Determine the location of the center of mass of a "L" whose thin vertical and horizontal members have the same length L and the same mass M. Use the formal definition to find the x and y coordinates, and check your result by doing the calculation with respect to two different origins, one in the lower left corner at the intersection of the horizontal and vertical members and one at the top of the vertical member.

(a) Origin at the lower left

x = ?

y = ?

(b) Origin at the top of the vertical member

x = ?

y = ?

The electrostatic force between two positively charged particles changes by a factor of 1/n² when they are moved further apart, based on Coulomb's Law.

The subject matter of this question involves gravitational force and electrostatic force, which makes it a physics question. When two positively charged particles are moved further apart, the gravitational force between them changes by a factor of n. The electrostatic force between the two objects, however, obeys Coulomb's Law, which states that the force between two charges is inversely proportional to the square of the distance between them. Therefore, when the particles are moved further apart, the magnitude of their electrostatic force changes by a factor of 1/n². So, the correct answer to this question would be Option A: 1/n².

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The magnitude of the mutual electrostatic force between two charged particles that are moved farther apart changes by a factor of 1/n²

When two positively charged particles are moved farther apart, we know that the force between them, according to Coulomb's law, is inversely proportional to the square of the distance between them.

Therefore, the magnitude of their mutual electrostatic force will change by a factor of 1/n²  when the distance between them is changed.

This is analogous to the gravitational force between two masses, which follows the same inverse-square law as electrostatic force does.

Answer:

The final angular velocity is [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]

Explanation:

From the question we are told that

     The mass of the first disk is  m

      The radius of the first  disk is  r

      The mass of  second disk is  M

      The radius of second disk is R

       The speed of rotation is w

       The moment of inertia of second disk is  [tex]I = \frac{1}{2} MR^2[/tex]

Since the first disk is at rest initially

        The initial angular momentum would be due to the second disk  and this is mathematically represented as

       [tex]L_i = Iw = \frac{1}{2} MR^2 w[/tex]

Now when the first disk is then dropped the angular momentum of the whole system now becomes

       [tex]L_f = (I_1 + I_2 ) w_f= ( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f[/tex]

This above is because the formula for moment of inertia is the same for every disk

       According to the law  conservation of  angular momentum

                [tex]L_f = L_i[/tex]

    [tex]( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2) w_f = \frac{1}{2} MR^2 w[/tex]

=>              [tex]w_f = \frac{\frac{1}{2} MR^2 w }{( \frac{1}{2} MR^2 + \frac{1}{2} m^2 r^2)}[/tex]

                  [tex]w_f = \frac{MR^2}{MR^2+ mr^2} w[/tex]

49.5 miles per hour will be the new speed.

This problem can be resolved using one-dimensional kinematics: v² = v₀² - 2 a d

Let's use this equation to find the beginning speed of the vehicle. In the first example, the distance is half (d´ = ½ d), the vehicle's acceleration is the same, and we find the fine velocity zero 0 = v₀² - 2 a d a = v₀² / 2d.

v₀´² = 2 a d = 2 (v₀² / 2d) 0 = v₀´² - 2 a d´ d´ v₀´ = v₀ √ 1/2

Let's compute: v₀ = 49.5 miles per hour; v₀´ = 70 ra 0.5.

Consequently, 49.5 miles per hour will be the new speed.

Answer:

The ocean absorbs a significant portion of carbon dioxide (CO2) emissions from human activities, equivalent to about one-third of the total emissions for the past 200 years from fossil fuel combustion, cement production and land-use change (Sabine et al., 2004). Uptake of CO2 by the ocean benefits society by moderating the rate of climate change but also causes unprecedented changes to ocean chemistry, decreasing the pH of the water and leading to a suite of chemical changes collectively known as ocean acidification. Like climate change, ocean acidification is a growing global problem that will intensify with continued CO2 emissions and has the potential to change marine ecosystems and affect benefits to society.

The average pH of ocean surface waters has decreased by about 0.1 unit—from about 8.2 to 8.1—since the beginning of the industrial revolution, with model projections showing an additional 0.2-0.3 drop by the end of the century, even under optimistic scenarios (Caldeira and Wickett, 2005).1 Perhaps more important is that the rate of this change exceeds any known change in ocean chemistry for at least 800,000 years (Ridgewell and Zeebe, 2005). The major changes in ocean chemistry caused by increasing atmospheric CO2 are well understood and can be precisely calculated, despite some uncertainty resulting from biological feedback processes. However, the direct biological effects of ocean acidification are less certain

image

1 “Acidification” does not mean that the ocean has a pH below neutrality. The average pH of the ocean is still basic (8.1), but because the pH is decreasing, it is described as undergoing acidification.

Page 2

Suggested Citation:"Summary." National Research Council. 2010. Ocean Acidification: A National Strategy to Meet the Challenges of a Changing Ocean. Washington, DC: The National Academies Press. doi: 10.17226/12904. ×

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and will vary among organisms, with some coping well and others not at all. The long-term consequences of ocean acidification for marine biota are unknown, but changes in many ecosystems and the services they provide to society appear likely based on current understanding (Raven et al., 2005).

In response to these concerns, Congress requested that the National Research Council conduct a study on ocean acidification in the Magnuson-Stevens Fishery Conservation and Management Reauthorization Act of 2006. The Committee on the Development of an Integrated Science Strategy for Ocean Acidification Monitoring, Research, and Impacts Assessment is charged with reviewing the current state of knowledge and identifying key gaps in information to help federal agencies develop a program to improve understanding and address the consequences of ocean acidification (see Box S.1 for full statement of task). Shortly after the study was underway, Congress passed another law—the Federal Ocean Acidification Research and Monitoring (FOARAM) Act of 2009—which calls for, among other things, the establishment of a federal ocean acidification program; this report is directed to the ongoing strategic planning process for such a program.

Although ocean acidification research is in its infancy, there is already growing evidence of changes in ocean chemistry and ensuing biological impacts. Time-series measurements and other field data have documented the decrease in ocean pH and other related changes in seawater chemistry (Dore et al., 2009). The absorption of anthropogenic CO2 by the oceans increases the concentration of hydrogen ions in seawater (quanti-

Explanation:

Answer:

Explanation:

wavelength of light λ = 502 x 10⁻⁹ m /s

screen distance D = 1.4 m

Slit separation d = ?

position of n the separation is given by the formula

x = n Dλ / d , n is order of fringe , x = distance of n th fringe

10.4 x 10⁻³ = 20 x 1.4 x  502 x 10⁻⁹ / d

d = 20 x 1.4 x  502 x 10⁻⁹ / 10.4 x 10⁻³

= 1351.54 x 10⁻⁶

= 1.35 x 10⁻³ m

1.35 mm.

Answer:

Kinetic energy = 0.145 J

Explanation:

See the attached file for explanation.

Rotational kinetic energy of a spinning cube can be calculated using the formula KErot = 0.5*Iw². This scenario involves kinetic energy gained by both the cube and the unwinding string's hanging mass. The specific energy amount depends on certain parameters that are currently absent.

The question involves the calculation of the rotational kinetic energy of a spinning cube. The rotational kinetic energy (KErot) of an object with moment of inertia (I) and angular velocity (w) is given by the formula KErot = 0.5*Iw².

In this case, the cube picks up kinetic energy as it spins up, and the hanging mass also gains kinetic energy as the string unwinds. The total distance the string unwinds (L) also plays a role in the final rotational kinetic energy of the spinning cube.

However, the lack of specific parameters like the mass and angular velocity of the cube, as well as the mass of the hanging object in your question, makes it challenging to give a numerical answer.

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Answer:

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

Explanation:

The relation that describes the pressure amplitude for a sound wave is  

P_MAX = B*k*A                                              (1)  

Where the bulk modulus of the air is B = 1.30 x 10^5 Pa and the displacement  amplitude of the waves produced by the machine is 1.00 μmμm.  

Using (1) we can calculate k then we can use k to determine the wavelength  A of the wave, and remember that λ = 2π/k.  

So, substitute into (1) with 10 Pa for P_max, (1.30 x 10^5 Pa) for B and  

1 x 10^-6 m for A  

10 Pa = (1.30 x 10^5 Pa) x k x (1 x 10^-6 m)  

k = 62.5 m^-1

We can use the following relation to calculate the wavelength  

λ = 2π/k.  

λ = 0.100 m

Finally, the relation between the wavelength and the frequency of a sound  

wave is given by the following equation  

f = v/ λ

 =344/0.100 m

f =3.4*10^3 hz

Since f is in the range of (20 Hz - 20,000 Hz] which is the range of audible frequencies, the frequency is audible.

Answer:

a) [tex]F = 660.576\,N[/tex], b) [tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex], c) [tex]v \approx 7255.423\,\frac{m}{s}[/tex], [tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex], d) [tex]T \approx 1.821\,h[/tex]

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

[tex]F = G\cdot \frac{m\cdot M}{r^{2}}[/tex]

[tex]F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}[/tex]

[tex]F = 660.576\,N[/tex]

b) The centripetal acceleration of the satellite is:

[tex]a_{c} = \frac{660.576\,N}{95\,kg}[/tex]

[tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex]

c) The speed of the satellite is:

[tex]v = \sqrt{a_{c}\cdot R}[/tex]

[tex]v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}[/tex]

[tex]v \approx 7255.423\,\frac{m}{s}[/tex]

Likewise, the angular speed is:

[tex]\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}[/tex]

[tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex]

d) The period of the satellite's rotation around the Earth is:

[tex]T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)[/tex]

[tex]T \approx 1.821\,h[/tex]

Answer:

E = 3.194 x 10⁹ J = 3.194 GJ

Explanation:

The formula for the absolute potential energy is:

U = - GMm/2r

where,

G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²

M = mass of Earth = 5.972 x 10²⁴ kg

m = mass of satellite = 470 kg

r = distance between the center of Earth and satellite

Thus, the energy required from engine will be difference between the potential energies.

E = U₂ - U₁

E = - GMm/2r₂ - (- GMm/2r₁)

E = (GMm/2)(1/r₁ - 1/r₂)

where,

r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m

r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m

therefore,

E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)

E = 3.194 x 10⁹ J = 3.194 GJ

Answer:

E₄ = - 0.85 eV

E₂ = - 3.4 eV

Ephoton = 2.55 eV

Explanation:

The sum of Kinetic Energy (K) and Potential Energy (U) of the Helium atom is equal to the total energy of Helium atom in the specified state N. From Bohr's atomic model, the energy of a hydrogen atom in state N is given as:

En = K + U = (-1/n²)(13.6 eV)

a)

Here,

n = 4

Therefore,

E₄ = (-1/4²)(13.6 eV)

E₄ = - 0.85 eV

b)

Here,

n = 2

Therefore,

E₂ = (-1/2²)(13.6 eV)

E₂ = - 3.4 eV

c)

The energy of photon emitted in the transition from level 4 to level 2 will be equal to the difference in the energy of both levels:

Ephoton = ΔE =  E₄ - E₂

Ephoton = - 0.85 eV - (- 3.4 eV)

Ephoton = 2.55 eV

The electrons energy will be:

As we know. the formula:

→ [tex]E_n = K+U[/tex]

        [tex]= (-\frac{1}{n^2} ) (13.6 \ eV)[/tex]

(a)

Given:

then,

→ [tex]E_4 = (-\frac{1}{4^2} )(13.6 \ eV)[/tex]

        [tex]= -0.85 \ eV[/tex]

(b)

Given:

then,

→ [tex]E_2 = (-\frac{1}{2^2} )(13.6 \ eV)[/tex]

        [tex]= -3.4 \ eV[/tex]

(c)

The energy of photon emitted will be:

→ [tex]E_{photon} = \Delta E[/tex]

               [tex]= E_4 -E_2[/tex]

               [tex]=-0.85 \ eV-(-3.4 \ eV)[/tex]

               [tex]= 2.55 \ eV[/tex]  

Thus the above answers are appropriate.

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Answer:

Assuming the mass of the garbage truck is [tex]1.6*10^{4}[/tex] kg, the speed of the garbage truck is [tex]approx. = 27.8503\frac{m}{s}[/tex].

Explanation:

This is conservation of momentum where both objects stick at the end so they have the same final velocity so our equation is:

[tex]m_{1}v_{1initial} + m_{2}v_{2initial} = v_{both final}(m_{1} + m_{2})[/tex]

To solve for the final velocity, just divide by the sum of both masses:

[tex]\frac{ m_{1}v_{1initial} + m_{2}v_{2initial}}{(m_{1} + m_{2})} = v_{both final}[/tex]

So, plug in the known values (remember initial velocity for the trash can is 0):

[tex]\frac{ 1.6*10^{4}*28 + 86*0}{1.6+10^{4} + 86} = v_{both final}[/tex]

Answer:

189.47nm

Explanation:

We can solve this problem by taking into account the time that light takes in crossing the distance between the laser and the photocell, and the time in crossing the slab.

By using the values of c and 16.5ns we can calculate the value of d

[tex]d=(3*10^{8}m/s)(16.5*10^{-9}s)=4.95m[/tex]

We have to compute the time that light takes in crossing d-0.820m:

[tex]4.95m-0.820m=4.13m\\\\t=\frac{4.13m}{3*10^8m/s}=6.22*10^{-8}s=13.7ns[/tex]

Now, we can calculate the speed of the light in the slab by using the time difference between 21.5 ns and 13.7ns:

[tex]\Delta t=21.5ns-13.7ns=7.8ns\\\\v_l=\frac{0.82m}{7.8ns}=1.05*10^8m/s[/tex]

Then, the index of refraction will be:

[tex]n=\frac{c}{v_l}=2.85[/tex]

Finally, we have that:

[tex]\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}\\\\\lambda_2=\lambda_1 \frac{n_1}{n_2}=(540nm)\frac{1.00}{2.85}=189.47nm[/tex]

hope this helps!!

Final answer:

In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump is 0, meaning it comes to a complete stop after the collision.

Explanation:

In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump can be determined by applying the principles of conservation of momentum and the relativistic addition of velocities.

Let's calculate the final speed of the combined lump using the given information:

Mass of each lump (m) = 1.500 kg
Speed of each lump (v) = 0.930c

Using the relativistic addition of velocities formula:

vf = (v1 + v2) / (1 + (v1 * v2 / c2))

Plugging in the given values:

vf = (0.930c + (-0.930c)) / (1 + (0.930c * (-0.930c) / c2))

vf = 0

Therefore, the final speed (vf) of the combined lump is 0, which means it comes to a complete stop after the inelastic collision.

Answer:

Explanation:

Moment of inertia of larger disk   I₁ = 1/2 MR²

Moment of inertia of smaller  disk   I₂ = 1/2 m r ²

Initial angular velocity

We shall apply law of conservation of angular momentum .

initial total momentum = final angular momentum

I₁ X ωi  = ( I₁ + I₂ )ωf

1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf

ωf =  ωi   / ( 1 + m r²/MR² )

Question: The question is incomplete. Diagram of the circuit was not added to your question. Find attached of the circuit diagram and the answer.

Answer:

For R1: Current moves from A to B

For R2: Current moves from B to E

For R3: Current moves from C to D

Explanation:

See the attached file for the explanation

Answer:   In order to walk on a floor (or any other surface), your foot must push backward on the floor (action force), so that the floor pushes you forward (reaction force).

Explanation:

Ans: 237.5J

Explanation;

WC= T*d

T* dcostheta. But theta=0

= 190* 1.25

= 237.5J

See attached file for diagram

Final answer:

The work done on the crate by the tension force when lifting it an additional 1.25 m with a force of 160 N is 200 joules.

Explanation:

The question is asking about the work done on a 14.5 kg crate by tension force when lifting the crate an additional 1.25 m with a tension force of 160 N.

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

Work = Force x Distance x cosθ

Since the force and displacement are in the same direction (theta = 0 degrees), the cosine term is 1, and the equation simplifies to:

Work = 160 N x 1.25 m

The work done on the crate by the tension force is:

Work = 200 joules (J)

Answer:

If you need to measure much longer lengths - for example the length of a football pitch - then you could use a trundle wheel. You use it by pushing the wheel along the ground. It clicks every time it measures one metre.

You could employ a trundle wheel if you need to estimate considerably greater distances, like the size of a football field. By moving the wheels along the surface, you can use it. It makes a click sound each time it measures a meter.

Football, commonly known as association soccer, is a sport in which two groups of 11 players attempt to advance the ball into the goal of the other team by using any component of their bodies other than their hands and arms.

Only the goalie is allowed to handle the ball, and only within the towns near the goal that is designated as the penalty area. The team with the most goals scored wins.

According to the number of players and spectators, football is the most watched sport in the world.

The sport may be played practically everywhere, from official football fields (pitches) to gyms, streets, school playground, parks, or beaches, thanks to its basic rules and necessary equipment. The Federation Internationale Football Association is the governing organization of football.

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Explanation:

Given that,

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1∘ from the vertical, [tex]\theta=15.1^{\circ}[/tex]

The magnitude of the magnetic field B changes in time according to the equation :

[tex]B(t)=3.75+2.75 t-7.05 t^2[/tex]

Radius of the loop, r = 0.21 m

We need to find the magnitude of the induced emf in the loop when t=5.63 s. The induced emf is given by :

[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA\cos \theta)}{dt}[/tex]

B is magnetic field

A is area of cross section

[tex]\epsilon=A\dfrac{-dB}{dt}\\\\\epsilon=\pi r^2\dfrac{-d(3.75+2.75 t-7.05 t^2)}{dt}\times \cos\theta\\\\\epsilon=\pi r^2\times(2.75-14.1t)\times \cos\theta[/tex]

At t = 5.63 seconds,

[tex]\epsilon=-\pi (0.21)^2\times(2.75-14.1(5.63))\times \cos(15.1)\\\\\epsilon=10.25V[/tex]

So, the magnitude of induced emf in the loop when t=5.63 s is 10.25 V.

The EMF generated at time t = 5.63 is 10.18V.

Given a horizontal circular wire loop with a radius, r = 0.21m.

A time-dependent magnetic field B(t) = 3.75 + 2.75t -7.05t².

At an angle of θ = 15.1° to the area of the loop.

The magnetic flux passing through the loop is given by:

Ф = B(t)Acosθ

where A = πr² is the are of the loop.

Since the magnetic field is time-dependent, the magnetic flux through the loop changes with time, therefore an EMF is generated in the loop, given by:

[tex]E=-\frac{d\phi}{dt}\\\\E =-\frac{dB(t)}{dt}Acos\theta\\\\E=-\pi r^2cos\theta\frac{d}{dt}[ 3.75 + 2.75t -7.05t^2] \\\\E=\pi r^2cos\theta[14.10t-2.75]\\\\[/tex]

At time t = 5.63s

[tex]E=3.14\times(021)^2\times cos15.1\times[14.1\times5.63-2.75][/tex]

E = 10.18V

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Answer:

0.1014 J

Explanation:

Considering the air resistance negligible kinetic energy is given by

K.E = (1/2) m v^2

K.E = 0.5 x 0.12 x (1.3)^2

K.E = 0.06 x 1.69

K.E = 0.1014 J

Final answer:

The 0.12 kg paper airplane flying at 1.3 m/s has a kinetic energy of 0.1014 joules, calculated using the formula KE = 0.5 × m × v^2.

Explanation:

The kinetic energy of a paper airplane can be calculated using the kinetic energy formula KE = 0.5 × m × v^2, where KE is kinetic energy, m is the mass of the object, and v is its velocity. For a 0.12 kg paper airplane flying at a speed of 1.3 m/s, the kinetic energy can be calculated as follows:

KE = 0.5 × 0.12 kg × (1.3 m/s)^2 = 0.1014 J

Therefore, the paper airplane has 0.1014 joules of kinetic energy.

Answer:

ΔE =  20 eV

Explanation:

In a Helium atom we have two electrons in the s layer, so they can accommodate one with the spin up and the other with the spin down, give us a total spin of zero (S = 0) this state is singlet, in general this very stable states,

  When you transition to the 1s state to complete the two electrons allowed by layers

    ΔE = -5 - (-25) = 20 eV

this is the energy of the transition,

It should be mentioned that there can also be transitions with the two spins of the same orientation, but in this case the energy is a little different due to the electron-electron repulsion, this state is called ortho helium S = 1

Answer:

5.9*10^{-4}m

Explanation:

to find the uncertainty of the displacement it is necessary to compute the uncertainty for the angular frequency:

[tex]\omega=\frac{2\pi}{T}=\frac{2\pi}{0.40s}=15.707rad/s\\\\\frac{d \omega}{\omega}=\frac{dT}{T}\\\\d\omega=\omega \frac{dT}{T}=(15.707rad/s)\frac{0.020s}{0.40s}=0.785rad/s[/tex]

then, you can calculate the uncertainty in angular displacement:

[tex]\theta=\omega t\\\\\frac{d\theta}{\theta}=\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}\\\\d\theta=\theta\sqrt{(\frac{d\omega}{\omega})^2+(\frac{dt}{t})^2}=0.0422[/tex]

finally, by using:

[tex]y=Acos(\omega t)\\\\dy=dAcos(\omega t)d(\omega t)=(dA)cos(\theta)d\theta=(0.002m)cos(0.785)(0.0422)\\\\dy=5.9*10^{-4}m[/tex]

The uncertainty of her displacement in mm is  : 0.59 mm

First step : determine the uncertaintiy of the angular frequency

w = [tex]\frac{2\pi }{T}[/tex]  = [tex]\frac{2\pi }{0.40} = 15.707 rad/s[/tex]

[tex]\frac{dw}{w} = \frac{dT}{T}[/tex]

therefore :

dw = 0.785 rad/s

Next step : determine the uncertainty of the angular displacement

θ = wt

dθ / θ = [tex]\sqrt{(\frac{dw}{w} )^2 + (\frac{dt}{t} )^2}[/tex]

therefore :

dθ = 0.0422

Final step : determine the uncertainty of displacement

y = Acos(wt)

dy = dAcos(wt)d(t) = (dA)cosθdθ

                              = ( 0.002m )cos (0.785)(0.0422)

                              = 5.9 * 10⁻⁴ m = 0.59 mm

Hence we can conclude that the  uncertainty of her displacement in mm is  : 0.59 mm

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Amplitude is a measurement of the magnitude of displacement (or maximum disturbance) of a medium from its resting state, as diagramed in the peak deviation example below (it can also be a measurement of an electrical signal's increased or decreased strength above or below a nominal state).

Answer:

21.5°

Explanation:

Given,

Refractive index of water, n₁ = 1.33

Refractive index of polystyrene, n₂ = 1.49

Angle of reflection = ?

Angle of refraction = 19.1°

Using Snell's law

n₁ sin θ₁ = n₂ sin θ₂

1.33 x sin θ₁ = 1.49 x sin 19.1°

sin θ₁ = 0.366

θ₁  = 21.5°

According to law of reflection angle of incidence is equal to angle of reflection.

Angle of reflection =  21.5°

Final answer:

The angle of reflection of the reflected ray in polystyrene that is submerged in water will be equal to the angle of refraction, so it is 19.1°.

Explanation:

The question is about the reflection and refraction of light as it strikes a flat surface of a polystyrene block that is in water. According to the Law of Reflection, the angle of reflection is equal to the angle of incidence.

Since we know the angle of refraction is 19.1°, and by Snell's Law (n₁ × sin(i) = n₂ × sin(r)), we can infer that the angle of incidence is also 19.1° because the boundary is between two different mediums (water and polystyrene), resulting in the light being split into reflected and refracted rays.

Therefore, the angle of reflection of the reflected ray is also 19.1°, identical to the angle of refraction given in the question, because the angle of incidence equals the angle of reflection.

Answer:

A) I_f3 = 27.58 W/cm²

B) I_f2 = 102.26 W/cm²

Explanation:

We are given;

-The angle of the second polarizing to the first; θ_2 = 21°

-The angle of the third  polarizing to the first; θ_1 = 61°

- The unpolarized light after it pass through the polarizing stack; I_u = I_3 = 60 W/cm² 

A) Let the initial intensity of the beam of light before polarization be I_p

Thus, when the unpolarized light passes through the first polarizing filter, the intensity of light that emerges would be given as;

I_1 = (I_p)/2

According to Malus’s law,

I = I_max(cos²Φ)

Thus, we can say that;

the intensity of light that would emerge from the second polarizing filter would be given as;

I_2 = I_1(cos²Φ1) = ((I_p)/2)(cos²Φ1)

Similarly, the intensity of light that will emerge from the third filter would be given as;

I_3 = I_2(cos²Φ1) = ((I_p)/2)(cos²Φ1)(cos²(Φ2 - Φ1)

Thus, making I_p the subject of the formula, we have ;

I_p = (2I_3)/[(cos²Φ1)(cos²(Φ2 - Φ1)]

Plugging in the relevant values, we have;

I_p = (2*60)/[(cos²21)(cos²(61 - 21)]

I_p = 234.65 W/cm²

Now, when the second polarizer is removed, the third polarizer becomes the second and final polarizer so the intensity of light emerging from the stack would be given as;

I_f3 = (I_p/2)(cos²Φ2)

I_f3 = (234.65/2)(cos²61)

I_f3 = 27.58 W/cm²

B) Similarly, when the third polarizer is removed, the second polarizer becomes the final polarizer and the intensity of light emerging from the stack would be given as;

I_f2 = (I_p/2)(cos²Φ1)

I_f2 = (234.65/2)(cos²21)

I_f2 = 102.26 W/cm²

Answer:

(a) g = 8.82158145[tex]m/s^2[/tex].

(b) 7699.990192m/s.

(c)5484.3301s = 1.5234 hours.(extremely fast).

Explanation:

(a) Strength of gravitational field 'g' by definition is

[tex]g = \frac{M_{(earth)} }{r^2} G[/tex] , here G is Gravitational Constant, and r is distance from center of earth, all the values will remain same except r which will be radius of earth + altitude at which ISS is in orbit.

r = 6721,000 meters, putting this value in above equation gives g = 8.82158145[tex]m/s^2[/tex].

(b) We have to essentially calculate centripetal acceleration that equals new 'g'.

[tex]a_{centripetal}=\frac{V^2}{r} =g[/tex] here g is known, r is known and v is unknown.

plugging in r and g in above and solving for unknown gives V = 7699.990192m/s.

(c)  S = vT,  here T is time period or time required to complete one full revolution.

S =  earth's circumfrence , V is calculated in (B) T is unknown.

solving for unknown gives T = 5484.3301s = 1.5234hours.

Answer: 4 kg

Explanation:

Given

Mass of the first shell, m1 = 1 kg

Diameter of the first shell, d1 = 2 m

Radius of the first shell, r1 = 1 m

Diameter of the second shell, d2 = 1 m

Radius of the second shell, r2 = 1/2 m

The moment of inertia of a spherical shell is given by the relation

I = mr²

This means that if two sphere's have the same moment of ineria:

I1 would be equal to I2. And thus

m1.r1² = m2.r2²

If we solve for the second mass m2

m2 = m1.r1²/r2²

m2 = m1 (r1 / r2)² and we substitute the values

m2 = 1 * (1 / 0.5)²

m2 = 2²

m2 = 4 kg

The needed mass of the second shell for their shells to have the same moment of inertia is 4 kg

Answer:

option D.

Explanation:

Given,

distance, d = 1 x 10⁵ km

speed , v = 0.217 c

time of dilation , T₀= ?

Using the formula of time dilation

[tex]T=\frac{T_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]

[tex]T_{0}=T \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]

[tex]=\left(\frac{d}{v}\right) \sqrt{1-\frac{v^{2}}{c^{2}}} [/tex]

[tex]=\left(\frac{1.00 \times 10^{8}}{0.217 \times 2.99 \times 10^{8}}\right) \sqrt{1-\frac{(0.217 c)^{2}}{c^{2}}} [/tex]

[tex]=1.54 \times 0.976 [/tex]

[tex]=1.50\ s[/tex]

Time he should be allowed between passing the buoy and ejecting is equal to 1.50 s.

The correct answer is option D.

Answer:

m = 41.39 m/s

Explanation:

For the source alone , we can write next equation.  

V_r = m/2πr

Stagnation point will occur where r=b, thus;  

V = m/2πb

Now we can find source strength, and therefore we have;  

m = 2*π*b*V

 b = 0.16/2*π

b = 0.0254

m = 2*π*0.32*20.6

m = 41.39 m/s

Final answer:

To simulate flow around a half-body with a thickness of 0.16 m in a 20.6 m/s air stream, the required source strength is 6.592 m^3/s, calculated using the formula Q = 2 * Vθ * d. = 6.592 m3/s

Explanation:

Calculation of Source Strength for Flow Simulation

To simulate flow around a streamlined half-body in a uniform air stream, a combination of a uniform flow and a source flow is used. In the given scenario, the thickness of the half-body is 0.16 m, and it is placed in an air stream moving at 20.6 m/s. The strength of the source, often represented by the symbol Q (and measured in m3/s), is the volume of fluid introduced into the flow per unit of time. For a two-dimensional flow over a half-body, the source strength required to simulate the flow pattern is proportional to the product of the freestream velocity and the thickness of the half-body.

To calculate the required source strength mathematically, you would use the formula:

Q = 2 * V∞ * d

Where V∞ is the freestream velocity, and d is the maximum thickness of the half-body. Substituting the given values:

Q = 2 * 20.6 m/s * 0.16 m

Q = 6.592 m^3/s

Thus, the source strength required to simulate flow around the half-body in uniform flow would be 6.592 m3/s. It is essential to ensure that the flow remains laminar and non-turbulent to maintain the accuracy of this simulation. Factors such as the Reynolds number would also be considered in a more complex analysis to determine the flow regime.

Answer: 1.3 ×10^-31

Explanation:

the required probability is P = e^(-2αL)

Firstly, evaluate (-2αL)

α= 1/hc √2mc^2 (U - E)

h= modified planck's constant

where,

(-2αL)

= -(2L)/(h/2π ) ×√2mc^2 (U - E)

= -(2L) / (hc^2/π )×√2mc^2 (U - E)

(hc^2/2pi) = 197*eV.nm (standard constant)

2*L = 8 nm

mc^2 = 0.511×10^6 eV

Where m = mass electron

C= speed of light

(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5

(-2αL) = -71.1

Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31

Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.