A 94.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 44.0 m/s. If both are initially at rest and if the ice is frictionless,

how far does the player recoil in the time it takes the puck to reach the goal 16.5 m away?

Answer:

Explanation:

On both sides of the film , the mediums have lower refractive index.

for interfering pattern from above , for constructive interference of reflected wave from both sides of the film , the condition is

2μt = ( 2n +1 ) λ / 2

μ is refractive index of film ,t is thickness of film λ is wavelength of light

n is order of fringe

for minimum thickness

n = 0

2μt =  λ / 2

t =  λ / 4μ

= 670 / 1.75 x 4

= 95.71 nm .

Answer:

The receptors that sense the Earth's magnetic field are probably located in the birds' eyes. Now, researchers at Lund University have studied different proteins in the eyes of zebra finches and discovered that one of them differs from the others: only the Cry4 protein maintains a constant level throughout the day and in different lighting conditions

Explanation:

Final answer:

The bird will be able to detect the power line at a much smaller distance than calculated due to the concentration of the magnetic field close to the wire.

Explanation:

The distance at which a migratory bird could detect the presence of a power line can be calculated using the equation for the magnetic field produced by a long straight wire. However, since the bird's sensitivity to magnetic fields is much smaller than the field produced by the power line, the real answer will be much smaller than the calculated distance. This is because power lines are typically arranged in such a way that the magnetic field they produce is concentrated close to the wire, diminishing rapidly with distance.

Answer:

a) 2.693*10^-4 C

b) 8.875*10^-5 s

c) 2.96 W

Explanation:

Given that

Inductance of the circuit, L = 4.24 mH

Capacitance of the circuit, C = 3.02 μF

Current in the circuit, I = 2.38 A

See attachment for calculations

Answer:

a) 0.269 mC

b) 0.355 ms

c) 1.39W

Explanation:

a) To find the charge off the capacitor you start by using the following expression for the charge in the capacitor:

[tex]q=Qsin(\omega t)[/tex]

next, you calculate the current I by using the derivative of q:

[tex]I=\frac{dq}{dt}=Q\omega cos(\omega t)\\\\for \ t= 0:\\\\I=Q\omega\\\\Q=\frac{I}{\omega}\\\\\omega=\frac{1}{\sqrt{LC}}\\\\Q=I\sqrt{LC}[/tex] ( 1 )

L: inductance = 4.24*10^{-3}H

C: capacitance = 3.02*10^{-6}F

I: current = 2.38 A

you replace the values of the parameters in (1):

[tex]Q=(2.38A)(\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)})=2.69*10^{-4}C=0.269mC[/tex]

b) to find the time t you use the following formula for the energy of the capacitor:

[tex]u_c=\frac{q^2}{2C}=\frac{Q^2sin^2(\omega t)}{2C}[/tex]

the maximum storage energy in the capacitor is obtained by derivating the energy:

[tex]\frac{du_c}{dt}=\frac{2\omega Q^2sin(\omega t)cos(\omega t)}{2C}=0\\\\\frac{du_c}{dt}=\frac{\omega Q^2 sin(2\omega t)}{2C}=0\\\\sin(2\omega t)=0\\\\2\omega t= 2\pi\\\\t=\frac{\pi}{\omega}=\pi\sqrt{LC}=\pi\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}F)}=3.55*10^{-4}s=0.355\ ms[/tex]

hence, the time is 0.355 ms

c) The greatest rate is obtained for duc/dt evaluated in t=0.355 ms:

[tex]\frac{du_c}{dt}=\frac{2Q^2sin(2\frac{t}{\sqrt{LC}})}{\sqrt{LC}}[/tex]

[tex]\frac{du_c}{dt}=\frac{(2.69*10^{-4}C)^2sin(2\frac{3.55*10^{-4}s}{\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}})}{2(3.02*10^{-6}C)\sqrt{(4.24*10^{-3}H)(3.02*10^{-6}C)}}=-1.39W[/tex]

Answer:

9.2 amperes

Explanation:

Ohm's law states that the voltage V across a conductor of resistance R is given by [tex]V = R I[/tex]

Here, voltage V is proportional to the current I.

For voltage, unit is volts (V)

For current, unit is amperes (A)

For resistance, unit is Ohms (Ω)

Put R = 12.5 and V = 115 in V=RI

[tex]115=12.5I\\I=\frac{115}{12.5}\\ =9.2\,\,amperes[/tex]

Answer:

a) v3 = 1 m/s

c) K3 < K1

d) p3 = p1

Explanation:

a) To solve this problem you use the conservation of the linear momentum in elastic collision.

In the first case you have:

[tex]p_i=p_f\\\\m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]

but the second block is at rest, then v2i = 0m/s:

[tex]m_1v_{1i}=m_1v_{1f}+m_2v_{2f}[/tex]

Furthermore, you can assume that the first object stops just after the collision with the second one. From this last expression you obtain the value of the second object:

[tex]v_{2f}=\frac{m_1v_{1i}}{m_2}\\\\m_2=2m_1\\\\v_{2f}=\frac{m_1(4m/s)}{2m_1}=2\ m/s[/tex]

Then, you use the conservation of momentum for the second case, in which the second objects impact the third one:

[tex]m_2v'_{2i}+m_3v_{3i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{3i}=0\\\\m_2v'_{2i}=m_2v'_{2f}+m_3v_{3f}\\\\v_{2f}=0\\\\m_2v'_{2i}=m_3v_{3f}\\\\v_{3f}=\frac{m_2v'_{2i}}{m_3}[/tex]

where again it has assumed that the second object stops, just after the impact with the third object. v'_2i = v_2f (in order to distinguish). BY using the fact m3 = 2m2 you obtain:

[tex]v_{3f}=\frac{m_2(2m/s)}{2m_2}=1\ m/s[/tex]

Then, you obtain that v3 < v2 < v1

c) The kinetic energy is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

you compute for all the three objects:

[tex]K_1=\frac{1}{2}m_1(4m/s)^2=8m_1\ m^2/s^2\\\\K_2=\frac{1}{2}m_2(2m/s)^2=\frac{1}{2}(2m_1)(4m^2/s^2)=4m_1\ m^2/s^2\\\\K_3=\frac{1}{2}m_3=(1m/s)^2=\frac{1}{2}(2m_2)(1\ m^2/s^2)=\frac{1}{2}(2(2m_1))(1 m^2/s^2)=2m_1\ m^2/s^2[/tex]

then, k3 < k2 < k1

d) For the momentum you have:

[tex]p_1=4m_1\ m/s\\\\p_2=m_2(2m/s)=(2m_1)(2m/s)=4m_1\ m/s\\\\p_3=m_3(1m/s)=(2m_2)(1m/s)=(2(2m_1))(1m/s)=4m_1\ m/s[/tex]

p1 = p2 = p3

Complete question:

A person uses 25.0 J of kinetic energy to push an object for 11.0 s How are work and power affected if the person uses the same amount of kinetic energy to push the object in less time ?

Answer:

The power will increase, and the amount of work will remain the same

Explanation:

Given;

Kinetic energy, K = 25.0 J

time of work, t = 11.0 s

Power = work / time = Energy / time

This equation shows that power is inversely proportional to time

Also, Energy is directly proportional to work (both are measured in Joules)

Since the person will use the same amount of kinetic energy to push the object in less time.

It means that energy will be constant (work done will not change) and the time will be reduced.

Power and time are inversely proportional, decrease in time means increase in power.

Thus, the power will increase, and the amount of work will remain the same

Answer: c

Explanation:

Answer:

Explanation:

Let θ be the inclination

downward acceleration on an inclined plane

= g sinθ

= 32 x sin6

a =  3.345 ft /s

a ) for knowing the speed at point B

v² = u² + 2 a s , v is final velocity , u is initial velocity , a is acceleration and s is distance travelled .

v² = 0 + 2 x 3.345 x 500

= 3345

v = 57.8 ft /s

from point B to C , the car decelerates so we shall find deceleration

v² = u² + 2 a s

0 = 3345 + 2 x a x 70      ( v becomes u here )

a = - 23.9 m /s²

net force on car during deceleration

= μmgcosθ - mg sinθ     where  μ is coefficient of static friction ,

= mg ( μcosθ -  sinθ )

deceleration = g ( μcosθ -  sinθ )

g ( μcosθ -  sinθ )  = 23.9

( μcosθ -  sinθ ) = .74

μcosθ = .74 + .104

= .8445

μ = .8445 / .9945

= .85 .

Final answer:

To calculate the maximum deceleration of a car heading down a 6° slope under different road conditions, we can use the coefficient of static friction. On dry concrete, the deceleration is approximately 2.12 m/s². On wet concrete, the deceleration is approximately 1.62 m/s². On ice, the deceleration is approximately 2.04 m/s².

Explanation:

To calculate the maximum deceleration of a car heading down a 6° slope, we need to consider the road conditions. Assuming the weight of the car is evenly distributed on all four tires, and that the tires are not allowed to slip during the deceleration, we can calculate the deceleration for different road conditions.

(a) On dry concrete, the coefficient of static friction can be calculated using the equation μs = tan(θ), where θ is the angle of the slope. In this case, the coefficient of static friction is approximately 0.105 and the maximum deceleration is approximately 2.12 m/s².

(b) On wet concrete, the coefficient of static friction is typically lower than on dry concrete. Let's assume a coefficient of 0.08. In this case, the maximum deceleration is approximately 1.62 m/s².

(c) On ice, assuming a coefficient of static friction of 0.100, the same as for shoes on ice, the maximum deceleration is approximately 2.04 m/s².

Answer:

Check the explanation

Explanation:

1) Pressure acting on the plug = Patm + P

Pressure = Patm + rho*g*h (Here h = D2)

Pressure = 101325 + 1000*9.8*7

Pressure = 169925 Pa

so, Force = PA

Force = 169925*pi*0.0152

Force = 120.1 N

Answer:

Explanation:

Given

Slope of inclination [tex]\theta =19^{\circ}[/tex]

Mass of cyclist and bicycle is [tex]m=85\ kg[/tex]

When cyclist is going up then there is two components of its weight , one is parallel to inclination and other is perpendicular to inclination

Weight of person is mg

it can be resolved in [tex]mg\cos \theta[/tex] and  [tex]mg\sin \theta [/tex]

[tex]mg\cos \theta[/tex] is perpendicular to the inclination

and [tex]mg\sin \theta[/tex] is parallel to inclination as shown in diagram

Parallel component [tex]=mg\sin \theta=85\times 9.8\times \sin 19=271.2\ N[/tex]

Normal reaction [tex]N=mg\cos \theta =85\times 9.8\times \cos 19=787.61\ N[/tex]

The normal reaction on the bicycle by the inclined slope is 787.62 N.

The parallel on the bicycle along the inclined slope is 271.2 N.

The given parameters;

The normal reaction on the bicycle by the inclined slope is calculated as; follows;

[tex]F_n = W cos \theta\\\\F_n = mg \ cos \theta \\\\F_n = 85 \times 9.8 \times cos \ (19)\\\\F_n = 787.62 \ N[/tex]

The parallel on the bicycle along the inclined slope is calculated as

[tex]F_x = Wsin\theta\\\\F_x = mg \ sin \ \theta \\\\F_x = 85 \times 9.8 \times sin(19) \\\\F_ x = 271 .2 \ N[/tex]

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Answer:

5,400 Newtons.

Explanation:

Force = 1,800 kg * 3 m/s^2 = 5,400 kg*m/s^2 = 5,400 N

Answer:

a) 423.64 KJ / mole

Explanation:

The pictures below explains it all in the calculation and i hope it helps you

The correct statements are: (b) All these technologies use radio waves, including high-frequency microwaves. (d) Microwave ovens emit in the same frequency band as some wireless Internet devices.

The correct statements that describe the various applications listed above are:

Statement a.) is incorrect because not all technologies listed use low-frequency microwaves. Statement c.) is incorrect because not all technologies listed use a combination of infrared waves and high-frequency microwaves. Statement e.) is incorrect because wireless Internet devices do not have the shortest wavelength among the technologies listed. Statement f.) is incorrect because the wavelengths of the technologies listed vary. Statement g.) is incorrect because the wavelengths of the technologies listed also vary.

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The correct answers are option (a) and option (d). All these technologies use radio waves, including low-frequency microwaves and Microwave ovens emit in the same frequency band as some wireless Internet devices.

The electromagnetic (EM) spectrum includes a wide range of frequencies, which are used in various modern technologies. Here are analyses correlating with the provided frequency ranges:

Answer:

Wavelength at which the light reflected by the film is brightest = 567.5 nm

Explanation:

We are given;

index of refraction n2 = 1.33

Thickness;(t) = 320 nm

Now the wavelength at which the light reflected by the film is brightest is gotten from the formula for path difference in critical interference as;

Path difference = (m + ½)(λ/n)

Where;

path difference = 2 x thickness = 2(320) = 640 nm

λ = Wavelength at which the light reflected by the film is brightest

n is Refractive index

m is an integer = 0,1,2,3...

Thus; at m = 0;

We have;

640 = (0 + ½)(λ/1.33)

640 = (λ/2.66)

λ = 640 x 2.66

λ = 1702.4 nm

at m = 1;

We have;

640 = (1 + ½)(λ/1.33)

640 = (3/2)(λ/1.33)

λ = 640 x 1.33 x 2/3

λ = 567.5 nm

at m = 2;

We have;

640 = (2 + ½)(λ/1.33)

640 = (5/2)(λ/1.33)

λ = 640 x 2 x 1.33/5

λ = 340.5 nm

Since we are told that the wavelength is between 400 – 690 nm.

Thus, the wavelength at which the light reflected by the film is brightest is the higher value gotten that is between 400nm and 690nm.

Thus, Wavelength at which the light reflected by the film is brightest = 567.5 nm

The light reflected by the water film will appear brightest to an observer at a wavelength of approximately 854.4 nm.

The wavelength of light in a medium is given by λn = λ/ʼn, where λ is the wavelength in vacuum and ʼn is the medium's index of refraction. In water, which has an index of refraction of n = 1.33, the range of visible wavelengths is 285 to 570 nm. When light is incident on the water film, it will reflect predominantly at the wavelengths where the film's thickness is an integer multiple of half the wavelength. The brightest reflection will occur when the film's thickness is such that the reflected wavelength is at its maximum in the range of visible light.

To find the wavelength of the brightest reflection, we can use the equation λn = 2nt, where λn is the wavelength in vacuum, n is the refractive index of the film, and t is the thickness of the film. Given the refractive index n2 = 1.33 and the thickness t = 320 nm, we can solve for λn.

Plugging in the values:

λn = 2nt = 2(1.33)(320 nm) ≈ 854.4 nm

Therefore, the wavelength of the light reflected by the film that appears brightest to an observer is approximately 854.4 nm.

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Answer:14.2 mV

Explanation:

Given

Length of steel beam [tex]L=17\ m[/tex]

Magnetic field [tex]B=30\ \muT[/tex]

speed of beam [tex]v=28\ m/s[/tex]

Now consider as beam slides it encloses an area of rectangle of

width [tex]w=17\ m and length L'=v\times t [/tex]

Area [tex]A=17v\cdot t[/tex]

and Induced EMF is [tex]V=-\frac{d(BA)}{dt}[/tex]

[tex]V=-B\frac{d(17vt)}{dt}[/tex]

[tex]V=-17Bv[/tex]

[tex]V=-17\times 30\times 10^{-6}\times 28[/tex]

[tex]V=-14.2\ mV[/tex]

Magnitude of EMF[tex]=14.2\ mV[/tex]

Answer:

Explanation:

We shall apply law of conservation of momentum to calculate the common velocity of pendulum and bullet after collision

m v = ( m + M ) V

m is mass of bullet M is mass of pendulum , v is velocity of bullet and V is their final velocity after collision.

V = m v / ( m + M )

= .030 x 185 / 3.18

= 1.745 m /s

The kinetic energy of bullet+pendulum  will be converted into potential energy.

1/2 (m+M) V² = ( m + M ) g h

h is height by which pendulum and bullet rises after collision

Putting the values

.5 x 3.18 x 1.745² = 3.18 x 9.8 x h

h = .155 m

If θ be the angle that the pendulum is making at deflected position

l - lcosθ = h , l is length of pendulum

2.85 ( 1 - cosθ) = .155

1 - cosθ = .054

cosθ = .946

θ = 19 degree

Horizontal displacement = 2.85 sin 19

= .9278 m

92.78 cm

Using the principle of conservation of momentum and kinetic energy transformation, we calculate the vertical displacement of the pendulum after the bullet embeds. This is then used to compute the horizontal displacement using the Pythagorean theorem. Finally, the angle of displacement is determined with the trigonometric principle of arcsine.

To solve this question, we will use the principle of conservation of momentum to the bullet and pendulum system. After the bullet embeds into the pendulum, they move together, and due to the conservation of momentum, the joint velocity will be lower than the initial velocity of the bullet. We can calculate this joint velocity using the formula: (Mass of bullet x bullet velocity + Mass of pendulum x pendulum velocity) / (Mass of bullet + Mass of pendulum).

Next, we calculate the kinetic energy of the bullet-pendulum system after the bullet embeds itself using the formula: 1/2 x Mass of system x Velocity of system². This kinetic energy is then transformed into potential energy as the pendulum swings upward, which can be calculated using the formula: Mass of system x gravitational acceleration (9.81 m/s²) x vertical displacement.

We can solve for vertical displacement by equating the kinetic and potential energies. The horizontal displacement can then be calculated by using the Pythagorean theorem, where the hypotenuse is the length of the pendulum string, and one side is the length of the string less the vertical displacement (calculated earlier). Using these, the horizontal displacement is calculated as the square root of (length of string² - vertical displacement²).

The angle of maximum displacement with the vertical can then be calculated using the principle of trigonometry, namely the Arcsine formula, where arcsin (vertical displacement/length of string).

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Answer:(b)

Explanation:

Optical fiber works on the principle of total internal reflection . Total internal reflection occurs when the following two conditions are met

(1) When the angle of incidence of light is greater than the critical angle and,

(2) When a ray of light travels from a optically denser medium to optically rarer medium .

We know that light bend away from normal when it travels from denser to rarer medium. If we we choose such an angle that the refraction angle is [tex]90^{\circ}[/tex] then the incident angle is called critical angle.

So, the refractive index of cladding must be must be less than the refractive index of core material.

Answer:

A town interested in installing wind turbines to generate electricity measures the speed of the wind in the town over the course of a year. They find that most of the time, the wind speed is pretty slow, while only rarely does the wind blow very fast during a storm. What would be the best measure of the central wind speed they should report to the mayor?

The median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.

Explanation:

Based on the scenario described in the question, the median is the best measure of the central wind speed they should report to the mayor because the distribution is not symmetric. The distribution of wind speeds is skewed.

Answer:

1*10^6 N/m^2

Explanation:

Coefficient of Linear Expansion for Concrete = α = 1.2 x 10^-5 (°C)^-1

Change in temperature = ΔT

ΔT= T2 - T1

ΔT = 33 - 21

ΔT = 12°C

ΔL = α * L(i) * ΔT

ΔL = (1.2 x 10^-5 (°C)^-1) * L(i) * (12°C)

ΔL = 1.44 x 10^-4

Stress = F / A

Strain = ΔL / L

Strain = (1.44*10^-4) * (L) / L

Strain = 1.44*10^-4

Y = Stress / Strain

Stress = Y * Strain

Stress = (7.00*10^9 N/m^2) * (1.44*10^-4)

Stress = 1*10^6 N/m^2

Thus, the stress in the cement on a hot day of 33° is 1*10^6 N/m^2

The speed of the train can be calculated using the formula for the Doppler effect. Using the given frequencies and the speed of sound in air, the student can calculate the speed of the train to be approximately 8.7 m/s.

To calculate the speed of the train, we can use the formula for the Doppler effect. The formula is given by:

Δf/f = v/c

Where Δf is the change in frequency, f is the frequency observed when the train is at rest, v is the speed of the train, and c is the speed of sound in air.

Using the given frequencies of 491 Hz and 472 Hz, and the speed of sound in air of 343 m/s, we can calculate the speed of the train:

v = (Δf/f) * c = (491 - 472) / 491 * 343 = 8.7 m/s

Therefore, the student finds that the speed of the train is approximately 8.7 m/s.

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Given the observed frequencies of 491 Hz when approaching and 472 Hz when receding, the calculated speed of the train is approximately 6.66 m/s. This calculation assumes the speed of sound in air is 343 m/s.

The question involves using the Doppler Effect to calculate the speed of the train. The Doppler Effect formula for a source moving towards a stationary observer is:

[tex]f'_{approach} = f \times (v + v_o)/(v - v_s)[/tex]

and when the source is moving away:

[tex]f'_{recede} = f \times (v - v_o)/(v + v_s)[/tex]

where:

Given:

First, solve for the source frequency (f) using both equations.

For the approaching train:

[tex]491 = f \times (343) / (343 - v_s)[/tex]

For the receding train:

[tex]472 = f \times (343) / (343 + v_s)[/tex]

Divide the two frequency equations to eliminate f:

(491 / 472) = (343 + [tex]v_s[/tex]) / (343 - [tex]v_s[/tex])

Cross-multiplying and solving for [tex]v_s[/tex]:

491 × (343 - [tex]v_s[/tex]) = 472 × (343 + [tex]v_s[/tex])

491 × 343 - 491 × [tex]v_s[/tex] = 472 × 343 + 472 × [tex]v_s[/tex]

168313 - 491 × [tex]v_s[/tex] = 161896 + 472 × [tex]v_s[/tex]

168313 - 161896 = 491 × [tex]v_s[/tex] + 472 × [tex]v_s[/tex]

6417 = 963 × [tex]v_s[/tex]

[tex]v_s[/tex] = 6417 / 963

[tex]v_s[/tex] ≈ 6.66 m/s

Therefore, the student calculates the speed of the train to be approximately 6.66 m/s.

Answer:

The number of molecules in the volume is  [tex]N_v = 2.27109* 10^{12}[/tex] molecules

Explanation:

From the question we are told that

    The pressure of the ultrahigh vacuum is [tex]P = 8.4*10^{-11} torr = 8.4*10^{-11} * 133 = 1.1172 *10^{-8}Pa[/tex]

     The molecular diameter of the gas molecules [tex]d = 2.2*10^{-10} m[/tex]

      The temperature is  [tex]T = 310 \ K[/tex]

      Avogadro's number is [tex]N = 6.02214 *10^{23}\ l/mol[/tex]

        The volume of the gas is [tex]V = 0.87 m^3[/tex]

From the ideal gas law[[tex]PV = nRT[/tex]] that the number of mole is mathematically represented as

           [tex]n = \frac{PV}{RT}[/tex]

Where R is the gas constant with a value  [tex]R = 8.314\ J/mol[/tex]

  Substituting values

              [tex]n = \frac{1.1172 *10^{-8} * 0.87}{8.314 * 310}[/tex]

             [tex]n = 3.771*10^{-12} \ mole[/tex]

The number of molecules is mathematically represented as

               [tex]N_v = n * N[/tex]

  Substituting values

              [tex]N_v = 3.771*10^{-12} * 6.02214 *10^{23}[/tex]

             [tex]N_v = 2.27109* 10^{12}[/tex] molecules

Answer:

The the elongated length is [tex]\Delta L = 0.4 \ mm[/tex]

The change in diameter is  [tex]\Delta d = - 0.0136\ mm[/tex]

Explanation:

From the question we are told that

   The diameter of  the cylindrical bar is [tex]d = 21.0 \ mm = \frac{21}{1000} = 0.021 \ m[/tex]

     The length of the cylindrical bar is  [tex]L= 210 \ mm = 0.21 \ m[/tex]

      The force that deformed it is [tex]F = 46800 \ N[/tex]

       Elastic modulus is  [tex]E = 60.9 \ GPa = 60.9 *10^{9}Pa[/tex]

       The Poisson's ratio is  [tex]\mu = 0.34[/tex]

Generally elastic modulus is mathematically represented as

             [tex]E = \frac{\sigma }{\epsilon}[/tex]

Where

[tex]\epsilon[/tex] is the strain which is mathematically represented  as

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

Where [tex]\Delta L[/tex] is the elongation length

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

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

Where F is the force and

 A is the area which is calculated as

               [tex]A = \frac{\pi} {4} d^2[/tex]

Substituting values

               [tex]A = \frac{\pi}{4} * (0.021)[/tex]

              [tex]A = 0.000346 \ m^2[/tex]

So the stress is

         [tex]\sigma = \frac{46800}{0.000346}[/tex]

        [tex]\sigma = 1.35 *10^{8} \ N \cdot m^2[/tex]

Thus the elastic modulus is  

        [tex]E = \frac{1.35 *10 ^{8}}{\frac{\Delta L}{L} }[/tex]

making [tex]\Delta L[/tex] the subject

       [tex]\Delta L = \frac{EL}{1.35 *10^{8}}[/tex]

Substituting values

      [tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{1.35 *10^{8}}[/tex]

      [tex]\Delta L = \frac{1.35 *10^{8} * 0.21}{60.9*10^{9}}[/tex]        

       [tex]\Delta L = 0.0004 \ m[/tex]

      Converting to mm

   [tex]\Delta L = 0.0004 * 1000[/tex]    

  [tex]\Delta L = 0.4 \ mm[/tex]    

Generally the poisson ratio is mathematically represented as

        [tex]\mu = - \frac{\frac{\Delta d }{d} }{\frac{\Delta L }{L} }[/tex]

The negative sign indicate a decrease in diameter as a result of the force

making [tex]\Delta d[/tex] the subject

        [tex]\Delta d = - \mu * \frac{\Delta L }{L } * d[/tex]

Substituting values

        [tex]\Delta d = - 0.34 * \frac{0.0004 }{0.210 } * 0.021[/tex]

       [tex]\Delta d = - 1.36 *10^{-5} \ m[/tex]

      Converting to mm      

 [tex]\Delta d = - 0.0136\ mm[/tex]

Answer:

the color is green

the observation is that there is a change of visible color

Explanation:

A) wavelength of visible light that is most strongly reflected from a point on a soap

refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ =  4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) /  3 = 1542.8 / 3 = 514.27 ( wavelength )

the color is green

B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

where m = 1

∝ = (4 * 1.33 * 340 ) / 3  = 1808.8 / 3 = 602.93 nm  ( orange color )

the observation is that there is a change of visible color

Final answer:

The wavelength of visible light most strongly reflected for a soap bubble wall thickness of 290 nm and index of refraction of 1.33 is 770 nm, corresponding to the red color. For a thickness of 340 nm, the wavelength is 904 nm, which falls outside the visible spectrum, in the infrared range. As the thickness increases, the reflected wavelength shifts toward longer wavelengths.

Explanation:

The phenomenon described in the question is known as thin film interference, which is a physical effect that occurs when light waves reflected off the top and bottom surfaces of a thin film interfere with each other. To find the wavelength (λ) of visible light that is most strongly reflected, we use the formula for constructive interference in thin films:

2nt = mλ,

where n is the index of refraction of the film, t is the thickness of the film, and m is the order of the fringe (which is an integer).

For part (a), we have a soap bubble thickness of 290 nm and an index of refraction n = 1.33. Assuming the light is perpendicular to the surface (m = 1 for the first order of constructive interference), we calculate the wavelength using the given formula as follows:

λ = 2nt/m = 2 * 1.33 * 290 nm / 1 = 770 nm.

The wavelength of 770 nm corresponds to red light in the visible spectrum.

For part (b), with a wall thickness of 340 nm, we calculate the wavelength in a similar fashion:

λ = 2 * 1.33 * 340 nm / 1 = 904 nm.

However, a wavelength of 904 nm falls outside the visible spectrum and cannot be seen as a color. It is in the infrared range.

Comparing answers from (a) and (b), we observe that as the thickness of the soap bubble increases, the wavelength of light most strongly reflected shifts towards the longer wavelengths, moving out of the visible range into the infrared.

Answer:

a) ∆x∆v = 5.78*10^-5

   ∆v = 1157.08 m/s

b) 4.32*10^{-11}

Explanation:

To solve this problem you use the Heisenberg's uncertainty principle, that is given by:

[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]

where h is the Planck's constant (6.62*10^-34 J s).

If you assume that the mass of the electron is constant you have:

[tex]\Delta x \Delta (m_ev)=m_e\Delta x\Delta v \geq \frac{\hbar}{2}[/tex]

you use the value of the mass of an electron (9.61*10^-31 kg), and the uncertainty in the position of the electron (50nm), in order to calculate ∆x∆v and ∆v:

[tex]\Delta x \Delta v\geq\frac{\hbar}{2m_e}=\frac{(1.055*10^{-34}Js)}{2(9.1*10^{-31}kg)}=5.78*10^{-5}\ m^2/s[/tex]

[tex]\Delta v\geq\frac{5.78*10^{-5}}{50*10^{-9}m}=1157.08\frac{m}{s}[/tex]

If the electron is a classical particle, the time it takes to traverse the channel is (by using the edge of the uncertainty in the velocity):

[tex]t=\frac{x}{v}=\frac{50*10^{-9}m}{1157.08m/s}=4.32*10^{-11}s[/tex]

Answer: when the wave encounters something, it can bounce (reflection) or be bent (refraction). In fact, you can "trap" waves by making them bounce back and forth between two or more surfaces. Musical instruments take advantage of this; they produce pitches by trapping sound waves.

Explanation:  Any bunch of sound waves will produce some sort of noise. But to be a tone - a sound with a particular pitch - a group of sound waves has to be very regular, all exactly the same distance apart. That's why we can talk about the frequency and wavelength of tones.

Final answer:

The human ear hears different instruments in music at once due to the superposition of sound waves and the ear's ability to process overlapping sounds, determining their pitch, loudness, and direction.

Explanation:

The human ear can hear all the different instruments in a musical piece simultaneously because of the principle of superposition of sound waves. Sound waves combine without affecting each other, so multiple sounds can reach our ears at the same time. Additionally, our ears and brain are capable of processing and interpreting these overlapping waves, enabling us to differentiate between the various sources of sound.

This is partly due to the design of the ear, which is sensitive to a wide range of frequencies, and the ear's ability to discern pitch, loudness, and direction of sound. The cochlea inside the ear uses temporal and place theories to help us perceive pitch, which allows us to enjoy the complexities of music and speech.

An impulse is the product of force and the time interval during which that force is applied, resulting in a change of an object's momentum. It can be described both mathematically (J = F⋅Δt) and visually (area under the force-time curve). An impulse leads to an object's acceleration or deceleration and affects both speed and direction.

An impulse acting on an object is a concept in physics that describes the effect of a force applied over a period of time. It is the product of the average force and the time duration during which the force acts, resulting in a change in the object's momentum. The impulse experienced by an object can result in acceleration or deceleration, dependent on the direction of the force. Moreover, impulse is not just about the magnitude of force, but also the duration over which it is applied. A key point is that an impulse can be delivered either by a large force over a short period or a smaller force over a longer period, tailored to the specifics of a situation.

Impulse is measured as the change in momentum, which is the mass of the object multiplied by its velocity (mv). The formula for impulse is typically represented as J = F⋅Δt, where J represents impulse, F the force, and Δt the change in time. If you graph force versus time, the area under the curve represents the impulse, visually demonstrating the relationship between force, time, and momentum change.

It is essential to understand that impulse not only influences the speed of an object but also its direction of motion. For example, when a tennis player hits a ball, the racket imparts an impulse to the ball changing its momentum. The total impulse given by multiple forces is considered the net impulse, which is the sum of all individual impulses over a specified time.

Answer:

Explanation:

Part A

- Work function ( ∅ ) is the minimum energy required by the photon to knock an electron out of the metal surface. That is the portion of energy of a photon transferred to an electron so that it can escape the metal.

- We can mathematically express it:

               ∅ = Ep - Ek

                    = [tex]\frac{h*c}{lamba} - 0.5*m_e*v^2_e\\[/tex]

Where,

            Planck's constant ( h ) = 6.6261*10^-34

            Speed of light ( c ) = 3*10^8 m/s

            mass of an electron ( m_e ) = 9.1094*10^-31 kg

Given:-

           Incident light's wavelength ( λ ) = 250*10^-9 m

           The maximum speed o electron ( v_e ) = 4*10^5 m/s

Solution:-

- Plug the values into the expression derived before:

               ∅ = [tex]\frac{(6.6261*10^-^3^4)*(3*10^8)}{250*10^-^9} \\\\[/tex]

               ∅ = [tex]7.22257*10^-^1^9 J * \frac{6.242*10^1^8 eV}{J}[/tex]

               ∅ = 4.508 eV  ... Answer

Part B

- 2.5% of the energy emitted by a 100-W light bulb was visible light with wavelength ( λ ) = 500*10^-9 m. In 1.0 min the amount of energy harbored by a stream of photons are:  

              [tex]E_p = n_p*\frac{h*c}{lambda} = P*t*e[/tex]

Where,

            Planck's constant ( h ) = 6.6261*10^-34

            Speed of light ( c ) = 3*10^8 m/s

Given:-

            visible light's wavelength ( λ ) = 500*10^-9 m

            Power of light bulb ( P ) = 100 W

            Time taken ( t ) = 1.0 min = 60 s

            Portion of energy as light ( e ) = 0.025

Solution:-

- Plug the values into the expression derived before:

             [tex]n_p = \frac{(lambda)*(P)*(t)*(e)}{h*c} \\\\n_p = \frac{(500*10^-^9)*(100)*(60)*(0.025)}{(6.6261*10^-^3^4)*(3*10^8)} \\[/tex]

             n_p = 3.773 * 10^20 ... Answer

Part C

- A blood vessel of radius ( r ) with length ( L ) carries blood with viscosity ( μ ). The pressure drop ( ΔP ) in the blood vessel was witnessed.

- Pressure loss ( ΔP ) in a cylindrical blood vessel is given by the Darcy's equation given below:

                    [tex]\frac{dP}{p} = f*\frac{L}{D}*\frac{v^2}{2}[/tex]

Where,

                  ρ: Density of blood

                  f: Friction factor

                  D: Diameter of vessel

                  v: Average velocity

- The friction factor is a function of Reynolds number and relative roughness of blood vessel. We will assume the blood vessel to be smooth, round and the flow to be laminar ( later verified ).

- The flow rate ( Q ) in a smooth blood vessel subjected to laminar flow conditions is given by the Poiseuille's Law. The law states:

                  [tex]Q = \frac{\pi*dP*r^4 }{8*u*L}[/tex]

- The velocity ( v ) in a circular tube is given by the following relation:

                 [tex]v = \frac{Q}{\pi*r^2 }[/tex]

Given:-

           dP ( Pressure loss ) = 2.5 Pa

           radius of vessel ( r ) = 10μm = 10*10^-6 m

           viscosity of blood ( μ ) = 0.0027 Pa.s

           Length of vessel ( L ) = 1μm = 10^-6 m

Solution:-    

- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood in the vessel:

               [tex]Q = \frac{\pi*(2.5)*(10*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]

               Q = 3.6361*10^-12 m^3 / s

- The corresponding velocity ( v ) of the blood flow would be:

               [tex]v = \frac{3.6361*10^-^1^2}{\pi*(10*10^-^6)^2 }[/tex]

              v = 0.01157 m/s

- Use the Poiseuille's Law to determine the flow rate ( Q ) of the blood blood in the enlarged vessel ( r = 12 μm = 10*10^-6 m ) :

               [tex]Q = \frac{\pi*(2.5)*(12*10^-^6)^4 }{8*(0.0027)*(10^-^6)}[/tex]

               Q = 7.54*10^-12 m^3 / s

- The corresponding velocity ( v ) of the blood flow would be:

               [tex]v = \frac{7.53982*10^-^1^2}{\pi*(12*10^-^6)^2 }[/tex]

              v = 0.01666 m/s

Part D

- A radioactive isotope of Cobalt ( Co - 60 ) undegoes Beta decay ( 0.31 MeV ) and emits two gamma rays of energy ( 1.17 & 1.33 ) MeV.

- The radioactive decay for the ( Cobalt - 60 ) can be expressed in form of an equation:

              [tex]_2_7Co ( 60 ) ---> _2_8 Ni ( 60 ) + e^- + v_e^- + gamma[/tex]

- The half life ( T_1/2 ) of the Co-60 can be used to determine the decay constant ( λ ):

             λ = [tex]lambda = \frac{Ln(2)}{T_1/2}\\[/tex]

Where,

             T_1/2 = 5.2 yrs = 1.68*10^8 s

Hence, the decay constant is

              λ = [tex]\frac{Ln ( 2 ) }{1.68*10^8}[/tex] = 4.22*10^-9 s^-1

- The activity ( A ) of any radioactive isotope is function of time ( t ) defined by negative exponential distribution:

            [tex]A = A_o*e^(^-^l^a^m^b^d^a^*^t^)[/tex]

Where,

             A_o: The initial activity ( Bq )

- The activity of the radioactive isotope Co-60 was A = 10 Ci after t = 30 months. The initial activity ( A_o ) can be determined:

            [tex]A_o= \frac{A}{e^(^-^l^a^m^b^d^a^*^t^)} \\\\A_o= \frac{(10Ci)*(3.7*10^1^0 Bq/Ci)}{e^(^-^4^.^2^2^*^1^0^-^9*^7^.^8^8^*^1^0^7^)} \\[/tex]

            A_o = 5.016 * 10^11 Bq

- The initial number of nuclei in the sample ( N_o ) is given by:

           [tex]N_o = (A_o) / (lambda)\\\\N_o = \frac{5.016*10^1^1}{4.22*10^-^9} = 1.22*10^2^2[/tex]

- The initial mass of Co-60 used as a sample can be determined:

           [tex]m_o = M_r*N_o\\\\m_o = (59.933822u)*(1.66*10^-^2^7)*(1.22*10^2^2)\\[/tex]

           m_o = 12.2 * 10^-6 kg  ... Answer

- The total energy ( E ) released from the beta decay transformation:

          E = E(β) + E(γ1) + E(γ2) = 0.31 + 1.17 + 1.33 = 2.81 MeV

- The rate at which the source emits energy after 30 months:

        P = E*A = [tex]( 2.81 MeV * \frac{1.6*10^-^1^3 J}{MeV} ) * ( 10 Ci * \frac{3.7*10^1^0 Bq}{Ci} )[/tex]

        P = 0.166 W  .. Answer

Answer:

A) i) at V_s = 5.6V

N_s = 6.53 turns

ii) at V_s = 12V, N_s = 14 turns

iii) at V_s = 480V, N_s = 560 turns

B) i) at V_s = 5.6V

I_s = 214.29A

ii) at V_s = 12V, I_s = 100A

iii) at V_s = 480V, N_s = 2.5A

Explanation:

A) The formula for calculating number of turns in the secondary coil is gotten from;

V_p/V_s = N_p/N_s

Making N_s the subject we have;

N_s =( V_s*N_p)/V_p

Where;

V_p is input voltage

V_s is output voltage

N_p is number of turns in primary coil

N_s is number of turns in secondary coil

We are given V_p = 240V and N_p = 280 turns

Thus;

i) at V_s = 5.6V,

N_s = (5.6*280)/240

N_s = 6.53 turns

ii) at V_s = 12V,

N_s = (12*280)/240

N_s = 14 turns

iii) at V_s = 480V,

N_s = (480*280)/240

N_s = 560 turns

B) The formula for calculating maximum output current in the secondary coil is gotten from;

I_s = (V_p*I_p)/V_s

Where;

I_s is maximum output current

V_p is input voltage

I_p is maximum input current

V_s is output voltage

We are given I_p = 5A

Thus;

i) at V_s = 5.6V,

I_s = (240*5)/5.6

I_s = 214.29 A

ii) at V_s = 12V

I_s = (240*5)/12

I_s = 100 A

iii) at V_s = 480V

I_s = (240*5)/480

I_s = 2.5 A

Final answer:

The number of turns in the secondary coil of the transformer to produce outputs of 5.60, 12.0, and 480 V are approximately 6.5, 14, and 560 turns respectively. The associated maximum output currents for each voltage, assuming 100% efficiency, are approximately 214.29 A for 5.60 V, 100.00 A for 12.0 V, and 2.50 A for 480 V.

Explanation:

The question involves a multipurpose transformer where the primary coil receives an input voltage of 240 V and has 280 turns, and we need to calculate the number of turns in the secondary to produce various output voltages of 5.60, 12.0, and 480 V. Secondly, we have to determine the maximum output currents for a maximum input current of 5.00 A using the transformer's power conservation principle.

(a) To find the number of turns in each part of the secondary coil, we can use the transformer equation:

VP / VS = NP / NS

For each output voltage, we can solve for NS (number of turns in the secondary):

For 12.0 V output: NS = (12.0 V / 240 V) × 280 turns = 14 turns

For 480 V output: NS = (480 V / 240 V) × 280 turns = 560 turns

(b) Power conservation in transformers (assuming 100% efficiency) indicates that PP = PS (power in primary equals power in secondary), so:

For 480 V: IS = (240 V × 5.00 A) / 480 V = 2.50 A

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is  [tex]g = 167.2 \ m/s^2[/tex]  

Explanation:

From the question we are told that

     The length of the simple pendulum is [tex]L = 1.081.08 \ m[/tex]

      The number of cycles is  [tex]N = 101[/tex]

       The time take is  [tex]t = 2.00 *10^{2 \ }s[/tex]

Generally the period of this oscillation is mathematically evaluated as

         [tex]T = \frac{N}{t }[/tex]

substituting values

         [tex]T = \frac{101}{2.0*10^2 }[/tex]

        [tex]T = 0.505 \ s[/tex]

The period of this oscillation is mathematically represented  as

               [tex]T = 2 \pi \sqrt{\frac{l}{g} }[/tex]

making g the subject of the formula we have

              [tex]g = \frac{L}{[\frac{T}{2 \pi } ]^2 }[/tex]

              [tex]g = \frac{4 \pi ^2 L }{T^2 }[/tex]

Substituting values

               [tex]g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }[/tex]

               [tex]g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }[/tex]  

              [tex]g = 167.2 \ m/s^2[/tex]  

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm