A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his ro- tation axis, how fast will he be spinning?

Answer:

Energy required = [tex]J=2.73\times10^7 \ J.[/tex]

Explanation:

We know according to zeroth law of Thermodynamics, all bodies in a close system remains in thermal equilibrium.

Means, Energy gain by one part of system = energy loss by one another part of system.

Here, energy is given to surrounding and it is lost by water.

Therefore, its temperature decreases.

Energy required to convert 82 kg liquid, E = [tex]m\times L[/tex].

Here, [tex]m=82 \ kg[/tex]

         [tex]L=larent \ heat \ of \ fusion=3.34\times 10^5J/kg.[/tex].

E=[tex]82 \times 3.34\times 10^5\ J=2.73\times10^7 \ J.[/tex]

Absorbance of a mixture solution is additive. That mean the total absorbance at a perpendicular wavelength is the sum of the absorbance of its components at that wavelength.

This can be demonstrated through the sum of the absorbance in the solutions for which it is necessary to add to each specific absorbance its respective wavelength. Mathematically this is:

[tex]A_{total,\lambda} = \sum\limit_i A_{i,\lambda}[/tex]

[tex]A_{total,\lambda} = A_{1,\lambda}+A_{2,\lambda}+A_{3,\lambda}+...[/tex]

Therefore the correct option is A.

Final answer:

The true statement for spectroscopy of a mixture is that the absorbance at a particular wavelength is the sum of the absorbances of the components that absorb at that wavelength, in line with Beer's Law.

Explanation:

When analyzing a mixtute using spectroscopy, it is true that the absorbance for a mixture at a particular wavelength is the sum of the absorbances for the components that absorb at that wavelength as stated by Beer's Law. Spectrophotometers cannot differentiate between mixture components that absorb at the same wavelength - distinction typically relies on different wavelengths where individual compounds absorb maximally. Each component in a mixture does not necessarily have the same molar absorptivity at the same wavelength, as molar absorptivity is a characteristic of the molecular structure and its interaction with specific wavelengths of light.

When an overlap in the spectra of the analytes is significant, it can be challenging to calculate their concentrations directly. In such cases, methods like multiwavelength linear regression analysis might be used to improve accuracy and precision. This involves comparing the absorbance of the mixture to that of standard solutions containing known concentrations of the analytes at multiple wavelengths.

Answer:

(a) P = 37.97 kN

(b) P = 35.62 kN

Explanation:

In the given problem, using the parameters and figure provided, we have:

A = b*h = 0.14*0.24 = 0.0336 m^2

S = (b*h^2)/6 = (0.14*0.24*0.24)/6 = 0.001344 m^3

q =  5400*0.0336 = 181.44 N/m

(a) The load P when σ (allowable bending stress) = 8.5 MPa = 8.5*10^6 Pa

σ = M/S

Thus: M = σ*S = 0.001344*8.5*10^6 = 11424 Nm

In addition,

M = (P*L/4) + (q*L^2)/8

11424 = (P*1.2/4) + (181.44*1.2^2)/8

11424 = 0.3*P + 32.6592

P = (11424-32.6592)/0.3 = 37.97 kN

(b)  The value of P if τ (allowable shear stress) = 0.8 MPa = 0.8*10^6 Pa

(2*A*τ/3) = P/2 + q*L/2

(2*0.0336*0.8*10^6)/3 = P/2 + 181.44*1.2/2

17920 = P/2 + 108.864

P = (17920 - 108.864)*2 = 35.62 kN

To calculate the maximum permissible value of the load P, we need to consider both the bending stress and the shear stress.

To calculate the maximum permissible value of the load P, we need to consider both the bending stress and the shear stress.

For (a), the maximum allowable bending stress is 8.5 MPa. Using the formula for bending stress, we can calculate the moment of inertia and then find the maximum permissible load P.

For (b), the maximum allowable shear stress is 0.8 MPa. Using the formula for shear stress, we can calculate the cross-sectional area and then find the maximum permissible load P.

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Answer:

24 hours

[tex]7\times 10^{-5}\ rad/s[/tex]

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

[tex]T=24\times 60\times 60\ s[/tex]

Angular velocity is given by

[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s[/tex]

The angular velocity of the satellite is [tex]7\times 10^{-5}\ rad/s[/tex]

A communications satellite in a geostationary orbit must have a period of 24 hours and an angular velocity of approximately 0.262 radians per hour.

To remain above the same point on Earth's surface, a communications satellite must be placed in a geostationary orbit. The period of such a satellite must be 24 hours, which is the same as one Earth day. This means that the satellite completes one orbit around the Earth in 24 hours. The angular velocity of the satellite depends on its position in the orbit and can be calculated using the formula:

Angular velocity = 2π / Period

Since the period is 24 hours, the angular velocity of the satellite is approximately 0.262 radians per hour.

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Answer:

[tex]|\vec{F}| =48.60\ N[/tex]

Explanation:

given,

Current in the power line = I = 5500 A

earth's magnetic field = 60.0 μT

inclination downward = 67°

Length = 160 m

magnetic force = ?

[tex]\vec{F} = I (\vec{L}\times \vec{B})[/tex]

[tex]|\vec{F}| = I |(\vec{L}\times \vec{B})|[/tex]

[tex]|\vec{F}| =I LB sin \theta [/tex]

[tex]|\vec{F}| = 5500 \times 160 \times 60 \times 10^{-6}\times sin 67^0[/tex]

[tex]|\vec{F}| =48.60\ N[/tex]

According to the right hand rule the direction of the force is perpendicular to the plane of the length and the magnetic field so, it is to west.

Answer:

Correct answer is (b) Revitalization movements always fail because they require too much change to be tolerated.

Example of a revitalization movement is the Ghost Dance that swept through western Native American cultures from 1870-1890.

Explanation:

Another example of Revitalization movements are mostly associated with religion. They often occur in disorganized societies due to warfare, revolution but not necessarily an animistic societies.

The incorrect statement is that 'Revitalization movements always fail because they require too much change to be tolerated.' Revitalization movements have led to significant societal changes and have even given rise to major religions like Christianity, Judaism, and Islam. The correct option is b.

The statement 'Revitalization movements always fail because they require too much change to be tolerated' is incorrect. Revitalization movements are deliberate, organized, conscious efforts by members of a society to construct a more satisfying culture. While it's true that they can face intense resistance because they often push for significant changes, it's not true that they always fail. Many have had long-lasting impacts. For example, the Protestant Reformation was a revitalization movement that sought to reform the Roman Catholic Church and led to the creation of Protestant churches. It instigated enormous change in society, and it certainly didn't fail.

All major religions, including Judaism, Christianity, and Islam, did indeed begin as revitalization movements. These movements were adapted to the changing needs of society, and they have resulted in lasting religions that are still prevalent today. Hence revitalization movements can have profound and lasting impacts on society, institutions and cultures, far from always being impractical or bound to fail.

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Answer:

B. The kinetic energy has to be greater than mgRp.

Explanation:

Hi there!

Since there is no friction and because of the conservation of energy, all the initial kinetic energy will be converted into gravitational potential energy. At a height equal to the radius of the planet, the gravitational potential energy will be:

EP = mgRp

Where:

m = mass of the projectile.

g = acceleration due to gravity.

Rp = height = Planet´s radius.

To reach that height, the initial kinetic energy has to be equal to that potential energy (remember that at the maximum height, the potential energy will be equal to the initial kinetic energy because there is no energy dissipation by heat because there is no air friction).

Then, to escape from the surface of the planet, the initial kinetic energy has to be greater than mgRp (Answer B).

To solve this problem it is necessary to apply the concepts related to the conservation of kinetic energy and potential energy.

As there is an increase in gravitational potential energy, there is a decrease in kinetic energy - which 'generates' the movement - on the particle.

Mathematically this can be expressed as

[tex]KE = PE_g[/tex]

[tex]KE = \frac{GMm}{R_p}[/tex]

Where,

G = Gravitational Universal Constant

M = Mass of Earth

m = Mass of object

R = Radius

Acceleration due to gravity we know that it is defined as

[tex]g = \frac{GM}{R_p^2}[/tex]

From the kinetic energy formula we can then re-adjust it mathematically as

[tex]KE = \frac{GMm}{R_p}[/tex]

[tex]KE = \frac{GMm}{R_p}*\frac{R_p}{R_p}[/tex]

[tex]KE = \frac{GM}{R_p^2}*(R_p)(m)[/tex]

[tex]KE = g R_p m[/tex]

Finally we can observe that the kinetic energy must be at least equivalent to [tex]mgR_p[/tex] (Correct Answer is B), in order to escape.

Answer:

h = 1.02 m

Explanation:

This is a fluid mechanics exercise, where the pressure is given by

       P = [tex]P_{atm}[/tex] + ρ g h

The gauge pressure is

      P - [tex]P_{atm}[/tex]  = ρ g h

In this case the upper part of the tube we have the atmospheric pressure. and the diver can exert a pressure 10 KPa below the outside pressure, this must be the gauge pressure

     [tex]P_{m}[/tex] =    P - [tex]P_{atm}[/tex]

     [tex]P_{m}[/tex] = ρ g h

     h =[tex]P_{m}[/tex] / ρ g

calculate

     h = 10 103 / (1000 9.8)

     h = 1.02 m

This is the depth at which man can breathe

A snorkeler can use a snorkel to go to a depth of approximately 0.099 atmospheres or 3.267 feet.

The pressure a snorkeler experiences while underwater increases with depth. To calculate the longest snorkel a person can use, we need to consider the pressure difference between the inside and outside of their body. If a snorkeler can develop a pressure in their lungs that is 10 kPa below the pressure outside their body, we can find the depth they can dive to by converting the pressure difference to atmospheres using the conversion 1 atm = 101.325 kPa. Then, we can use the relationship that every 33 feet of seawater represents 1 atmosphere of pressure in addition to the 1 atmosphere of pressure from the atmosphere at sea level. By dividing the pressure difference in atmospheres by this value, we can determine the maximum depth a snorkeler can go.

Let's calculate:

(10 kPa * 1 atm/101.325 kPa) / (33 ft * 1 atm / (33 ft))

≈ 0.099 atm

So, a snorkeler can use a snorkel to go to a depth of approximately 0.099 atmospheres or 3.267 feet.

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Answer:

option B

Explanation:

height, h = 145 m

cw = 4186 J/kg °C

g = 9.8 m/s^2

According to the conservation of energy

Potential energy = thermal energy

m x g x h = m x c x ΔT

where, ΔT is the rise in temperature

9.8 x 145 = 4186 x ΔT

ΔT = 0.34°C

Answer:[tex]\Delta T=0.339^{\circ}C[/tex]                    

Explanation:

Given

height from which water is falling [tex]h=145 m[/tex]

heat capacity of water [tex]c_w=4186 J/kg-^{\circ}C[/tex]

here Potential Energy is converted to heat the water

i.e. [tex]P.E.=mc_w\Delta T[/tex]

[tex]mgh=mc_w\Delta T[/tex]

[tex]9.8\times 145=4186\times \Delta T[/tex]

[tex]\Delta T=0.339^{\circ}C[/tex]                      

Answer:

a. [tex]m=\frac{2f_{s}tan\theta}{g}[/tex]

b. m=5.99kg

Explanation:

a.

In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).

In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any  movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.

So we start with the sum of forces about y, so we get:

[tex]\sum F_{y}=0[/tex]

N-W=0

N=W

N=mg

Next we can do the sum of forces about x, so we get:

[tex]\sum F_{x}=0[/tex]

which yields:

[tex]f_{s}-F_{W}=0[/tex]

so:

[tex]f_{s}=F_{W}[/tex]

Next the torque about point B, so we get:

[tex]\sum M_{B}=0[/tex]

so:

[tex]f_{s} L sin\theta - NLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:

[tex]f_{s} L sin\theta - WLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]

We can now combine like terms, so we get:

[tex]f_{s} L sin\theta -\frac{WL}{2}cos\theta = 0[/tex]

we know that W=mg, so we can substitute that into our equation, so we get:

[tex]f_{s} L sin\theta -\frac{mgL}{2}cos\theta = 0[/tex]

which can now be solved for the mass m:

[tex]\frac{mgL}{2}cos\theta = f_{s} L sin\theta[/tex]

If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:

[tex]\frac{mg}{2}cos\theta = f_{s}sin\theta[/tex]

we can now divide both sides of the equation into g so we get:

[tex]\frac{m}{2}cos\theta = \frac{f_{s}sin\theta}{g}[/tex]

next we can divide both sides of the equation into cos θ so we get:

[tex]\frac{m}{2}= \frac{f_{s}sin\theta}{g cos\theta}[/tex]

and finally we can multiply both sides of the equation by 2 so we get:

[tex]m=\frac{2f_{s}sin\theta}{g cos\theta}[/tex]

we know that:

[tex]tan \theta=\frac{sin \theta}{cos \theta}[/tex]

so we can simplify the equation a little more, so we get:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

b.

So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:

θ=32° and [tex]f_{s}=47N[/tex]

we also know that [tex]g=9.8m/s^{2}[/tex]

so we can use our equation now:

[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]

so we get:

[tex]m=\frac{2(47N)tan (32^{o})}{9.8m/s^{2}}[/tex]

which yields:

m=5.99kg

Answer:

a) m = 2fs(tanθ)/g

b) m = 5.99 kg

Explanation:

a) Expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.

Data

m₁ : mass of the lader

g: acceleration due to gravity

L : ladder length

θ  : angle that makes  the  ladder  with the floor

µ = 0 : coefficient of friction between the ladder and the wall

fs : friction force between the ladder and the floor

Forces acting on the ladder

W =m*g : Weight of the ladder (vertical downward) , m: mass of the lader 

FN :Normal force that the floor exerts on the ladder (vertical upward) (point A)

fs : friction force that the floor exerts on the ladder (horizontal to the left) (point A)

N : Forces that the wall exerts on the ladder (horizontal to the right)

Equilibrium  of the forces in X

∑Fx=0

N -fs = 0

N = fs

The equilibrium equation of the moments at the point contact point of the ladder with the floor:

∑MA = 0  

MA = F*d  

Where:  

∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)  

MA : moment in the point A ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point A ( m )

Calculation of the distances of the forces at the point A

d₁ = (L/2)*cosθ : Distance from W to the point A

d₂ = L*sinθ : Distance from N to the point A

Equilibrium of the moments at the point A

∑MA = 0  

N(d₂)-W(d₁) = 0

W( (L/2)*cosθ)= N(L*sinθ )

mg( (L/2)*cosθ)= fs(L*sinθ )

We divided by L both sides of the equation

mg (cosθ/2) =fs(sinθ)

m=2fs(sinθ)/ g( cosθ)

m = 2fs(tanθ)/ g

b) If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?

m = 2fs(tanθ)/ g

m = 2(47)(tan32°)/(9,8)

m = 5.99 kg

Answer:

The ice is cooler than the table.

Explanation:

The difference in temperature is a necessary condition for the transfer of heat between two bodies.

Here if the heat is being transferred from the table to an ice block this means that the temperature of the table is greater than the ice which drives the heat energy from the table to the ice block and the heat will continue to flow until they both attain a common temperature.

However,

Final answer:

Heat flow from a table to a block of ice enables the ice to undergo a phase change from solid to liquid, known as the latent heat of fusion, without a change in temperature .

Explanation:

If heat is flowing from a table to a block of ice moving across the table, which of the following must be true? The correct answer from the choices provided is that the ice is changing phase. Heat flow from the table to the ice suggests that the ice is absorbing energy. When ice absorbs heat, it undergoes a phase change, where the ice melts and becomes liquid water without any change in temperature. This process is known as the latent heat of fusion, where heat is transferred to cause a phase change without altering the temperature of the system. Therefore, while the table may be rough and friction may exist, and the ice might indeed be cooler than the table, the guaranteed phenomenon occurring is a phase change from solid to liquid as the ice absorbs heat.

The smallest possible value for the total energy of an electron with l=5 in a hydrogen atom is -0.37778 eV.

The smallest possible value for the total energy of an electron in a hydrogen atom can be found by using the energy formula for Bohr's model. The energy is given by:

E = -13.6 eV / n^2

where n is the principal quantum number. In this case, since l=5, the smallest possible value for the total energy is when n=6. Plugging this value into the energy formula, we get:

E = -13.6 eV / 6^2 = -0.37778 eV

Answer:

The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

[tex]\phi=BA[/tex]

[tex]\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2[/tex]

Put the value into the formula

[tex]\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2[/tex]

[tex]\phi=11.486\times10^{-8}\ Wb[/tex]

Hence, The average magnetic flux through each turn of the inner solenoid is [tex]11.486\times10^{-8}\ Wb[/tex]

The common angular speed is 9090.91 rad/h.

To find the common angular speed, we need to first find the initial and final radii of the full reel. The initial radius is 35 mm and the final radius is 11 mm. The angular speed can be found using the equation v = rω, where v is the linear speed, r is the radius, and ω is the angular speed. We can solve for ω by dividing the linear speed by the radius. Since both reels will have the same angular speed, we can use the same equation to find the angular speed when the tape has unspooled a distance of 191 m.

First, let's convert the radii from millimeters to meters:

Next, we'll find the linear speeds at the initial and final radii. Since the length of the tape is 191 m and it plays for 1.9 hours, the linear speed can be calculated by dividing the length of the tape by the playtime:

Finally, we can find the common angular speed by dividing the linear speed by the radius:

Therefore, the common angular speed is 9090.91 rad/h.

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Answer:

a.  μ[tex]_{95%} = [/tex] 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Explanation:

a. The population mean can be determined using a confidence interval which is made up of a point estimate from a given sample and the calculation error margin. Thus:

μ[tex]_{95%} = x_[/tex]±(t*s)/sqrt(n)

where:

μ[tex]_{95%} = [/tex] = is the 95% confidence interval estimate

x_ = mean of the sample = 3

s = standard deviation of the sample = 5.8

n = size of the sample = 41

t = the t statistic for 95% confidence and 40 (n-1) degrees of freedom = 2.021

substituting all the variable, we have:

μ[tex]_{95%} = [/tex] 3 ± (2.021*5.8)/sqrt(41) = 3 ± 1.8 = [1.2,4.8]

b. The correct answer is option D. No, because the sample size is large enough.

Using the the Central Limit Theorem which states that regardless of the distribution shape of the underlying population, a sampling distribution of size which is ≥ 30 is normally distributed.

Answer:

at n= 3     λ = 656 nm

at n= 2     λ = 121.58 nm

Explanation:

Given details

transition of hydrogen atom from n = 2 to n = 3 state

Difference in energy between n = 3 state and n = 2 state :

      [tex]= 2.18*10^{-18} \times [1/4 - 1/9] J = 3.03 \times 10^{-19} J[/tex]

    so, energy of photon is given as [tex]=\frac{h\timesc}{\lambda} [/tex]

[tex]E = 3.03*10^{-19}

[/tex] So solve for  wavelength

so, λ [tex]= \frac{6.626*10^{-34}*3*10^8}{3.03*10^{-19}}m[/tex]

   =[tex] 6.56*10^{-7} m[/tex]

   = 656 nm

for second transition,

energy transmitted is given asΔE [tex]=\frac{h\times c}{\lambda}[/tex]

and it is calculated as [tex]= 2.18*10^{-18}*[1/1 -1/4] J[/tex]

E = 1.635*10-18 J  solving for wavelength in ENERGY equation we get

so, [tex]\lambda' = 1.2158*10^{-7}[/tex] m

      = 121.58 nm

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or  it is filled with water , in both the cases pressure centre will lie at the centre of the car .

The net force acting on the door (normal to its surface) and the location of the pressure should be  [tex]2.01356 \times 10^5 N[/tex]

The position of center of mass of door from the surface of water should be

[tex]= 10 + 1.1 \div 2[/tex]

= 10.55 m

Now

Pressure on center of mass

= atmospheric pressure + pressure due to water column

[tex]= 10^5 + 10.55 \times 1000 \times 9.8\\\\= 2.0339 \times 10^5 Pa[/tex]

Now

the net force acting on the door (normal to its surface)

[tex]= .9 \times 1.1 \times 2.0339 \times 10^5\\\\= 2.01356 \times 10^5 N[/tex]

So,

pressure centre will be at 10.55 m below the surface.

Therefore,

When the car is filled with air or  it is filled with water , in both the cases pressure center should be lie at the centre of the car .

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1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

[tex]a = -\omega^2 A[/tex]

Where,

a = Acceleration

A = Amplitude

[tex]\omega[/tex]= Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

[tex]a = -g[/tex]

[tex]-\omega^2 A = -g[/tex]

[tex]\omega = \sqrt{\frac{g}{A}}[/tex]

From the definition of frequency and angular velocity we have to

[tex]\omega = 2\pi f[/tex]

[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}[/tex]

[tex]f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}[/tex]

[tex]f = 2.5Hz[/tex]

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

Final answer:

The penny loses contact with the piston at the highest point in the cycle, and the maximum frequency for which the penny remains in place is at its lowest point.

Explanation:

In simple harmonic motion, the penny loses contact with the piston at the highest point in the cycle. This is because at the highest point, the acceleration due to gravity is greater than the restoring force provided by the piston. Therefore, the correct answer is C. highest point.

The maximum frequency for which the penny just barely remains in place for the full cycle is when the acceleration due to gravity is equal to the restoring force provided by the piston. At this frequency, the penny is in equilibrium and does not leave the surface. Therefore, the maximum frequency is when the penny is at its lowest point in the cycle. So, the correct answer is D. lowest point.

Answer:

Induced emf in the coil, [tex]\epsilon=4.05\ volts[/tex]

Explanation:

Given that.

Number of turns in the coil, N = 200

Side of square, d = 18 cm = 0.18 m

The field changes linearly from 0 to 0.50 T in 0.80 s.

To find,

The magnitude of the induced emf in the coil while the field is changing.

Solution,

We know that due to change in the magnetic field, an emf gets induced in the coil. The formula of induced emf is given by :

[tex]\epsilon=\dfrac{d\phi}{dt}[/tex]

[tex]\phi[/tex] = magnetic flux

[tex]\epsilon=-\dfrac{d(NBA)}{dt}[/tex]

A is the ares of square

[tex]\epsilon=AN\dfrac{d(B)}{dt}[/tex]

[tex]\epsilon=AN\dfrac{B_f-B_i}{t}[/tex]

[tex]\epsilon=(0.18)^2\times 200 \times \dfrac{0.5-0}{0.8}[/tex]

[tex]\epsilon=4.05\ volts[/tex]

So, the induced emf in the coil is 4.05 volts. Hence, this is the required solution.

To calculate the induced emf in a 200-turn coil with a changing magnetic field, use Faraday's Law of Induction. The magnitude of the induced emf is 4.05 V. This result is obtained through calculations involving the number of turns, area of each turn, and the rate of change of the magnetic field.

The problem involves calculating the induced emf in a coil with 200 turns of wire, where each turn is a square of side d = 18 cm (or 0.18 m). The magnetic field (B) changes linearly from 0 to 0.50 T over a time interval of 0.80 s.

We use Faraday's Law of Induction, which states that the induced emf (ε) in a coil is given by:

ε = -N(dΦ/dt)

where:

The magnetic flux (Φ) through one turn is calculated as:

Φ = B × A

where A is the area of one turn of the coil:

A = d × d = 0.18 m × 0.18 m = 0.0324 m²

Since the magnetic field changes linearly, the rate of change of the magnetic field (dB/dt) is:

dB/dt = (0.50 T - 0 T) / 0.80 s = 0.625 T/s

Now, we can calculate the rate of change of the magnetic flux (dΦ/dt):

dΦ/dt = A × dB/dt = 0.0324 m² × 0.625 T/s = 0.02025 Wb/s

Finally, we use Faraday's Law to find the induced emf:

ε = -N(dΦ/dt) = -200 (0.02025 Wb/s) = -4.05 V

The magnitude of the induced emf in the coil while the field is changing is 4.05 V.

Complete Question:

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to

Answer:

t= 16.7 sec.

Explanation:

As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:

γ = (ωf -ω₀) / t

If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:

γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².

When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:

γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec

Answer:

a) dh/dt = -44.56*10⁻⁴ cm/s

b) dr/dt = -17.82*10⁻⁴ cm/s

Explanation:

Given:

Q = dV/dt = -35 cm³/s

R = 1.00 m

H = 2.50 m

if h = 125 cm

a) dh/dt = ?

b) dr/dt = ?

We know that

V = π*r²*h/3

and

tan ∅ = H/R = 2.5m / 1m = 2.5  ⇒ h/r = 2.5

⇒  h = (5/2)*r

⇒  r = (2/5)*h

If we apply

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = 3*35/π = 105/π   ⇒   d(r²*h)/dt = -105/π

a) if   r = (2/5)*h

⇒  d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π

⇒  (4/25)(3*h²)(dh/dt) = -105/π

⇒  dh/dt = -875/(4π*h²)

b) if  h = (5/2)*r

Q = dV/dt = -35 = d(π*r²*h/3)*dt

⇒  d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π

⇒  (5/2)*(3*r²)(dr/dt) = -105/π

⇒  dr/dt = -14/(π*r²)

Now, using h = 125 cm

dh/dt = -875/(4π*h²) = -875/(4π*(125)²)

⇒  dh/dt = -44.56*10⁻⁴ cm/s

then

h = 125 cm  ⇒  r = (2/5)*h = (2/5)*(125 cm)

⇒  r = 50 cm

⇒  dr/dt = -14/(π*r²) = - 14/(π*(50)²)

⇒  dr/dt = -17.82*10⁻⁴ cm/s

The rate at which the depth of the water is changing is -0.178 cm/sec, and the rate at which the radius of the top of the water is changing is -0.14 cm/sec.

The subject of this question is related to Calculus and the specific topic is related rates. For this problem, in addition to the rate at which water is leaking, we need to consider the geometric relationship within the cone-shaped water tank.

First, we are given that V' = -35 cm^3/sec (we make it negative because the volume is decreasing) and we know that the volume of a cone is V = (1/3)πr²h. Our tank parameters are r = 1m and h = 2.5m, but we need everything in the same units, so we convert our radius to 100 cm.

We know through similar triangles that r/h = R/H, where r and R are the radii at any given time, and h and H are the heights at any given time respectively. Thus, r = (Rh)/H. Substituting r in our volume equation, we get: V = (1/3)π((Rh)^2)/H=h²πR²/H, and hence V=hπR². Differentiating this implicitly with t gives V' = h'πR²+2hπRr'.

Substituting for V', h and r from our given information, we get: -35=h'πR²+2(1.25)πRr'. We can solve for h' and r' separately.

(a) To find h' we set r' = 0, because we only want to know how depth h is changing. Solving for h' we find it to be -0.178 cm/sec.

(b) To find r', we set h' = 0, because we are only interested in how radius r is changing. Solving for r' we get -0.14 cm/sec.

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Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

Explanation:

This resonant system can be simulated by a system with a closed end, the tile wall and an open end where it is being sung

In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

With 3 nodes       λ₃ = 4L / 5

The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

         f = v / λ

         f = v  n / 4L

Let's consider each length independently

L = 0.75 m

        f₁ = 340 1/4 0.75 = 113.33 n

         f₁ = 113.33 Hz

        f₃ = 113.33   3

       f₃ = 340 Hz

L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

       f₃ = 56.67 3

       f₃ = 170 Hz

The lowest two frequencies of resonance for the shower enclosure are 229 Hz and 114.33 Hz. The second harmonics for these lengths would be 458 Hz and 228.67 Hz respectively.

The lowest two frequencies that correspond to resonances in your shower enclosure, which is 0.75 m wide and 1.5 m long, can be found using the formula for the fundamental frequency of a standing wave: f = v/2L, where v is the speed of sound in the air (~343 m/s), and L is the length or width of the enclosure, respectively.

For the axis parallel to the width (0.75 m), the lowest frequency is f = 343/(2*0.75) = 229 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*229 = 458 Hz.

For the axis parallel to the length (1.5 m), the lowest frequency is f = 343/(2*1.5) = 114.33 Hz. For the second harmonic along this axis, the frequency would be twice the fundamental, so 2*114.33 = 228.67 Hz.

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Answer:

The minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm.

Explanation:

The diameter of person's eye is 2.5 cm. The close point or the near point of the eye is 25 cm and the far point is infinity. We need to determine the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye. Let f is the minimum effective focal length. It can be calculated using lens formula as :

[tex]\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{2.5}[/tex]

f = 2.27 cm

So, the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm and it is at the nearest point. Hence, this is the required solution.

Answer:

The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.

We will use conversation of energy.

[tex]K_A_1 + U_A_1 = K_A_2 + U_A_2\\\frac{1}{2}I\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_A[/tex]

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.

For the ball B;

[tex]K_B_1 + U_B_1 = K_B_2 + U_B_2[/tex]

[tex]\frac{1}{2}I_B\omega^2 + \frac{1}{2}mv^2 + 0 = 0 + mgH_B[/tex]

The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.

Hence, both balls climb the same point.

Explanation:

Both balls A and B will climb up to the same height according to the conservation of energy.

Since the balls are identical, let both have mass m and velocity v.

Also the moment of inertia I and the angular speed ω will be the same.

Ball A encounters a frictionless ramp:

On a frictionless ramp, the ball slides down the ramp since it cannot roll as there is no friction present. Since there is o frictional force, there is no dissipation of energy. Energy is conserved.

According to the law of conservation of energy, the total energy of the system must be conserved.

KE(initial) + PE(initial) = KE(final) + PE(final)

[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_A\\\\H_A=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2][/tex]

initial KE has rotational and translational kinetic energy and the initial PE is zero since the ball is on the ground, also the final KE is zero since the velocity at the highest point will be zero.

Ball B encounters a regular ramp:

On a regular ramp, the ball can roll without sliding due to friction.

[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2+0=0+mgH_B\\\\H_B=\frac{1}{mg} [\frac{1}{2}I\omega^2+\frac{1}{2}mv^2][/tex]

The height will be the same since the velocity is the same as ball A.

Also, the translational or rotational velocity will be zero at the highest point.

Hence, both balls climb the same height.

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To solve this problem we will use the concept of the Doppler effect applied to the speed of blood, the speed of sound in the blood and the original frequency. This relationship will also be extrapolated to the frequency given by the detector and measured the change in frequencies through the beat frequency. So:

[tex]f_{blood} = f (1-\frac{v_{blood}}{v_{snd}})[/tex]

Where

[tex]f_{blood}[/tex] = Frequency of the blood flow

f = Frequency of the original signal

[tex]v_{blood}[/tex] = Speed of the blood flow

[tex]v_{snd}[/tex] = Speed of sound in blood

[tex]f''_{detector} = \frac{f_{blood}}{(1+\frac{v_{blood}}{v_{snd}})}[/tex]

[tex]f''_{detector} = f (\frac{(1-\frac{v_{blood}}{v_{snd}})}{(1+\frac{v_{blood}}{v_{snd}})})[/tex]

[tex]f''_{detector} = f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

Now calculating the beat frequency is

[tex]\Delta f = f-f''_{detector}[/tex]

Replacing this latest value we have that,

[tex]\Delta f = f-f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]

[tex]\Delta f = f \frac{2v_{blood}}{v_{snd}+v_{blood}}[/tex]

Replacing we have,

[tex]\Delta f = (3.5*10^6)(\frac{2*(3*10^{-2})}{1.54*10^3+3*10^{-2}})[/tex]

[tex]\Delta f = 136.36Hz[/tex]

Therefore the beat frequency is 136.36Hz

Using the beat frequency relation, the expected beat frequency observed at the flow meter would be 136.36 Hz

Given the Parameters :

The expected beat frequency observed can be calculated uisng the relation :

Substituting the values into the formula :

[tex] \delta F = 3.5 \times 10^{3} \frac{2 \times 0.03}{(1540 + 0.03}[/tex]

[tex] \delta F = 3.5 \times 10^{3} \frac{0.06}{(1540.03}[/tex]

[tex] \delta F = 136.36 Hz [/tex]

Therefore, the expected beat frequency observed at the flow meter will be 136.36 Hz

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Answer:C

Explanation:

There are two types of cells in the eyes rod and cone cells. Rod cells Provide vision during the night or dim light also called of scotopic vision whereas cone cells provide vision during day time or at bright light also called photopic vision.  

Rod cells do not support the color vision that is why it is difficult to differentiate between colors.

Also, the amount of light entering the eyes is low as result cone cells are unable to stimulate.

It is hard to see color at night because cones, which are responsible for color vision, need more light to be stimulated than rods, which allow us to see in low light but only in grayscale.

The best explanation for why it is difficult to discriminate the color of an object at night is option C. At night, the amount of light entering the eye is insufficient to stimulate the cone cells but is sufficient to stimulate the rod cells. The rods are highly sensitive to light, allowing us to see in low light conditions but do not provide color information. Cones require brighter light to function and are responsible for our color vision. Since cones do not react to low-intensity light, our vision at night is predominantly in shades of gray, serviced by the activity of rods.

Answer:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop

Explanation:

The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines.

Mathematically:

[tex]\phi_B=\vec{B}.\vec{A}[/tex]

where:

[tex]\vec{A}=[/tex] area vector directed normal to the surface

[tex]\vec{B}=[/tex] magnetic field vector

Final answer:

Rotating the conducting loop about an axis in the z direction will not change the magnetic flux through the loop because the angle between the magnetic field and the normal to the plane of the loop remains constant. so the correct option is D

Explanation:

The question pertains to the change in magnetic flux through a conducting loop when subjected to different actions. According to Faraday's Law of Electromagnetic Induction, the magnetic flux through a loop is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field direction and the normal to the loop. In options A, B, and C, changes to the area or the strength of the magnetic field alter the magnetic flux since these factors directly affect the calculation of flux. Option D involves rotating the loop about an axis in the z direction; such a rotation does not change the angle between the magnetic field and the normal to the loop's plane, therefore would not change the flux. In contrast, option E, where the loop is rotated about an axis in the y direction, changes this angle and thus the flux.

Therefore, the action that will not change the magnetic flux through the loop is:

D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop.

Answer:

the demonstrations are in the description.

Explanation:

First of all, reference is made to a picture that may be the one shown attached.

when the string hits the pin, the sphere will continue to swing to the right until a height h'.

a) Initially, the system has only potential energy because it is released with zero velocity from a height h.

if we consider that the system has no energy losses, then all that potential energy must be conserved. Thus, this energy is transformed in principle into kinetic energy and once the kinetic energy is maximum, it will be transformed again into potential energy.

The potential energy is given by:

[tex]E_p = mgh[/tex]

As we can see, potential energy only depends on mass, height and gravity. Since mass and gravity are constant, then for potential energy to be conserved, heights must be the same.

b) the statement probably has a transcription error in question b. Corrected, it would look like:

Show that if the pendulum is released from the horizontal position (∡ = 90) and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5.

podemos asumir que la altura cero está dada en el punto de reposo del péndulo, por lo cual la energía inicial (que es potencial) está dada por:

[tex]E_0=MGL[/tex]

For the pendulum to swing, the tension must always be positive, and as a consequence, the centripetal force must be greater or at least equal to the weight. Mathematically

[tex]mV^2 /(L-d) = mg[/tex]  (1)

Where:

[tex]mV^2 /(L-d) [/tex] It is the centripetal force that the sphere experiences.

Now, for the sphere to describe the circular motion, the minimum kinetic energy must be:

[tex]\frac{mV^2}{2} = mgL-2mg(L-d)[/tex]   (2)

Then, replacing equation 1 in 2:

[tex]\frac{mg(L-d)}{2} = mgL-2mg(L-d)[/tex]

Simplifying:

[tex]L-2L+2d= \frac{L}{2} - \frac{d}{2}[/tex]

[tex]\frac{5}{2}d= \frac{3}{2}L[/tex]

Which means

[tex]d=\frac{3L}{5}[/tex]

Final answer:

When a pendulum hits a peg, it will return to its original height if released from below the peg. The minimum value of d, in order for a pendulum to swing in a complete circle centered on a peg, must be 3L/5.

Explanation:

A pendulum comprising a light string and a small sphere swings in the vertical plane. When the string hits a peg located below the point of suspension, the sphere will return to its original height if it is released from a height below that of the peg. The reason for this is that when the string hits the peg, the tension in the string abruptly changes direction, causing the sphere to swing back upwards.

If the pendulum is released from rest at the horizontal position and is to swing in a complete circle centered on the peg, the minimum value of d must be 3L/5. This is because when the pendulum is at the lowest point, the tension in the string must be equal to or greater than the weight of the sphere in order to maintain circular motion. The minimum value of d ensures that this condition is satisfied.

Answer:

If the period of the wave is increased by the factor of 2.70, the wavelength of the wave is also increased by a factor of 2.70. So,

[tex]\lambda_2 = 235\times 2.70 = 634.5 ~nm[/tex]

The magnetic field component can be written as

[tex]\vec{B} = \frac{E_{max}}{c}e^{i(\vec{k}\vec{z}-\omega t)}\^{z}[/tex]

The magnetic field is in the z-direction, because the E-field is directed towards +y and the wave is propagating in the +x-direction. The right-hand rule gives us the direction of the B-field.

[tex]\vec{E} \times \vec{B} = \vec{S}[/tex]

S is the Poynting vector which gives us the propagation of the wave.

We will use the following relationships

[tex]k = 2\pi / \lambda\\f = \omega / 2\pi\\c = \lambda f = \lambda \omega / 2\pi\\\omega = 2\pi c/\lambda[/tex]

[tex]\vec{B} = \frac{7.7\times 10^{-3}}{3\times 10^8}e^{(\frac{2\pi}{3\times 10^8}z - \frac{2\pi\times 3\times 10^8}{634.5})}\\\vec{B} = 2.56\times10^{-11} e^{(2.09\times10^{-8}z - 2.96\times10^{6}t)}\^{z}[/tex]

To find the resulting magnetic field equation of an electromagnetic wave after the period is increased, we use relationships between the wave's wavelength, frequency, period, and electric and magnetic fields, applying the changes to the wave's properties based on Maxwell's theory of electromagnetism.

The question is asking for the equation of the resulting magnetic field component of an electromagnetic wave after its period has been increased by a factor of 2.70. Given the initial wavelength λ of 235 μm (micrometers) and the maximum electric field Emax in the +y direction of 7.70×10⁻³ V/m, we can find the initial frequency ν using the relationship ν = c / λ, where c is the speed of light. Since the period T is the reciprocal of the frequency (ν = 1/T), when the period is increased by a factor of 2.70, the frequency decreases by the same factor. The maximum magnetic field Bmax is related to the maximum electric field Emax by Bmax = Emax / c. The resulting magnetic field B can be described by the function B(x, t) = Bmax sin(kx - ωt + φ), where k is the wave number, ω is the angular frequency (2πν), and φ is the phase constant.

Answer:

d. 37°C

Explanation:

When Equilibrium is reached,

Heat lost = heat gain.

Heat lost by the hot metal = c₁m₁(T₁-T₂)............. equation 1

Where c₁ = specific heat capacity of the metal = 0.25 cal/g⋅°C, m₁ = mass of the metal = 250 g, T₁ = initial Temperature of the metal = 70°C, T₂ final temperature of the metal .

Substituting this values into equation 1,

Heat lost by the metal = 62.5(70-T₂)

Also,

Heat gain by the water = c₂m₂(T₂-T₁)................. equation 2

c₂ = 1.00 cal/g⋅°C., m₂ = 75 g, T₁ = 20°C

Substituting this values into equation 2,

Heat gain by water = 1× 75 (T₂ - 20)

Heat gained by water = 75(T₂ - 20)

Also,

Heat gained by the calorimeter = c₃m₃(T₂-T₁)............. equation 3

Where c₃ = 0.10 cal/ g⋅°C, m =500 g, T₁ =20°C.

Substituting this values into equation 3

Heat gained by the calorimeter = 0.1 × 500(T₂ - 20)

Heat gained by the calorimeter = 50(T₂ - 20)

Heat lost by the metal = heat gained by water + heat gained by the calorimeter.

62.5(70-T₂) = 50(T₂ - 20) + 75(T₂ - 20)

4375 - 62.5T₂ = 50T₂ - 1000 + 75T₂ - 1500

Collecting like terms,

-62.5T₂ - 50T₂ - 75T₂ =  - 1000 - 1500 - 4375

  -187.5T₂ = -6875

Dividing both side by the coefficient of T₂

-187.5T₂ / -187.5 = -6875 /-187.5

 T₂ = 36.666°C

T₂ ≈ 37°C

The final temperature = 37°C