Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm^ 2 after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?

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:

d. a2 = 9a1

Explanation:

We can apply the following equation of motion to calculate the angular acceleration:

[tex]\omega^2 - \omega_0^2 = 2\alpha\theta[/tex]

Since both wheel starts from rest, their [tex]\omega_0 = 0 rad/s[/tex]

[tex]\omega^2 = 2\alpha\theta[/tex]

We can take the equation for the 1st wheel, divided by the equation by the 2nd wheel:

[tex]\frac{\omega_1^2}{\omega_2^2} = \frac{2\alpha_1\theta_1}{2\alpha_2\theta_2}[/tex]

As they were rotating through the same angular displacement [tex]\theta_1 = \theta_2[/tex], these 2 cancel out

[tex]\left(\frac{\omega_1}{\omega_2}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\left(\frac{1}{3}\right)^2 = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\frac{1}{9} = \frac{\alpha_1}{\alpha_2}[/tex]

[tex]\alpha_2 = 9\alpha_1[/tex]

So d is the correct answer

The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.

It can be defined as the rate of change in the angular velocity of an object or body. It can be calculated by the equation of motion:

[tex]\omega ^2 - \omega _0^2 = 2\alpha \theta[/tex]

Since initial angular rotation is zero for both the wheels,

[tex]\omega ^2 = 2\alpha \theta[/tex]

Compare the angular acceleration of both wheels,

[tex]\dfrac {\omega_1^2}{\omega_2^2} = \dfrac {2\alpha_1 \theta}{2\alpha_2\theta }[/tex]

Put the values,

[tex]\begin{aligned} (\dfrac 13)^2&= \dfrac {\alpha_1 }{\alpha_2 }\\\\ \dfrac 19 &= \dfrac {\alpha_1 }{\alpha_2 }\\\\\alpha_2 &= 9\alpha_1 \end {aligned}[/tex]

Therefore, The angular acceleration of the first wheel is four times higher than that of the second. Option D is correct.

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Answer:

λ = 3.2 x 10⁻⁷ m = 320 nm

Explanation:

The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:

v = fλ

where,

v = c = speed of the electromagnetic waves (UV rays) = speed of light

c = 3 x 10⁸ m/s

f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz

λ = wavelength of the electromagnetic waves (UV rays) = ?

Therefore, substituting the values in the relation, we get:

3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)

λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)

λ = 3.2 x 10⁻⁷ m = 320 nm

So, the radiation of 320 nm wavelength is absorbed by Ozone.

Answer:

0.03052 nm

Explanation:

From c = f¥

Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s

f = frequency of the wave

¥ = wavelenght of the wave.

¥ = c/f = (3x10^8)/(9.38×10^14)

= 3.052x10^-7 m

= 0.03052 nm

Answer:

A) The thickness of the glass is 868 nm

B) The wavelength is 3472 nm

Explanation:

We are given;

Refractive index = 1.30

Wavelength = 496 nm

Next wavelength = 386 nm

Now, we need to calculate the thickness of the glass

Using formula for constructive interference which is given as;

2nt = (m + ½)λ

Where;

m is the order of the interference

λ is wavelength

t is thickness

Now, for the first wavelength, we have;

2nt = (m + ½)496 - - - - eq1

for the second wavelength, we have;

2nt = (m + 1 + ½)386

2nt = (m + 3/2)386 - - - - eq2

Thus, combining eq1 and eq2, we have;

(m + ½)496 = (m + 3/2)386

496m + 248 = 386m + 579

496m - 386m = 579 - 248

110m = 331

m = 331/110

m = 3

Put 3 for m in eq 1;

2nt = (3 + ½)496

2nt = 1736

t = 1736/(2 x 1)

t = 868 nm

B) now we need to calculate the longest wavelength.

From earlier, we saw that ;

2nt = (m + ½)λ

Making wavelength λ the subject, we have;

λ = 2nt/(m + ½)

The longest wavelength will be at m = 0

Thus,

λ = (2 x 1 x 868)/(0 + ½)

λ = 3472 nm

Answer:

a) [tex]v = 2.886\,\frac{m}{s}[/tex], b) [tex]\mu_{k} = 0.014[/tex]

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

[tex](62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v[/tex]

[tex]v = 2.886\,\frac{m}{s}[/tex]

b) The deceleration experimented by the system person-sled is:

[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}[/tex]

[tex]a = -0.139\,\frac{m}{s^{2}}[/tex]

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

[tex]-\mu_{k}\cdot m\cdot g = m\cdot a[/tex]

[tex]\mu_{k} = -\frac{a}{g}[/tex]

[tex]\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }[/tex]

[tex]\mu_{k} = 0.014[/tex]

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           [tex]Em_{f}[/tex] = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

Answer:

a) [tex]\mu_{k} = 0.704[/tex], b) [tex]R = 0.312\,m[/tex]

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s[/tex]

[tex]\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}[/tex]

[tex]\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}[/tex]

[tex]\mu_{k} = 0.704[/tex]

b) The speed of the block is determined by using the Principle of Energy Conservation:

[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]

[tex]v = x\cdot \sqrt{\frac{k}{m} }[/tex]

[tex]v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }[/tex]

[tex]v \approx 9.829\,\frac{m}{s}[/tex]

The radius of the circular loop is:

[tex]\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}[/tex]

[tex]\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}[/tex]

[tex]R = 0.312\,m[/tex]

The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi

It is given that l = 12 in

The force F = 60 lb

The radiuses are as follows:

outer radius, R = 3.90 in/2

R = 1.95 in

Inner radius, r = 3.65 in/2

r = 1.825 in

The angle between the force applied and the distanced from the axis is 30°

So we get the torque:

T = l×Fsin 30°

T = 12 × 60 × (0.5)

T = 360 lb-in

Now, the angle of rotation or angular displacement ωt is given as

ωt = π(R⁴ - r⁴)/(2R)

ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)

ωt = 2.7114 in³

Then the principal stress in the pipe at point A is:

principal stress = T/ωt

principal stress = (360 lb-in)/(2.7114 in³)

principal stress = 132.773 lb/in²

principal stress = 132.773 psi

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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]

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:

 k = 11,564 N / m,   w = 6.06 rad / s

Explanation:

In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;

 let's apply the equilibrium condition at this point

Axis y

          W_{y} - Fr = 0

          Fr = k y

let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal

             sin 46 = [tex]W_{y}[/tex] / W

             W_{y} = W sin 46

 we substitute

           mg sin 46 = k y

           k = mg / y sin 46

If the length of the bar is L

          sin 46 = y / L

           y = L sin46

we substitute

           k = mg / L sin 46 sin 46

           k = mg / L

for an explicit calculation the length of the bar must be known, for example L = 1 m

           k = 1.18 9.8 / 1

           k = 11,564 N / m

With this value we look for the angular velocity for the point tea = 30º

let's use the conservation of mechanical energy

starting point, higher

          Em₀ = U = mgy

end point. Point at 30º

         [tex]Em_{f}[/tex] = K -Ke = ½ I w² - ½ k y²

          em₀ = Em_{f}

          mgy = ½ I w² - ½ k y²

          w = √ (mgy + ½ ky²) 2 / I

the height by 30º

           sin 30 = y / L

           y = L sin 30

           y = 0.5 m

the moment of inertia of a bar that rotates at one end is

          I = ⅓ mL 2

          I = ½ 1.18 12

          I = 0.3933 kg m²

let's calculate

          w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)

          w = 6.06 rad / s

Answer:

3.78V

Explanation:

The resistance in the circuit is

The potential difference across the 4.00 V battery, Vbc, in a circuit where it is in parallel with a series combination of 10.30 V and 8.00 V batteries, is equal to the total emf of the two batteries in series minus the potential drop due to their current through the 0.50 Ω internal resistance.

To determine the potential difference Vbc across the terminals of the 4.00 V battery, we first need to understand the circuit's behavior. The circuit now consists of two batteries connected in series - one with 10.30 V (internal resistance 0.50 Ω) and another of 8.00 V - and the 4.00 V battery is in parallel with this series combination.

Because the batteries in series share the same current, we can calculate the total current (I) using Ohm's law and considering the total emf and resistance:

I= (10.30 V + 8.00 V)/( R + 0.50 Ω)

Note that R includes any circuit resistance (which hasn't been mentioned) and the 4.00 V battery's own internal resistance. Therefore, Vbc, the potential difference across the 4.00 V battery will be equal to the total potential provided by the series batteries minus the potential drop due to their current through the 0.50 Ω internal resistance.

Vbc = (10.30 V + 8.00 V) - I * 0.50 Ω

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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:

The object strikes the surface of the Earth with a speed of approximately 11.2 km/s.

Ignoring atmospheric friction, the object's total mechanical energy (the sum of its gravitational potential energy and kinetic energy) is conserved throughout its fall. We can use this principle to calculate the object's speed at the Earth's surface.

Here's how:

Calculate the gravitational potential energy (PE) at the initial altitude:

PE_i = -G * M_earth * m / (R_earth + h)

where:

G is the gravitational constant (6.67 × 10^-11 N m²/kg²)

M_earth is the mass of the Earth (5.97 × 10^24 kg)

m is the mass of the object (910 kg)

R_earth is the Earth's polar radius (6357 km)

h is the initial altitude (1200 km)

Set the final PE (PE_f) to 0:

Since all the PE will be converted to kinetic energy when the object reaches the surface, PE_f = 0.

Apply the conservation of mechanical energy:

E_i = PE_i + KE_i = PE_f + KE_f = KE_f (as PE_i = PE_f = 0)

where:

KE_i is the initial kinetic energy (0 J, as the object is at rest)

KE_f is the final kinetic energy

Solve for the final velocity (v_f):

KE_f = 1/2 * m * v_f^2

v_f = sqrt(2 * E_i / m)

Plug in the values and calculate v_f:

v_f = sqrt(2 * (-G * M_earth * m / (R_earth + h)) / m)

v_f ≈ 11,180 m/s

Convert m/s to km/s:

v_f ≈ 11.2 km/s

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.

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!!

The energy differences between the levels of the quantum objects and the photons in the visible spectrum determine the dark spectral lines in an absorption spectrum.

In order to determine the energies of the dark spectral lines in an absorption spectrum, we need to consider the energy differences between the energy levels of the quantum objects and the energy levels of the photons in the visible spectrum. The energy levels of the quantum objects are -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. The range of photon energies for visible light is 1.63 to 3.26 eV. Only photons with energies matching the energy differences between the levels of the quantum objects will be absorbed, resulting in dark lines in the spectrum.

Based on the given information, the dark spectral lines within the visible spectrum would be:

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Answer:

35.35 is the final intensity

Explanation:

See attached file for calculation

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:

(D) 3

Explanation:

The angular momentum is given by:

[tex]\vec{L}=\vec{r}\ X \ \vec{p}[/tex]

Thus, the magnitude of the angular momenta of both solar systems are given by:

[tex]L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}[/tex]

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

[tex]L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1[/tex]

but L1=L2=L:

[tex]L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3[/tex]

Hence, the answer is (D) 3

HOPE THIS HELPS!!

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:

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:

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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Answer:

Check attachment

The question have two distance

I decided to use the one in the question "a" in attachment and I will use the other one here

Explanation:

Given that,

Speed of particle relative to the earth is

V = 0.99543c

Where c is speed of light

c = 3 × 10^8 m/s

Particle height as detected by scientist is 43km

The initial length is 43km

Lo = 43km

Lo = 43,000m

A. Time taken for the particle to reach the earth surface?

Speed = distance / Time

Time = distance / speed

t = L / V

t = 43,000 / 0.99543c

t = 43,000 / (0.99543 × 3 × 10^8)

t = 1.4399 × 10^-4 seconds

b. Initial Lenght is given as Lo = 43km

Using length contraction formula

L = Lo√(1 — u² / c²)

L = 43√[1 — (0.99543c)² / c²]

L = 43√[1 — 0.990881c² / c²]

L = 43√[1 — 0.990881]

L = 43 × √(9.1191 × 10^-3)

L = 43 × 0.095494

L = 4.1062 km

c. Using time dilation formula

∆to = ∆t√(1 — u² / c²)

∆t is gotten from question a

∆t = 1.4399 × 10^-4 seconds

∆to = ∆t√[1 — (0.99543c)² / c²]

∆to = ∆t√[1 — 0.990881c² / c²]

∆to = ∆t√[1 — 0.990881]

∆to = ∆t × √(9.1191 × 10^-3)

∆to = 1.4399 × 10^-4 × 0.095494

∆to = 1.375 × 10^-5 seconds

To check if the time dilation agree

t = L / V

t = 4.1062 × 1000 / 0.99543c

t = 4.1062 × 1000 / 0.99543 × 3 × 10^8

t = 1.375 × 10^-5 seconds

The time dilation agreed

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:

See explanation

Explanation:

Solution:-

Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.

Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.

( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

                                  [tex]B = \frac{uo*I}{2\pi *r}[/tex]

Where,

I : The current,

r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7  T. m/A

-  Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.

- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:

                             [tex]B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}[/tex]      

Where the integral sums over,

 1) The wire length where vector dl = direction of current (in or out of plane)

 2) r is the distance between the location of dl and the location at which the magnetic field is being calculated

 3)  r^ is a unit vector in the direction of r.

Answer:

B

Explanation:

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

The required time is: 0.395 s. Weight and wavelengths are 0.013 and 1.5. The equation for waves traveling down the string is: y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

a) To find the time it takes for a pulse to travel the full length of the string, we need to determine the velocity of the wave. The velocity of a wave on a string is given by the equation [tex]v = \sqrt{(T/\mu)[/tex], where T is the tension and μ is the linear mass density of the string.  

Given that the weight is 0.0125 N and the length is 1.50 m.

Here, mass = w/g = 0.0125/9.8 = 1.275 x 10⁻³ kg

we can calculate μ as follows: μ = m/L = (1.275 x 10⁻³)/(1.50 m) = 0.00085 kg/m.

Substituting the values into the equation:

we get v = [tex]\sqrt{W/0.00085}[/tex]. As it is given that the tension of the string is constant and equal to W.

Thus, v = [tex]\sqrt{14.7}[/tex] = 3.8 m/s

The time it takes for a pulse to travel the full length of the string can be calculated using the formula time = distance/velocity = 1.50/3.8. So, time = 0.395 s.

b) In case of string, V² = F/μ

μ x V² = F

This implies:

F = 0.00085 x [tex](3.8)^2[/tex] = 0.0123

C)  We know, f (frequency) = 1/t = 1/0.395 = 2.53

Also, v = λf. This implies:

λ = v/f = 3.8/2.53 = 1.5

D) Finally, to find the equation for waves traveling down the string, we need to adjust the sign inside the cosine function to account for the wave traveling in the opposite direction. It is given that:

y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

So, the equation becomes y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

Complete Question:

Question

A 1.50-m string of weight 0.0125 N is tied to the ceil-ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.

y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).

(a) How much time does it take a pulse to travel the full length of the string?

(b) What is the weight W?

(c) How many wavelengths are on the string at any instant of time?

(d) What is the equation for waves traveling ?down? the string?

Answer:

Explanation:

The magnitude of electric field = 8 x 10⁴ V /m

there is a potential difference of 8 x 10⁴ V on a separation of 1 m

so on a separation of .5 m , potential drop or change in potential will be equal to

.5 x 8 x 10⁴

= 4 x 10⁴ V .

The increase in kinetic energy of proton = V X Q

V is potential drop x Q is charge on proton

= 4 x 10⁴ x 1.6 x 10⁻¹⁹

= 6.4 x 10⁻¹⁵ J

potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of

-  6.4 x 10⁻¹⁵ J .

The change in electric potential is 4.0 x 10^4 volts. The change in potential energy of the proton-field system, given the charge of a proton, is calculated to be 6.4 x 10^-15 joules.

The change in electric potential (or voltage) is calculated by multiplying the electric field strength by the displacement, so in this case it is (8.0 x 10^4 V/m) * (0.50 m) = 4.0 x 10^4 volts.

The change in potential energy of the proton-field system can be found by multiplying the change in electric potential by the charge of a proton. The charge of a proton is 1.6 x 10^-19 C, so the change in potential energy is (4.0 x 10^4 volts) * (1.6 x 10^-19 C), which equals to 6.4 x 10^-15 joules.

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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:

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]