According to Little's Law, which of the following can reduce the number of skiers waiting in line to ride a chairlift at a ski resort? a. Using a process layout b. Decreasing the chairlift's bandwidth c. Increasing the chairlift's throughput rate d. Increasing the skiers' average flow time e. Decreasing the skiers' average flow time

Answer:

Force in the rocket will be 4166.64 N

Explanation:

We have given mass of the rocket m = 3 kg

Rocket acquires a speed of 50 m sec so final speed v = 50 m/sec

Initial speed u = 0 m/sec

Distance traveled s = 90 cm = 0.9 m

From third equation of motion we know that [tex]v^2=u^2+2as[/tex]

[tex]50^2=0^2+2\times a\times 0.9[/tex]

[tex]a=1388.88m/sec^2[/tex]

From newton's law we F = ma

So force will be [tex]F=1388.88\times 3=4166.64N[/tex]

So force in the rocket will be 4166.64 N

The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The average force that acted on the rocket is 166.8 Newtons.

The average force that acted on the rocket can be calculated using Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the rocket is 3.00 kg and the acceleration can be determined by dividing the change in velocity by the distance traveled. The change in velocity is 50.0 m/s and the distance traveled is 90.0 cm, which is equivalent to 0.90 m.

So, the acceleration of the rocket is 50.0 m/s divided by 0.90 m, which gives us 55.6 m/s². Now, we can use Newton's second law to calculate the force: F = m * a, where F is the force, m is the mass, and a is the acceleration.

Plugging in the values, we get F = 3.00 kg * 55.6 m/s² = 166.8 N. Therefore, the average force that acted on the rocket is 166.8 Newtons.

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Answer:

 x = 6.58 m

Explanation:

For this exercise let's use the concept of conservation of mechanical energy. Let's look for energy at two points the highest and when the spring is compressed

Initial

    Em₀ = U = m g and

Final

    [tex]E_{mf}[/tex] = ke = ½ k x2

As there is no friction the mechanical energy is conserved

    Em₀ = [tex]E_{mf}[/tex]

     m g y = ½ k x²

Let's use trigonometry to face height

     sin θ = y / L

     y = L sin θ

     x = √ 2mg (L synth) / k

     x = √ (2 11.8 9.8 2.17 sin35.6 / 3.11 104)

      x = 6.58 m

Final answer:

The new pressure drop p2 will be equal to the original pressure drop p1 at the original mass flow rate.

Explanation:

The relation between the new pressure drop p2 and the original pressure drop p1 at the original mass flow rate can be determined using Poiseuille's equation. Poiseuille's equation states that the pressure drop p2 - p1 is equal to the product of the resistance R and the flow rate Q.

When the length of the pipe is doubled, the resistance R remains constant because the friction factor is constant at high Reynolds numbers. Therefore, the new pressure drop p2 will be equal to the original pressure drop p1.

Answer:

Torque on the rocket will be 1.11475 N -m

Explanation:

We have given that muscles generate a force of 45.5 N

So force F = 45.5 N

This force acts on the is acting on the effective lever arm of 2.45 cm

So length of the lever arm d = 2.45 cm = 0.0245 m

We have to find torque

We know that torque is given by [tex]\tau =F\times d=45.5\times 0.0245=1.11475N-m[/tex]

So torque on the rocket will be 1.11475 N -m

Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, [tex]\lambda = 652\ nm = 652\times 10^{- 9}\ m[/tex]

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

[tex]R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}[/tex]

[tex]R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm[/tex]

The problem presents a collision scenario where principles of momentum conservation and kinetic energy are applied to determine the final velocity of a striking puck and the fraction of kinetic energy lost during the collision.

The question describes a collision scenario in physics involving two pucks, where conservation of momentum and kinetic energy principles are applied. The question can be resolved by dividing it into two sections: (a) Determination of the velocity of the 0.30-kg puck after the collision, and (b) Determination of the fraction of kinetic energy lost in the collision.

For part (a), we use the law of conservation of momentum, which states that the total linear momentum of an isolated system remains constant, regardless of whether the objects within the system are at rest or in motion. To find the final velocity of the 0.30-kg puck, we subtract the momentum vector of the 0.20-kg puck after the collision from the momentum vector before the collision. For part (b), we first find the initial and final total kinetic energy of the system. The fraction of kinetic energy lost is given by the difference between the initial and final kinetic energies, divided by the initial kinetic energy.

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The principle of conservation of momentum and kinetic energy principles are employed to figure out the final velocity of the 0.30-kg puck and the fraction of kinetic energy lost in the collision. Initial and final velocities of the pucks, as well as their masses, are used in the process.

To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a system before a collision is equal to the total momentum after the collision. To find out the velocity of the 0.30-kg puck after the collision (a), the momentum equation (m₁v₁_initial + m₂v₂_initial = m₁v₁_final + m₂v₂_final) is applied. It is known that initial velocities of 0.30-kg and 0.20-kg pucks are 0 and 9.1 m/s respectively; the final velocity of the 0.20-kg puck is 5.5 m/s at 53° to the x-axis.

To calculate the fractional kinetic energy lost in the collision (b), firstly, the kinetic energy before and after the collision is calculated using the equation KE = 1/2mv₂. The fraction of kinetic energy lost is then (KE_initial - KE_final)/KE_initial.

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Answer:

[tex]r=5.278\times 10^{-4}\ m[/tex]

Explanation:

Given that:

Now, form the mathematical expression of Kinetic Energy:

[tex]KE= \frac{1}{2} m.v^2[/tex]

[tex]1.92\times 10^{-19}=0.5\times 9.1\times 10^{-31}\times v^2[/tex]

[tex]v^2=4.2198\times 10^{11}[/tex]

[tex]v=6.496\times 10^6\ m.s^{-1}[/tex]

from the relation of magnetic and centripetal forces we have the radius as:

[tex]r=\frac{m.v}{q.B}[/tex]

[tex]r=\frac{9.1\times 10^{-31}\times 6.496\times 10^6 }{1.6\times 10^{-19}\times 0.07}[/tex]

[tex]r=5.278\times 10^{-4}\ m[/tex]

The radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

The magnetic force and centripetal force of the electron is calculated as follows;

qvB = mv²/r

qB = mv/r

r = mv/qB

where;

The kinetic energy of the electron is given as;

K.E = ¹/₂mv²

mv² = 2K.E

v² = 2K.E/m

v² = (2 x 1.2 x 1.6 x 10⁻¹⁹)/(9.11 x 10⁻³¹)

v² = 4.215 x 10¹¹

v = √4.215 x 10¹¹

v = 6.5 x 10⁵ m/s

The radius of their orbit is calculated as follows;

r = mv/qB

r = (9.11 x 10⁻³¹ x 6.5 x 10⁵) / (1.6 x 10⁻¹⁹ x 70 x 10⁻³)

r = 5.287 x 10⁻⁵ m

Thus, the radius of orbit of the cyclotrons in the given field is 5.287 x 10⁻⁵ m.

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Answer:

a) T = 0.579 s , b)  T = 0.579 s

Explanation:

The great advantage of wave mechanics is that the general equations for different systems are the same, what changes are the physical parameters involved, the equation of motion is

    x = A cos (wt +φ)

Where w is the angular velocity that is this case for being a solid body is

    w = √ (mg d / I)

Where I is the moment of inertia and d the distance to the pivot point

The moment of inertia for a ruler hold one end is

     I = 1/12 M L²

The lost is related to the frequency and is with the angular velocity

     T = 1 / f

    w = 2π f

    w = 2π / T

    T = 2π / w

    T = 2π √ I / mgd

For our case

    d = L

    T = 2π √(1/12 M L²) / M g L)

    T = 2π √(L/(12 g))

a) wood suppose it is one meter long (L = 1m)

     T = 2π √ (1 / (12 9.8))

     T = 0.579 s

b) metal length (L = 1m)

    T = 2pi RA (1 / (12 9.8))

    T = 0.579 s

The period does not depend on mass but on length

Answer:

Explanation:

N = 65

Area, A = 0.1 x 0.2 = 0.02 m^2

R = 10 ohm

ω = 29.5 rad/s

B = 1 T

(a) at t = 0

e = N x B x A x ω

e = 65 x 1 x 0.02 x 29.5

e = 38.35 V

(b) The maximum rate of change of magnetic flux is equal to the maximum value of induced emf.

Ф = 38.35 Wb/s

(c) e = NBAω Sinωt

e = 65 x 1 x 0.02 x 29.5 x Sin (29.5 x 0.05)

e = 38.174 V

(d) Maximum torque

τ = M B Sin 90

τ = N i A B

τ = N e A B / R

τ = 65 x 38.35 x 0.02 x 1 / 10

τ = 5 Nm

To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as

[tex]P = P_{atm} +\rho gh[/tex]

Where,

[tex]P_{atm} =[/tex] Atmospheric Pressure

[tex]\rho =[/tex]Density, water at this case

g = Gravity

h = Height

The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.

Our values are given as

[tex]P_{atm} = 1.013*10^5Pa[/tex]

[tex]\rho = 1000Kg/m^3[/tex]

[tex]g = 9.8m/s^2[/tex]

[tex]h = 0.78m[/tex]

Replacing at the equation we have,

[tex]P = P_{atm} +\rho gh[/tex]

[tex]P = 1.013*10^5 +(1000)(9.8)(0.78)[/tex]

[tex]P = 108944Pa[/tex]

[tex]P = 0.1089Mpa[/tex]

Therefore the pressure inside the hose is 0.1089Mpa

Answer:

amount of work done is[tex] W  =  \frac{-4}{3}[/tex]      

Explanation:

Formula for work done by force field

[tex]W = \int F. dr = \int_{t_o}^{t_1} f(r(t)) r'(t) dt[/tex]

where

r(t) is parametrization of line

as it is straight line so

[tex]r(t) = 1- t) r_o + tr_1   0 \leq t \leq 1[/tex]

thus,

r(t) = (1-t)(-1,1) + t(3,-3)

    = (-1+t,1-t) + (3t - 3t)

   = (-1+t +3t, 1-t-3t)

r(t) = (4t -1, 1- 4t)

r'(t) = (4,-4)

putting value in above integral

[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \int_{0}^{1} (-(4t -1)^2 , 2-8t).(4,-4) dt[/tex]

                                                  [tex]= \int_{0}^{1} (-16 t^2 + 8t -1,2-8t) .(4,-4) dt[/tex]

                                                  [tex] =\int_{0}^{1} (4(-16t^2 +8t -1) -4(2-8t)) dt[/tex]

                                                   [tex]= 4[ -16 \frac{t^3}{3} + 16\frac{t^2}{2} - 3t]_{0}^{1}[/tex]

[tex]\int_{0}^{1} ((4t -1,1-4t)). (4,4) dt = \frac{-4}{3}[/tex]                                                        

To find the work done by the force F = xyi +(y-x)j over the straight line from (-1,1) to (3,-3), we compute the line integral of the dot product of the force and displacement vectors along the path. Performing the integrations using the limits set by the line's endpoints gives us the work done.

The task involves physics concepts in vector calculus. Specifically, we're looking at the work done by a force vector over a path in two-dimensional space. Here, the force is described by the vector function ​F = xyi +(y-x)j. We need to calculate the work done by this force as it moves along a straight line path from ​ (-1,1) to (3,-3).

Work done by a force in moving an object is given by the line integral of ​force · displacement. In mathematical terms, this definition reads as W = ∫F.dr. The computation requires us to take the dot product of the force and differential displacement vectors along the path.

The direction of displacement as we go from (-1,1) to (3,-3) is simply the path's tangent. Therefore, dr = dx i + dy j, where (dx,dy) is the differential displacement vector along the line. Since the line is straight, (dx, dy) = (3 - (-1), -3 - 1) = (4, -4) up to a constant. Then dot product becomes F · dr = (xy - (y-x))*(4).

Now, we take the line integral of this product over the path. The limits of integration, derived from the line's endpoints, are x = -1 to x = 3. We carry out the integration and simplify to get the work done by the force over the straight line. Note that the straight-line path simplifies the calculation. For paths with more complex shapes or forces that vary nonlinearly with displacement, such integrations may require specialist mathematical techniques.

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To solve this exercise it is necessary to apply the concepts related to Magnetic Flow which is defined through the Gaus Law of Magnetism as the measure of the total magnetic field that passes through a given area.

Vectorially it can be defined as

[tex]\Phi = \vec{A}\vec{B}[/tex]

Where,

A = Area

B = Magnetic Field

A scalar mode can also be expressed as

[tex]\Phi = A*B Cos\theta[/tex]

Where,

[tex]\theta[/tex] is the angle between B and A, at this case the direction are perpendicular then

Our values are given as,

[tex]A = 0.19m^2[/tex]

[tex]B_1 = 0.5T[/tex]

[tex]B_2 = 0.8T[/tex]

The magnetic field component that interests us is the perpendicular therefore

[tex]\Phi = 0.19* 0.5cos(90)[/tex]

[tex]\Phi = 0.095Tm^2[/tex]

Therefore the magnetic flux through this coil is 0.095[tex]Tm^2[/tex]

D. the gravitational force increases by a factor of 4

Explanation:

The magnitude of the gravitational force between two objects is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

Here let's call F the initial gravitational force between the two objects. Later, the distance between the objects is halved, so the new distance is:

[tex]r'=\frac{r}{2}[/tex]

Substituting into the equation, we find the new force:

[tex]F'=G\frac{m_1 m_2}{(r/2)^2}=4(G\frac{m_1 m_2}{r^2})=4F[/tex]

Therefore, the gravitational force increases by a factor of 4.

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Final answer:

The gravitational force between two objects doubles if the distance between them is cut in half.

Explanation:

The gravitational force between two objects is inversely proportional to the square of the distance between them. This means that if the distance between the objects is cut in half, the force will increase by a factor of 4.

For example, let's say the original distance is 10 units. If we cut it in half, the new distance would be 5 units. The force would increase by a factor of (10/5)² = 4, meaning the gravitational force would quadruple.

So the correct answer is, A. the gravitational force doubles.

Answer:

Part A:

[tex]n=8.85*10^{28}m^{-3}[/tex]

Part B:

[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]

Explanation:

Part A:

To calculate the number of free electrons n we use the following formula::

n=1.5N-Au

Where N-Au is number of gold atoms per cubic meter

[tex]N-Au=\frac{Density*Avogadro Number}{atomic weight}[/tex]

[tex]Density = 19.32g/cm^3[/tex]

[tex]Avogadro Number=6.02*10^{23} atoms/mol[/tex]

[tex]Atomic weight=196.97g/mol[/tex]

So:

[tex]n=1.5*\frac{Density*Avogadro Number}{atomic weight}[/tex]

[tex]n=1.5*\frac{19.32*6.02*10^{23}}{196.97}[/tex]

[tex]n=8.85*10^{28}m^{-3}[/tex]

Part B:

[tex]Electron Mobility=\frac{Elec-conductivity}{n * charge on electron}[/tex]

n is calculated above which is 8.85*10^{28}m^{-3}

Charge on electron=1.602*10^{-19}

Elec- Conductivity= 4.3*10^{7}

[tex]Electron Mobility=\frac{4.3*10^{7}}{ 8.85*10^{28} * 1.602*10^{-19}}[/tex]

[tex]Electron Mobility=3.03*10^{-3} m^2/V[/tex]

To develop this problem it is necessary to apply the concepts related to the definition of absolute, manometric and atmospheric pressure.

The pressure would be defined as

[tex]P = P_0 + P_w +P_m[/tex]

Where,

[tex]P_0 =[/tex] Standard atmosphere pressure

[tex]P_W =[/tex]Pressure of Water

[tex]P_m =[/tex]Pressure of mercury

Therefore we have that

[tex]P = P_0 \rho_w g \frac{h}{2} + \rho_m g\frac{h}{2}[/tex]

Here g is the gravity, \rho the density at normal conditions and h the height of the cylinder.

Replacing with our values we have that,

[tex]P = 1.013*10^5+10^3*9.81*\frac{0.25}{2}+13.6*10^3*9.81*\frac{0.25}{2}[/tex]

[tex]P = 1.21*10^5Pa \approx 0.121Mpa[/tex]

Therefore the pressure at the Bottom of a graduated cylinder is 0.121Mpa

The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.

A graduated cylinder is half full of mercury and half full of water. If the height of the cylinder is 0.25 m, the height of each column of liquid is:

[tex]\frac{0.25m}{2} = 0.13 m[/tex]

We can calculate the pressure at the bottom of the cylinder (P) using the following expression.

[tex]P = P_{atm} + P_{Hg} + P_w[/tex]

where,

We can calculate the pressure exerted by each liquid (P) as:

[tex]P = \rho \times g \times h[/tex]

where,

The pressure exerted by the column of Hg is:

[tex]P_{Hg} = \frac{13.6 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 17 \times 10^{3} Pa[/tex]

The pressure exerted by the column of water is:

[tex]P_{w} = \frac{1.00 \times 10^{3} kg}{m^{3} } \times \frac{9.81m}{s^{2} } \times 0.13 m = 1.3 \times 10^{3} Pa[/tex]

The pressure at the bottom of the cylinder is:

[tex]P = P_{atm} + P_{Hg} + P_w = 101325 Pa + 17 \times 10^{3} Pa + 1.3 \times 10^{3} Pa = 1.2 \times 10^{5} Pa[/tex]

The pressure at the bottom of a 0.25 m-graduated cylinder half full of mercury and half full of water is 1.2 × 10⁵ Pa.

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The wavelength of the wave is 0.055 m

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by the wave equation:

[tex]v=f\lambda[/tex]

where

v is the speed

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the sound wave in this problem we have

v = 340 m/s is the speed

f = 6,191 Hz is the frequency

Solving for [tex]\lambda[/tex], we find the wavelength:

[tex]\lambda=\frac{v}{f}=\frac{340}{6191}=0.055 m[/tex]

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The work function of the metal can be calculated using Einstein's photoelectric equation and the given information about the wavelength of light, stopping potential, Planck's constant, and speed of light. The initial step is to calculate the energy of the photon; thereafter, calculate the kinetic energy of the photoelectron with the stopping potential, and finally find the work function by subtracting the kinetic energy from the photon energy.

To calculate the work function of the metal, we need to first figure out the energy of the photon hitting the metal surface. This can be found using Einstein's photoelectric equation: E = hc/λ. Here, h is Planck's constant (6.626 x 10-34 J · s), c is the speed of light (3.00 x 108 m/s), and λ is the wavelength of the light (350 nm or 350 x 10-9 m).

After calculating the energy of the photon, we can determine the energy required to eject the photoelectrons, known as the stopping potential, which in this case is 0.500 V. Note this is the extra energy required to stop the photoelectron and is equal to the kinetic energy of the photoelectron. Hence, 0.500V = 0.500 eV because 1V = 1eV for a single electron with charge e. To convert this to joules, we use the conversion 1eV = 1.60 x 10-19J.

Finally, in order to find the work function, we use Einstein's photoelectric equation again which states that the energy of the incident photon is equal to the work function (φ) plus the kinetic energy of the photoelectron. That is, E = φ + KE. From this, we can rearrange the equation to solve for φ, the work function of the metal where: φ = E - KE.

By substituting the appropriate values into this equation, the work function of the metal can be found out. But remember, work function is a property of the metal only and is independent of the incident radiation. Also, it determines the threshold frequency or cutoff wavelength of the metal beyond which photoelectric effect will not occur.

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Answer:

The hill should be not less than 0.625 m high

Explanation:

This problem can be solved by using the principle of conservation of mechanical energy. In the absence of friction, the total mechanical energy is conserved. That means that

[tex]E_m=U+K[/tex] is constant, being U the potential energy and K the kinetic energy

[tex]U=mgh[/tex]

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

When the car is in the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

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

We can solve for h:

[tex]h=\frac{v^2}{2g}=\frac{3.5^2}{2(9.8)}=0.625m[/tex]

The hill should be not less than 0.625 m high

The hill should have a height of 0.625 m and above.

This is defined as the measurement of the vertical position of a body.

Total mechanical energy = Potential energy + kinetic energy.

Potential energy = mgh

Kinetic energy = 1/2mv²

We can infer that:

mgh = 1/2mv²

h = v² / 2g

= (3.5)² / 2(9.8)

= 0.625m.

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Final answer:

The angular acceleration of the yo-yo is 850 rad/s², the angular velocity after 0.750 s is 637.5 rad/s, and the tangential acceleration of a point on its edge is 28.05 m/s².

Explanation:

To solve the yo-yo problem:
(a) The angular acceleration (α) can be found by using the linear acceleration (a) of the yo-yo and the radius (r) of its center shaft. The formula is α = a/r. Given that the yo-yo accelerates at 1.70 m/s2 and the radius of the center shaft is 0.200 cm (0.002 m), the angular acceleration is α = 1.70 m/s2 / 0.002 m = 850 rad/s2.
(b) The angular velocity (ω) after a certain time can be calculated using the formula ω = αt, assuming it starts from rest. So after 0.750 s, the angular velocity is ω = 850 rad/s2 * 0.750 s = 637.5 rad/s.
(c) To find the tangential acceleration (at) of a point on the edge of the yo-yo, with an outside radius of 3.30 cm (0.033 m), we use the formula at = α * R. Therefore, the tangential acceleration at the edge is at = 850 rad/s2 * 0.033 m = 28.05 m/s2.

Answer:

75 W/m²

Explanation:

[tex]I_0[/tex] = Unpolarized light intensity = 800 W/m²

[tex]\theta[/tex] = Angle between filters = 30°

Intensity of light after passing through first polarizer

[tex]I=\frac{I_0}{2}\\\Rightarrow I=\frac{800}{2}\\\Rightarrow I=400\ W/m^2[/tex]

Intensity of light after passing through second polarizer

[tex]I_1=Icos^2\theta\\\Rightarrow I_1=400\times cos^2(30)\\\Rightarrow I_1=300\ W/m^2[/tex]

Intensity of light after passing through third polarizer

[tex]I_2=I_1cos^2\theta\\\Rightarrow I_2=300\times cos^2(90-30)\\\Rightarrow I_1=75\ W/m^2[/tex]

The intensity of the light passing through the stack of polarizing sheets is 75 W/m²

The intensity of the light passing through three polarizing sheets, with the first and the third being perpendicular and the second making an angle of 30 degrees with the first, is zero, due to the rules of light polarization.

Unpolarized light is composed of many rays having random polarization directions. When it passes through a polarizing filter, it decreases its intensity by a factor of 2, thus the intensity after the first polarizer is 800 W/m^2 /2 = 400 W/m^2. According to Malus's Law, the intensity of polarized light after passing through a second polarizing filter is given as I = Io cos² θ, where Io is the incident intensity (from the first polarizer) and θ is the angle between the direction of polarization and the axis of the filter.

The second polarizer is making an angle of 30 degrees with the first polarizer. Therefore, the intensity after the second polarizer is I = 400 W/m^2 * cos²(30) ≈ 346 W/m^2. The third polarizer is perpendicular to the first polarizer (or, equivalently, it makes an angle of 90 degrees with the second polarizer), so the final intensity is I = 346 W/m^2 * cos²(90), which is 0 because the cosine of 90 degrees is zero. So, no light (intensity=0) passes through the stack of these polarizing sheets.

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When a doubled rope, i.e. two pieces of rope used together, is used, it stretches half as much as a single rope would under the same load. Correspondingly, a rope's stretch is directly proportional to its length. So, it will stretch twice as much when its length is doubled.

The stretchiness of a rope under a load is a property called elasticity. When there are two ropes, they share the load between them. Therefore, each carries only half the load. Hence, a doubled rope would stretch half as much as a single rope when supporting the same weight. In this case, the single rope stretches by 20 cm under a load. Thus, the doubled rope would stretch by 10 cm.

Similarly, the amount a rope stretches is proportional to its length. If a 10 m rope stretches 60 cm under a certain load, a 20 m rope -- twice as long -- would stretch twice as much under the same load, or 120 cm.

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In the context of elasticity in physics, a doubled rope that halves each individual rope's load would stretch half as much. Therefore, if a 10m rope stretches by 20cm, a doubled rope would stretch by 10cm. If a 10m rope stretches by 60cm, a 20m rope under the same load would stretch twice as much, or 120cm.

These questions refer to the concepts of strain and stress in physics, specifically within the context of elasticity. Strain refers to the deformation of a body due to the applied stress, in this case, the length of rope extending under the weight of the climber.

a. If a 10m climbing rope stretches by 20cm under a certain load, its strain is 0.02 (20cm/10m). When the load is shared by two ropes, each rope bears half of the load. Hence, each rope is likely to stretch half as much. So, the doubled rope would stretch by 10cm.

b. Stretching is directly proportional to the length of the rope. Therefore, if a 10m rope stretches by 60cm under a load, a 20m rope under the same load would stretch twice as much, which is 120cm or 1.2m.

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Answer:

v = 300.6 m/s

Explanation:

given,

mass of bullet(m)  = 10 g = 0.01 kg

mass of block of wood (M)= 5 Kg

speed of the block plus bullet after collision(V) = 0.6 m/s

speed of wood(u) = 0 m/s

original speed of the bullet(v) = ?

using conservation of momentum

m v + Mu = (M + m)V

0.01 x v + 5 x 0 = (5 + 0.01) x 0.6

0.01 x v = (5.01) x 0.6

[tex]v = \dfrac{3.006}{0.01}[/tex]

v = 300.6 m/s

the original speed of the bullet before collision is equal to v = 300.6 m/s

The original speed of the bullet can be determined using the law of conservation of momentum. Applying the formula and solving for the original bullet speed, the bullet was initially traveling at 3 m/s.

This problem is an application of the law of conservation of momentum. The principle states that the total momentum before a collision is equal to the total momentum after the collision. In this case, before the collision, only the bullet has momentum as the block is stationary. After the collision, the momentum is shared between the block and the lodged bullet.

To find the original speed of the bullet, we can set up this equation representing the law of conservation of momentum: m1v1 (before collision) = (m1 + m2) v2 (after collision). Here, m1 = mass of the bullet, v1 = original speed of the bullet, m2 = mass of the block, v2 = final speed of the block and bullet together.

Substituting the given values: (10 g)x = (10 g + 5.00 kg)(0.600 m/s). Solving for x (original bullet speed) yields a value of x = 3 m/s.

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Answer:

(a) check attachment

(b)5 m/s²

Explanation:

Given: radius = 2.00mm: density = 2500kg/m³: viscosity of glycerin = 1.5pa: decity of glycerin = 1250kg/m³: g = 10N/kg = 10m/s²: Fdrag = 6πnrv

(a) for answer check attachment.

(b) For the magnitude of the balls initial acceleration:

     Initial net force(f) = mg - upthrust

                                  = [tex]mg - (\frac{m}{p} )pg.g[/tex]

     acceleration (a) = [tex]Acceleration(a)=\frac{f}{m}\\=g - (\frac{pg}{p})g\\=g(1-\frac{pg}{p} )\\=10(1-\frac{1250}{2500} )\\a=10(1-0.5)\\a=5 m/s²[/tex]

c.) fromthe force diagram in the attachment; when the ball attains terminal velocity the net force will be zero(0)

                                [tex]mg=6πnrv + upthrust[/tex]

d.) For the magnitude of terminal velocity:

                                             [tex]mg=6πnrv + (\frac{m}{p})pg.g\\\\(\frac{4}{3}πr^{3} p)g=6πnrv +\frac{4}{3}πr^3pg.g\\\\V = \frac{2}{9}.\frac{(2*10^{-3})^{2}*(2500-1250)*10}{1.5}\\\\=0.79cm/s[/tex]

e.) when the ball reaches terminal velocity, the acceleration is zero (0)

Explanation:

Moment of Inertia in physics is the measure of the rotational inertia of an object. It is nothing but the opposition that the body offers to having its speed of rotation about an axis changed by application some external torque.

I= MR^2

M= mass of object R= radius of rotation.

Therefore, MOI depends upon

a)total mass

b)shape and density of the object

c)location of the axis of rotation

The moment of inertia of an object depends on its total mass, the distribution of this mass in relation to the rotation axis (shape and density), and the location of the rotation axis. The higher the moment of inertia, the greater the object's resistance to changes in rotational rate. The parallel axis theorem provides a tool to calculate the moment of inertia for different rotation axes.

The moment of inertia of an object depends on certain factors. Specifically, its total mass, the shape and density of the object, and the location of the axis of rotation. The moment of inertia is defined as the measure of the object's resistance to changes in its rotation rate. The moment of inertia varies depending on how mass is distributed in relation to the axis of rotation.

For instance, an object with more mass distributed farther from its axis of rotation will have a higher moment of inertia. This is seen in a hollow cylinder versus a solid cylinder of the same mass and size. The hollow cylinder has a greater moment of inertia because more of its mass is distributed at a greater distance from its axis of rotation.

A principle known as the parallel axis theorem can be used to find the moment of inertia about a new axis of rotation once it is known as a parallel axis. This theorem involves the center of mass and the distance from the initial axis to the parallel axis.

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Answer:

37.11231 m/s

Explanation:

u = Initial velocity

v = Final velocity = 0

s = Displacement = 78 m

g = Acceleration due to gravity = 9.81 m/s²

[tex]\mu[/tex] = Coefficient of kinetic friction = 0.9

Acceleration is given by

[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=-2\mu gs\\\Rightarrow -u^2=2(-\mu g)s-v^2\\\Rightarrow u=\sqrt{v^2-2(-\mu g)s}\\\Rightarrow u=\sqrt{0^2-2\times (-0.9\times 9.81)\times 78}\\\Rightarrow u=37.11231\ m/s[/tex]

The initial speed of that car is 37.11231 m/s

Answer: a) 2.33V b) Ip = 0.0035A, Is = 0.179A

Explanation:

The relationship formula between the Voltage, Number of turns and current in a transformer is given as

Vp/Vs = Np/Ns = Is/Ip

Vp is voltage in the primary coil

Vs is voltage in the secondary coil

Ns is number of turns in the secondary coil

Np is number of turns in the primary coil

Is is current in the secondary coil

Ip is current in the primary coil

a) substituting the values in the formula

Vp/Vs = Np/Ns

120/Vs = 500/9.7

Vs = 120×9.7/500

Vs = 2.33V

b) since secondary now has a resistive load of 13 Ω, Using ohm's law to get current in the secondary

V=IR

Is = Vs/R

Is =2.33/13

Is = 0.179A

To get Ip, we will use the relationship Np/Ns = Is/Ip

500/9.7= 0.179/Ip

Ip = 9.7×0.179/500

Ip = 0.0035A

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section of a given toroid, calculated using Ampere's Law, is approximately 0.00269 T.

The magnetic field B produced by a current in a toroidal solenoid is described by Ampere's Law. The formula to calculate magnetic field (B) in a toroid is B = μ₀IN / (2πr), where:

Substituting these values in the formula, we get:

B = (4π x 10⁻⁷ T m/A) * (600) * (3 A) / (2π * 0.266 m) = 0.00269 T.

So the strength of the magnetic field at the center of the square cross section is approximately 0.00269 T.

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To develop this problem it is necessary to apply the kinematic equations that describe displacement, velocity and clarification.

By definition we know that velocity is defined as the change of position due to time, therefore

[tex]V = \frac{d}{t}[/tex]

Where,

d = Distance

t = Time

Speed can also be expressed in vector form through its components [tex]V_x[/tex] and [tex]V_y[/tex]

In the case of the horizontal component X, we have to

[tex]V_x = \frac{d}{t}[/tex]

Here d means the horizontal displacement, then

[tex]t = \frac{d}{V_x}[/tex]

[tex]t = \frac{67}{V_x}[/tex]

At the same time we have that the vertical component of the velocity is

[tex]V_y = gt[/tex]

Here,

g = Gravity

Therefore using the relation previously found we have that

[tex]V_y = g \frac{67}{V_x}[/tex]

The relationship between the two velocities and the angle can be expressed through the Tangent, therefore

[tex]tan\theta = \frac{V_y}{V_x}[/tex]

[tex]tan \theta = \frac{g \frac{67}{V_x} }{V_x}[/tex]

[tex]tan 3 = \frac{9.8\frac{67}{V_x} }{V_x}[/tex]

[tex]tan 3 = \frac{9.8*67}{V_x^2}[/tex]

[tex]V_x^2 = \frac{9.8*67}{tan 3}[/tex]

[tex]V_x= \sqrt{ \frac{9.8*67}{tan 3}}[/tex]

[tex]V_x = 111.93m/s \hat{i}[/tex]

This is the horizontal component, we could also find the vertical speed and the value of the total speed with the information given,

Then [tex]V_y,[/tex]

[tex]V_y = g \frac{67}{V_x}[/tex]

[tex]V_y = 9.8*\frac{67}{111.93}[/tex]

[tex]V_y = 5.866m/s\hat{j}[/tex]

[tex]|\vec{V}| = \sqrt{111.93^2+5.866^2}[/tex]

[tex]|\vec{V}| = 112.084m/s[/tex]

Final answer:

To calculate the initial speed of the arrow in a projectile motion problem, one needs to find the time of flight using the horizontal range and the final impact angle and then solve for the initial horizontal and vertical velocity components.

Explanation:

To determine the initial speed of the arrow shot from the bow, we'll use the principle of projectile motion. First, we need to find the time of flight. The arrow makes a 3.00° angle with the ground upon impact, and since it was shot parallel to the ground, it maintains a constant horizontal velocity throughout its flight. So, we can calculate the initial horizontal speed using this final impact angle.

The horizontal range (R) is given by R = v0x * t, where v0x is the initial horizontal velocity and t is the time of flight. The arrow landed 67.0 m away, so R is 67.0 m.

To find t, we will use the vertical motion equation under constant acceleration due to gravity (g = 9.8 m/s2). We know the final vertical velocity component (vfy) can be related to the initial vertical velocity component (v0y = 0 since it was shot parallel to the ground) by vfy = v0y + g*t. We can also calculate vfy using the impact angle: vfy = tan(impact angle) * v0x. Combining these, we can solve for t and subsequently for the initial speed v0.

Answer:

[tex]a =29.54\ m/s^2[/tex]

Explanation:

given,

radius of CD player = 12 cm

assume rate of acceleration = 100 rad/s²

times = 0.15 s

now,

tangential acceleration  

  [tex]a_t = \alpha r[/tex]

  [tex]a_t = 100 \times 0.12[/tex]

  [tex]a_t = 12 m/s^2[/tex]

now using equation

v = v₀ + a_t x t

v =0+ 12 x 0.15

v = 1.8 m/s

now, radial acceleration

[tex]a_r = \dfrac{v^2}{r}[/tex]

[tex]a_r = \dfrac{1.8^2}{0.12}[/tex]

[tex]a_r =27\ m.s^2[/tex]

now

acceleration

[tex]a = \sqrt{a_r^2+a_t^2}[/tex]

[tex]a = \sqrt{27^2+12^2}[/tex]

[tex]a =29.54\ m/s^2[/tex]

The angular momentum of the ice skater spinning at 6.00 rev/s is 15.82 kg · m²/s. When the skater reduces their rate of spin to 1.95 rev/s, the moment of inertia is 1.29 kg · m². If the skater keeps their arms in and allows friction with the ice to slow them to 3.00 rev/s in 23.0 seconds, the average torque exerted is -0.343 N · m (indicating the direction of the torque).

(a) The angular momentum of the ice skater can be calculated by multiplying the angular velocity (in radians per second) by the moment of inertia (in kg · m²). First, we need to convert the angular velocity from rev/s to rad/s. Since 1 rev = 2π rad, the angular velocity in rad/s is 6.00 rev/s * 2π rad/rev = 37.70 rad/s. Now, we can calculate the angular momentum: Angular momentum = angular velocity * moment of inertia = 37.70 rad/s * 0.420 kg · m² = 15.82 kg · m²/s.

(b) To find the new moment of inertia when the angular velocity decreases to 1.95 rev/s, we can rearrange the formula for angular momentum to solve for moment of inertia. Angular momentum = angular velocity * moment of inertia. Rearranging the equation: Moment of inertia = angular momentum / angular velocity = 15.82 kg · m²/s / (1.95 rev/s * 2π rad/rev) = 1.29 kg · m².

(c) To calculate the average torque exerted when the skater slows to 3.00 rev/s, we can use the equation torque = change in angular momentum / time. First, we need to calculate the change in angular momentum by subtracting the initial angular momentum from the final angular momentum. The initial angular momentum is 6.00 rev/s * 2π rad/rev * 0.420 kg · m² = 15.82 kg · m²/s, and the final angular momentum is 3.00 rev/s * 2π rad/rev * 0.420 kg · m² = 7.91 kg · m²/s. The change in angular momentum is 7.91 kg · m²/s - 15.82 kg · m²/s = -7.91 kg · m²/s. Finally, we can calculate the average torque: Torque = change in angular momentum / time = -7.91 kg · m²/s / 23.0 s = -0.343 N · m (negative sign indicates the direction of the torque).