm to achieve an angular magnification of magnitude 5.53.

To determine the required focal length of the eyepiece for a refracting telescope, one can use the formula for the angular magnification  M  of the telescope, which is given by:

[tex]\[ M = \frac{f_o}{f_e} \][/tex]

where  [tex]f_o[/tex] is the focal length of the objective lens and[tex]\( f_e \)[/tex] is the focal length of the eyepiece.

Given that the focal length of the objective lens [tex]\( f_o \)[/tex] is 89.7 cm, and the desired angular magnification  M  is 5.53, we can rearrange the formula to solve for the focal length of the eyepiece[tex]\( f_e \)[/tex]:

[tex]\[ f_e = \frac{f_o}{M} \][/tex]

Substituting the given values:

[tex]\[ f_e = \frac{89.7 \text{ cm}}{5.53} \][/tex]

[tex]\[ f_e \ = 16.22 \text{ cm} \][/tex]

Therefore, the eyepiece should have a focal length of approximately 16.22 cm to achieve the desired angular magnification of 5.53 when used with an objective lens of focal length 89.7 cm.

Answer:[tex]3800\ W[/tex]

Explanation:

Given

Lawn mover running for [tex]t=20\ min [/tex]

and does [tex]W=4560\times 10^3\ J[/tex]

We know Power is rate of work i.e.

[tex]P=\frac{\text{Work}}{\text{time}}[/tex]

[tex]P=\frac{4560\times 10^3}{20\times 60}[/tex]

[tex]P=3800\ W[/tex]

Thus Power output is [tex]3800\ W[/tex]

Answer:

Explanation:

The specific heat of gold is 129 J/kgC

It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

The question involves a thermodynamic analysis using heat transfer and phase change principles to determine if a mixture of solid and liquid gold will become entirely solid or remain mixed upon reaching thermal equilibrium.

To determine whether the mixture of 2.0 kg of solid gold at 1000K and 1.5 kg of liquid gold at 1336K will be entirely solid or in a mixed solid/liquid phase upon reaching thermal equilibrium, a thermodynamic analysis involving the principles of heat transfer and phase change is required. This analysis will utilize the concept of conservation of energy, accounting for the specific heats of solid and liquid gold as well as the latent heat of fusion if a phase change occurs. The calculations will involve setting the heat lost by the liquid gold equal to the heat gained by the solid gold until both reach the same temperature. If the final temperature is below the melting point of gold without all the solid gold melting, the result will be a mixture. If all the solid melts before reaching the equilibrium temperature, the final state will be all liquid. It's critical to remember that the process must comply with the melting point of gold, which is 1064K.

Answer:

a) 0.063m

b) 2.72°

c) 3151 fringes

d) 1.87*10^-6m

Explanation:

a) To find the screen distance you use the following formula:

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

D: screen distance

d: distance between slits

m: order of the fringes

λ: wavelength

By replacing the values of the parameters you obtain:

[tex]D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m[/tex]

b) The condition for dark fringes is given by:

[tex]\lambda(m+\frac{1}{2})=dsin\theta[/tex]

for the first dark fringe the angle is:

[tex]\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°[/tex]

c) the visible number of fringes is given by:

[tex]N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes[/tex]

d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:

[tex]y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m[/tex]

Answer:

z = 1.16m

Explanation:

The electric field in a point of the axis of a charged ring, and perpendicular to the plane of the ring is given by:

[tex]E_z=\frac{Qz}{(z^2+r^2)^{\frac{3}{2}}}[/tex]

z: distance to the plane of the ring

r: radius of the ring

Q: charge of the ring

you have that:

E_{z->0} = 0

E_{z->∞} = 0

To find the value of z that maximizes E you use the derivative respect to z, and equals it to zero:

[tex]\frac{dE_z}{dz}=Q[\frac{1}{(z^2+r^2)^{3/2}}+z(-\frac{3}{2})\frac{1}{(z^2+r^2)^{5/2}}(2z)]=0\\\\(z^2+r^2)^{5/2}=3z^2(z^2+r^2)^{3/2}\\\\(z^2+r^2)^2=3z^4\\\\z^4+2z^2r^2+r^4=3z^4\\\\2z^4-2z^2r^2-r^4=0\\\\z^2_{1,2}=\frac{-(-2)+-\sqrt{4-4(2)(-1)}}{2(2)}=\frac{2\pm 3.464}{4}\\\\[/tex]

you take the positive value:

[tex]z^2=\frac{2+3.464}{4}=1.366\\\\z=1.16m[/tex]

hence, the distance in which the magnitude if the electric field is maximum is 1.16m

Answer:

A massive object (like a galaxy cluster) bends the light from an object (like a quasar) that lies behind it.

Explanation:

A massive object, like a galaxy cluster, is able to deform the space-time shape as a consequence of its own gravity, so the light that it is coming from a source that is behind it in the line of sight will be bend or distorts in a way that will be magnified, making small arcs around the cluster with the image of the background object.

This technique is useful for astronomers since they make research of faraway objects (at hight redshift) that otherwise will difficult to detect with a telescope.

Answer:

The coefficient of static friction is : 0.36397

Explanation:

When we have a box on a ramp of angle [tex]\alpha[/tex] , and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.

In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:

[tex]F_g=W\,*\, sin(\alpha)\\F_g=m*g\,*\, sin(\alpha)[/tex]

Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:

[tex]N=m*g*cos(\alpha)[/tex]

and the force of static friction (f) is given as the static coefficient of friction ([tex]\mu[/tex]) times the normal N:

[tex]f=\mu *m*g*cos(\alpha)[/tex]

When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:

[tex]f=F_g\\\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)[/tex]

Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:

[tex]\mu *\,m*g*cos(\alpha)=m*g\,*\, sin(\alpha)\\\mu=\frac{sin(\alpha)}{cos(\alpha)}\\\mu = tan(\alpha)\\\mu=tan(20^o)\\\mu=0.36397[/tex]

The coefficient of static friction (μ) for the box on the 20-degree incline can be found using the tangent of the angle, resulting in μ ≈ 0.3640.

To find the value of the coefficient of static friction (μ) when the box begins to slide at a 20-degree angle, we first need to understand the relationship between the normal force (Fn), the weight of the box (W), and the angle of the incline (θ).

Since the box is sliding at an angle of 20 degrees, we can use the component of the gravitational force perpendicular to the ramp to find Fn. The normal force is given by Fn = Wcos(θ), where W = mg and g is the acceleration due to gravity (9.8 m/s²). In this case, Fn = (12 pounds)(cos(20°)) × 4.44822, converting pounds to Newtons.

At the point of sliding, the static friction force is at its maximum, and it equals the component of the weight of the box parallel to the ramp, which is Wsin(θ). Thus, Fstatic_max = Wsin(θ), and since Fstatic_max = μFn, we can solve for μ.

μ = Fstatic_max / Fn = (Wsin(θ)) / (Wcos(θ)) = tan(θ). Substituting the given values, we get μ = tan(20°) ≈ 0.3640, which is the coefficient of static friction.

Answer:

468 m

Explanation:

So the building and the point where the laser hit the water surface make a right triangle. Let's call this triangle ABC where A is at the base of the building, B is at the top of the building, and C is where the laser hits the water surface. Similarly, the submarine, the projected submarine on the surface and the point where the laser hit the surface makes a another right triangle CDE. Let D be the submarine and E is the other point.

The length CE is length AE - length AC = 284 - 234 = 50 m

We can calculate the angle ECD:

[tex]tan(\hat{ECD}) = \frac{ED}{EC} = \frac{100}{50} = 2[/tex]

[tex]\hat{ECD} = tan^{-1} 2 = 63.43^o[/tex]

This is also the angle ACB, so we can find the length AB:

[tex]tan(\hat{ACB}) = \frac{AB}{AC} = \frac{AB}{234}[/tex]

[tex]2 = \frac{AB}{234}[/tex]

[tex]AB = 2*234 = 468 m[/tex]

So the height of the building is 468m

Answer:

B

Explanation:

i had a test on this and got it correct

Answer:

Explanation:

Given

[tex]N=3680 cm^{-1}[/tex]

therefore slit spacing [tex]d=\frac{1}{N}=\frac{1}{3680}=2.717\times 10^{-4}\ cm[/tex]

since [tex]d\sin \theta =n\lambda [/tex]

for [tex]n=1[/tex]

[tex]d\sin \theta =\lambda [/tex]

Now,at [tex]\theta _1=12.9^{\circ},\Rightarrow \lambda _1=6.0657\times 10^{-7}\ m=606.57\ nm[/tex]

at [tex]\theta _2=14.2^{\circ}\Rightarrow \lambda_2=666.5\ nm[/tex]

at [tex]\theta _3=15^{\circ}\Rightarrow \lambda_3=703.21\ nm[/tex]

Answer:

Explanation:

Given that,

The capacitance of the capacitor is

C = 4 µF

Discharging through a resistor of resistance

R = 2 MΩ

Time the energy stored in the capacitor be one-third of its initial energy

i.e. u(E)_ final = ⅓u(E)_initial

U / Uo = ⅓

Energy stored in a capacitor (discharging) can be determined using

U = Uo•RC•exp(-t/RC)

U / Uo = -RC exp(-t/RC)

U / Uo = ⅓

RC = 2 × 10^6 × 4 × 10^-6 = 8s

⅓ = 8 exp( -t / 8)

Divide both side by -8

1/24 = exp(-t/8)

Take In of both sides

In(1/24) = In•exp(-t/8)

-3.1781 = -t / 8

t = -3.1781 × -8

t = 25.42 seconds

It will take 25.42 seconds for he capacitor to be ⅓ of it's initial energy

Answer:

0.631 m

6.53315 J

Explanation:

m = Mass = 2.47 kg

v = Velocity = 2.30 m/s

k = Spring constant = 32.8 N/m

A = Amplitude

In this system the energy is conserved

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{2.47\times 2.3^2}{32.8}}\\\Rightarrow A=0.631\ m[/tex]

The amplitude is 0.631 m

Mechanical energy is given by

[tex]E=\dfrac{1}{2}mv^2\\\Rightarrow E=\dfrac{1}{2}2.47\times 2.3^2\\\Rightarrow E=6.53315\ J[/tex]

The mechanical energy is 6.53315 J

Answer:

Torque = 35.60 N.m (rounded off to 3 significant figures.

Explanation:

Given details:

The mass of the rock on the left, ms = 2.25 kg

The total mass of the rocks, mp = 10.1 kg

The distance from the fulcrum to the center of the pile of rocks, rp = 0.360 m

(a) The torque produced by the pile of rock, T = F*rp = m*g*rp

Torque = 9.8*0.360*10.1 = 35.6328

Torque = 35.60 N.m (rounded off to 3 significant figures).

Final answer:

To calculate the number of loops for an inductor in a resonant circuit with a given capacitor and frequency, the resonant frequency formula can be used in conjunction with the inductance formula for a solenoid. These formulas involve calculations that may require additional information not provided in the question, such as the length of the solenoid.

Explanation:

Calculating the Required Number of Loops for the Inductor

To find the number of loops needed for a solenoid inductor to resonate with a 286 nF capacitor at 18.0 kHz, we must use the formula for the resonant frequency (f) of an LC circuit:

f = 1 / (2π√(LC))

Where:

C is the capacitance (286 nF or 286 × 10-9 F).

L is the inductance we need to calculate.

f is the resonant frequency (18.0 kHz or 18,000 Hz).

The inductance (L) of a solenoid inductor is given by:

L = (μ0× N2× A) / l

Where:

N is the number of loops.

A is the cross-sectional area of the coil (π × radius2).

l is the length of the solenoid.

μ0 is the permeability of free space (4π × 10-7 T·m/A).

By rearranging the resonant frequency formula to solve for L and then equating it to the inductance formula of a solenoid, we can determine the number of loops required for the solenoid to meet the resonance condition with the given capacitor at the specified frequency.

Answer:

0.8024

Explanation:

From the given question; we can say that the angular momentum of the system is conserved if the net torque is  is zero.

So; [tex]I_o \omega _o = I_2 \omega_2[/tex]

At the closest distance ; the friction is :

[tex]f_s = \mu_s (mg)[/tex]

According to Newton's Law:

F = ma

F = mrω²

From conservation of momentum:

[tex]I_o \omega _o = I_2 \omega_2[/tex]

[tex]\omega_2= \frac { I_o \omega _o}{ I_2 }[/tex]

[tex]\omega_2=( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })* \omega_o[/tex]

However ; since the static friction is producing the centripetal force :

[tex]\mu_s (mg) = mr \omega_2^2[/tex]

[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]

The coefficient of static friction between the bottoms of your feet and the surface of the turntable can now be calculated by using the formula :

[tex]\mu _s = \frac{ \omega^2_2 *r }{g}[/tex]

=  [tex]( \frac { I_1+ mr_o^2}{ I_1 + mr^2 })^2 * \frac{ \omega^2_o *r }{g}[/tex]

= [tex][\frac{1200+(73(6)^2)}{((1200)+(73)(3)^2)} ]^2*[\frac{(\frac{\pi}{4})^2(3)} {9.8}][/tex]

= 0.8024

The friction force between the bottom of the feet and the surface of the

turntable balance the centrifugal force due to rotation.

Reasons:

Radius of the turntable, r₀ = 6.00 m

The period of rotation of the turntable, T₀ = 8.00s

Moment of inertia of the turntable, [tex]I_t[/tex] = 1,200 kg·m²

Mass of the person, m = 73.0 kg

The point at which the feet starts to slip, r₁ = 3.00 m

Required:

The friction between the bottom of the feet and the surface of the turntable.

Solution:

The moment of inertia of the person and the turntable combined, J, can be

found by considering the person as a point mass.

Therefore;

At the rim, J₀ = [tex]I_t[/tex] + m·r² = 1,200 + 73×6² = 3828

J₀ = 3,828 kg·m²

At 3.00 m from the center, J₁ = [tex]I_t[/tex] + m·r² = 1,200 + 73×3² = 1,857

J₁ = 1,857 kg·m²

Angular momentum, L = J·ω

Whereby the angular momentum is conserved, we have;

L₀ = L₁

J₀·ω₀ = J₁·ω₁

[tex]\displaystyle \omega_1 = \mathbf{ \frac{J_0 \cdot \omega_0}{J_1}}[/tex]

Which gives;

To remain at equilibrium, the friction force, [tex]F_f[/tex] = The centrifugal force, [tex]F_c[/tex]

[tex]\displaystyle F_c = \frac{m \cdot v^2}{r} = \frac{m \cdot (\omega \cdot r)^2}{r} = m \cdot r \cdot \omega^2[/tex]

Where;

[tex]\mu_s[/tex] = The coefficient of static friction

g = Acceleration due to gravity which is approximately 9.81 m/s²

r = The radius of rotation

ω = The angular speed

Therefore;

Therefore, at 3.00 m from the center of the turntable, we have;

[tex]\displaystyle F_f = \mu_s \cdot m \cdot g = \mathbf{ m \cdot r_1 \cdot \omega_1^2}[/tex]

[tex]\displaystyle\mu_s \cdot g = r_1 \cdot \omega_1^2[/tex]

[tex]\displaystyle\mu_s = \frac{r_1 \cdot \omega_1^2}{g} = \mathbf{\frac{r_1 \cdot \left(\displaystyle \frac{J_0 \cdot \omega_0}{J_1}\right)^2}{g}}[/tex]

Which gives;

[tex]\displaystyle\mu_s =\frac{3 \times \left(\displaystyle \frac{3,828 \times \frac{2 \cdot \pi}{8} }{1,857}\right)^2}{9.81} \approx \mathbf{0.802}[/tex]

The coefficient of static friction between the bottom of the feet and the surface of the turntable, at 3.00 m from the center, [tex]\mu_s[/tex] ≈ 0.802.

Learn more about the conservation of angular momentum here:

https://brainly.com/question/11259397

The total heat supplied to the gas in the process is calculated by considering the two expansion processes separately.

The total heat supplied to the gas in the process can be calculated by considering the two expansion processes separately.

In the first expansion, which is isobaric, the gas volume doubles. Since the pressure remains constant, the work done on the gas is given by W = PΔV, where P is the pressure and ΔV is the change in volume. The work done is equal to the heat supplied to the gas.

In the second expansion, which is adiabatic, the temperature returns to its initial value. In an adiabatic process, the work done on the gas is given by W = (γ - 1)ΔU, where γ is the ratio of specific heats and ΔU is the change in internal energy. Since the temperature returns to its initial value, the change in internal energy is zero, and therefore, no heat is supplied to the gas in this process.