Three polarizing filters are stacked, with the polarizing axis ofthe second and third filters at angles of 22.2^\circ and 68.0^\circ, respectively, to that of thefirst. If unpolarized light is incident on the stack, the light hasan intensity of 75.5 W/cm^2 after it passes through thestack.
a) If the incident intensity is kept constant, what is theintensity of the light after it has passed through the stack if thesecond polarizer is removed?

a. The angle between the reflected part of the beam and the surface of the glass is [tex]\(43.5^\circ\).[/tex]

b. The angle between the refracted beam and the surface of the glass is [tex]\(64.5^\circ\).[/tex]

To solve this problem, we need to apply the laws of reflection and refraction. Let's address each part separately.

Part (a): Angle Between the Reflected Beam and the Surface of the Glass

The law of reflection states that the angle of incidence is equal to the angle of reflection. The angle of incidence is given as 43.5° with respect to the surface of the glass. However, angles in optics are typically measured with respect to the normal (a line perpendicular to the surface).

So, the angle of incidence with respect to the normal (which we'll call [tex]\(\theta_i\)[/tex] ) is:

[tex]\[ \theta_i = 90^\circ - 43.5^\circ = 46.5^\circ \][/tex]

Since the angle of incidence equals the angle of reflection:

[tex]\[ \theta_r = \theta_i = 46.5^\circ \][/tex]

Therefore, the angle between the reflected part of the beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_r = 90^\circ - 46.5^\circ = 43.5^\circ \][/tex]

So, the angle between the reflected beam and the surface of the glass is:

[tex]\[ 43.5^\circ \][/tex]

Part (b): Angle Between the Refracted Beam and the Surface of the Glass

For the refracted beam, we need to apply Snell's Law, which is:

[tex]\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \][/tex]

Where:

- [tex]\( n_1 \)[/tex] is the refractive index of the first medium (air), [tex]\( n_1 = 1.00 \)[/tex],

- [tex]\( \theta_i \)[/tex] is the angle of incidence with respect to the normal, [tex]\( \theta_i = 46.5^\circ \),[/tex]

- [tex]\( n_2 \)[/tex] is the refractive index of the second medium (glass), [tex]\( n_2 = 1.68 \)[/tex],

- [tex]\( \theta_t \)[/tex] is the angle of refraction with respect to the normal.

Using Snell's Law, we can solve for [tex]\(\theta_t\):[/tex]

[tex]\[ 1.00 \sin(46.5^\circ) = 1.68 \sin(\theta_t) \][/tex]

[tex]\[ \sin(\theta_t) = \frac{\sin(46.5^\circ)}{1.68} \][/tex]

Calculating [tex]\(\sin(46.5^\circ)\):[/tex]

[tex]\[ \sin(46.5^\circ) \approx 0.723 \][/tex]

So,

[tex]\[ \sin(\theta_t) = \frac{0.723}{1.68} \approx 0.430 \][/tex]

Now we find [tex]\(\theta_t\):[/tex]

[tex]\[ \theta_t = \sin^{-1}(0.430) \approx 25.5^\circ \][/tex]

The angle between the refracted beam and the surface of the glass is:

[tex]\[ 90^\circ - \theta_t = 90^\circ - 25.5^\circ = 64.5^\circ \][/tex]

So, the angle between the refracted beam and the surface of the glass is:

[tex]\[ 64.5^\circ \][/tex]

Answer:

Explanation:

The problem relates to Compton Effect in which electrons are scattered due to external radiation . The electron is scattered out and photons relating to radiation also undergo scattering at angle θ .

The formula relating to Compton Effect is as follows

[tex]\lambda_f-\lambda_i=\frac{h}{m_0c} (1-cos\theta)[/tex]

Here [tex]\lambda_i[/tex]  = 3 0 x 10⁻¹¹

For longest [tex]\lambda_f[/tex] θ =180°

[tex]\lambda_f[/tex] = [tex]\lambda_i + \frac{2\times h}{m_0c}[/tex]

= .3 x 10⁻⁹ + [tex]\frac{2\times6.6\times 11^{-34}}{9\times10^{-31}\times3\times10^8}[/tex]

= .348 nm

For shortest wavelength θ = 0

Putting this value in the given formula

[tex]\lambda_f=\lambda_i[/tex]

[tex]\lambda_f[/tex] = .3 nm

Answer:

2.771208%

Explanation:

[tex]\Delta V[/tex] = Change of volume

[tex]V_0[/tex] = Initial volume

[tex]\Delta T[/tex] = Change in temperature = (0.3-1050)

[tex]\beta[/tex] = Volumetric coefficient of thermal expansion = [tex]-26.4\times 10^{-6}\ /K[/tex]

Volumetric expansion of heat is given by

[tex]\frac{\Delta V}{V_0}=\beta \Delta T\\\Rightarrow \frac{\Delta V}{V_0}=-26.4\times 10^{-6}\times (0.3-1050)\\\Rightarrow \frac{\Delta V}{V_0}=0.02771208[/tex]

Finding percentage

[tex]\frac{\Delta V}{V_0}=0.02771208\times 100=2.771208\%[/tex]

The ratio of change of volume to initial volume is 2.771208%

To calculate the ratio ΔV/V0 for zirconium tungstate, the volume change can be determined using the volumetric coefficient of thermal expansion. The formula for the ratio is ΔV/V0 = (Volume change/Initial volume) × 100.

The ratio ΔV/V0 can be calculated using the formula:

ΔV/V0 = (Volume change/Initial volume) × 100

Given that the volumetric coefficient of thermal expansion for zirconium tungstate is -26.4 × 10^(-6)/K, we can use this value to calculate the volume change. The volume change can be found by multiplying the coefficient of thermal expansion by the change in temperature:

Volume change = (-26.4 × 10^(-6)/K) × (1050 K - 0.3 K)

Using this value, we can calculate the ratio ΔV/V0:

ΔV/V0 = (Volume change/Initial volume) × 100

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To solve this exercise it is necessary to use the concepts related to Difference in Phase.

The Difference in phase is given by

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

Where

[tex]\delta =[/tex] Horizontal distance between two points

[tex]\lambda =[/tex] Wavelength

From our values we have,

[tex]\lambda = 500nm = 5*10^{-6}m[/tex]

[tex]\theta = 1\°[/tex]

The horizontal distance between this two points would be given for

[tex]\delta = dsin\theta[/tex]

Therefore using the equation we have

[tex]\Phi = \frac{2\pi \delta}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(dsin\theta)}{\lambda}[/tex]

[tex]\Phi = \frac{2\pi(5*!0^{-6}sin(1))}{500*10^{-9}}[/tex]

[tex]\Phi= 1.096 rad \approx = 1.1 rad[/tex]

Therefore the correct answer is C.

The phase difference for waves from the top and bottom of the slit can be calculated by using the formula for calculating phase difference. With the provided values for wavelength, angle and slit width, the calculated phase difference is 0.55 rad.

The phase difference of the waves is directly related to the path difference between them and can be calculated by using the formula:

Φ = 2 π × (d / λ) × sin(θ).

Where φ is the phase difference, π is Pi, d is the slit width, λ is the wavelength and θ is the incident angle.

Let's plug the provided numbers into our formula:

Φ = 2 π × (5.0x10-6 m / 500x10-9 m) × sin(1.0°).

Φ = 2 π × 10 × sin(1.0°)= 0.55 rad.

So, the correct answer is (B) 0.55 rad.

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Answer:

1.6 penny

Explanation:

[tex]V_0[/tex] = Original volume = 1 gal (Assumed)

[tex]\Delta T[/tex] = Change in temperature

[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]

Change in volume is given by

[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]

New volume would be

[tex]1+0.008036=1.008036\ gal[/tex]

The amount of money earned extra would be

[tex]0.008036\times 2=0.016072\ \$[/tex]

1.6 penny more would be earned if the temperature is 32°C

Final answer:

By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.

Explanation:

To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.

The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.

The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.

Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:

ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons

To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:

ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters

Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:

Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies

Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.

Answer: The silver pellet will release heat

Explanation:

Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.

A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.

A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.

Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.

Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:

Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:

(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))

(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)

(1/f) = 0.5 * 0.1667cm⁻¹

(1/f) = 0.08335cm⁻¹

Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:

f ≈ 12cm

Answer:

e. All of these statements are false.

Explanation:

As we know that heat transfer take place from high temperature to low temperature.

It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.

The first law of thermodynamics is the energy conservation law.

Second law of thermodynamics  states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.

By using heat pump ,heat can transfer from cooler body to the hotter body.

Therefore all the answer is False.

The true statement among the given options is that the entropy of a system can be reduced by cooling it.

The correct answer to the student's question regarding true statements about thermodynamics is option (c) It is always possible to reduce the entropy of a system, for instance, by cooling it. This statement aligns with the principles of thermodynamics, which affirm that entropy, a measure of disorder or randomness, can decrease in a system if energy is removed from the system, such as by lowering its temperature. However, this does not violate the second law of thermodynamics because entropy may still increase in the overall process when considering the surroundings.

In contrast, options (a) and (b) are false because they wrongly imply irreversibility in scenarios where reversibility is possible. Specifically, it is possible to reverse an entropy increase by cooling a system and it is also possible to convert some amount of thermal energy back into mechanical energy, although not with 100% efficiency due to inherent thermodynamic losses.

The second law also articulates that heat transfer occurs spontaneously from a higher to a lower temperature body and not in the reverse direction without external work, implying that a spontaneous flow of heat from a colder to a warmer body is impossible, as is complete conversion of heat to work in a cyclical process.

Answer:

[tex]D_{s}[/tex] ≈ 2.1 R

Explanation:

The moment of inertia of the bodies can be calculated by the equation

     I = ∫ r² dm

For bodies with symmetry this tabulated, the moment of inertia of the center of mass

Sphere               [tex]Is_{cm}[/tex] = 2/5 M R²

Spherical shell   [tex]Ic_{cm}[/tex] = 2/3 M R²

The parallel axes theorem allows us to calculate the moment of inertia with respect to different axes, without knowing the moment of inertia of the center of mass

    I = [tex]I_{cm}[/tex] + M D²

Where M is the mass of the body and D is the distance from the center of mass to the axis of rotation

Let's start with the spherical shell, axis is along a diameter

     D = 2R

    Ic = [tex]Ic_{cm}[/tex] + M D²

    Ic = 2/3 MR² + M (2R)²

    Ic = M R² (2/3 + 4)

    Ic = 14/3 M R²

The sphere

    Is =[tex]Is_{cm}[/tex] + M [[tex]D_{s}[/tex]²

    Is = Ic

    2/5 MR² + M [tex]D_{s}[/tex]² = 14/3 MR²

    [tex]D_{s}[/tex]² = R² (14/3 - 2/5)

    [tex]D_{s}[/tex] = √ (R² (64/15)

    [tex]D_{s}[/tex] = 2,066 R

The correct option can be seen in Option A.

The diagrammatic expression of the question can be seen in the image attached below.

From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.

Mathematically;

[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]

Given that the mass of the woman = 60 kg

[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]

[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]

[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]

Equating both (1) and (2) together, we have:

[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]

Dividing like terms on both side

mass (m) = 60 kg

As such, the correct option can be seen in Option A.

Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.

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Answer:v=3.08 m/s

Explanation:

Given

mass of student [tex]m=21 kg[/tex]

distance moved [tex]d=10 m[/tex]

Force applied [tex]F=10 N[/tex]

acceleration of system during application of force is a

[tex]a=\frac{F}{m}=\frac{10}{21}=0.476 m/s^2[/tex]

using [tex]v^2-u^2=2 as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex]v^2-0=2\times 0.476\times 10[/tex]

[tex]v=\sqrt{9.52}[/tex]

[tex]v=3.08 m/s[/tex]

Answer:

[tex]\Delta P=1060184.8946\ Pa[/tex]

[tex]P_1=124651.2383\ Pa[/tex]

Explanation:

Given:

We know the Bernoulli's equation of in-compressible flow:

[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)

Cross sectional area of pipe at location 2:

[tex]A_2=\pi \frac{d_2^2}{4}[/tex]

[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]

[tex]A_2=0.0246\ m^2[/tex]

Cross sectional area of pipe at location 1:

[tex]A_1=\pi \frac{d_1^2}{4}[/tex]

[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]

[tex]A_1=0.0115\ m^2[/tex]

Using continuity equation:

[tex]A_1.v_1=A_2.v_2[/tex]

[tex]0.0115\times 9.83=0.0246\times v_2[/tex]

[tex]v_2=4.5953\ m.s^{-1}[/tex]

Now apply continuity eq. on both the locations:

[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]

[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]

[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]

[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)

Now the mass flow rate at location 1:

[tex]\dot{m_1}=\rho\times \dot{V}[/tex]

[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]

[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]

Now pressure at location 1:

[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]

[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]

[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)

The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as

dP = 114 kPa

Question Parameter(s):

Generally, the Bernoulli's equation   is mathematically given as

P + ρ*g*y + v² =pipe  constant

Where

A1*v1 = A2*v2

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

v2 = 3.9 m/s

Therefore

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

dP = 1290*9.8*9.01 + 9.91² - 3.9²

dP = 114 kPa

In conclusion, difference between fluid pressure is

dP = 114 kPa

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To solve this problem it is necessary to apply Snell's law and thus be able to calculate the angle of refraction.

From Snell's law we know that

[tex]n_1sin\theta_1 = n_2 sin\theta_2[/tex]

Where,

n_i = Refractive indices of each material

[tex]\theta_1[/tex] = Angle of incidence

[tex]\theta_2[/tex] = Refraction angle

Our values are given as,

[tex]\theta_1 = 38\°[/tex]

[tex]n_1 = 1[/tex]

[tex]n_2 = 1.4[/tex]

Replacing

[tex]1*sin38 = 1.4*sin\theta_2[/tex]

Re-arrange to find [tex]\theta_2[/tex]

[tex]\theta_2 = sin^{-1} \frac{sin38}{1.4}[/tex]

[tex]\theta_2 = 26.088°[/tex]

Therefore the  angle will the beam make with the normal in the glass is 26°

To solve the problem it is necessary to apply the concepts related to the flow rate of a fluid.

The flow rate is defined as

[tex]Q = Av[/tex]

Where,

[tex]Q = Discharge (m^3/s)[/tex]

[tex]A = Area (m^2)[/tex]

v = Average speed (m / s)

And also as

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

Where,

V = Volume

t = time

Let's start by finding the total volume according to the given dimensions, that is to say

[tex]V = 5*11*2.4[/tex]

[tex]V = 132m^3[/tex]

The entire cycle must be completed in 50 min = 3000s

In this way we know that the [tex]132m ^ 3[/tex] must be filled in 3000s, that is to say that there should be a flow of

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

[tex]Q = \frac{132}{3000}[/tex]

[tex]Q = 0.044m^3/s[/tex]

Using the relationship to find the speed we have to

[tex]Q = Av[/tex]

[tex]v = \frac{Q}{A}[/tex]

Replacing with our values,

[tex]v = \frac{0.044}{900*10^{-4}m^2}[/tex]

[tex]v = 0.488m/s[/tex]

Therefore the air speed in the duct must be 4.88m/s

Answer:

-0.31563 m/s

1.16437 m/s

Explanation:

[tex]m_1[/tex] = Mass of girl = 45 kg

[tex]m_2[/tex] = Mass of plank = 166 kg

[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s

[tex]v_2[/tex] = Velocity of the plank relative to ice surface

In this system the linear momentum is conserved

[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]

Velocity of the plank relative to ice surface is -0.31563 m/s

Velocity of the girl relative to the ice surface is

[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]

Answer:

safe speed for the larger radius track u= √2 v

Explanation:

The sum of the forces on either side is the same, the only difference is the radius of curvature and speed.

Also given that r_1= smaller radius

r_2= larger radius curve

r_2= 2r_1..............i

let u be the speed of larger radius curve

now, [tex]\sum F = \frac{mv^2}{r_1} =\frac{mu^2}{r_2}[/tex]................ii

form i and ii we can write

[tex]v^2= \frac{1}{2} u^2[/tex]

⇒u= √2 v

therefore, safe speed for the larger radius track u= √2 v

Answer:

[tex]v_s=27.8m/s[/tex]

Explanation:

If the person hearing the sound is at rest, then the equation for the frequency heard [tex]f[/tex] given the emitted frequency [tex]f_0[/tex], the speed of the truck [tex]v_s[/tex] and the speed of sound [tex]c[/tex] will be:

[tex]f=f_0\frac{c}{c+v_s}[/tex]

Where [tex]v_s[/tex] will be positive if the truck is moving away from the person, and negative otherwise. We then do:

[tex]\frac{f}{f_0}=\frac{c}{c+v_s}[/tex]

[tex]\frac{f_0}{f}=\frac{c+v_s}{c}=1+\frac{v_s}{c}[/tex]

[tex]v_s=c(\frac{f_0}{f}-1)=(340m/s)(\frac{238Hz}{220Hz}-1)=27.8m/s[/tex]

Answer:

Angular velocity will be 2.843 rad/sec

Explanation:

We have given the radius r = 5.17 m

Centripetal acceleration [tex]a_c=1.5g=1.5\times 9.8=14.7m/sec^2[/tex]

We know that centripetal acceleration is given by

[tex]a_c=\frac{v^2}{r}[/tex]

And linear velocity is given by [tex]v=\omega r[/tex]

[tex]a_c=\frac{(\omega r)^2}{r}=\omega ^2r[/tex]

[tex]14.7=\omega ^2\times 5.17[/tex]

[tex]\omega =2.843rad/sec[/tex]

Answer:

[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

Explanation:

Normal reaction from 40 kg slab on 10 kg block

M × g  = 10 × 9.8 = 98 N  

Static frictional force = 98 × 0.7 N

Static frictional force = 68.6 N is less than 100 N applied  

10 kg block will slide on 40 kg slab and net force on it  

= 100 N - kinetic friction  

[tex]=100-(98 \times 0.4)\left(\mu_{\text {kinetic }}=0.4\right)[/tex]

= 100 - 39.2

= 60.8 N

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with, } \frac{\mathrm{Net} \text { force }}{\text { mass }}[/tex]

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=\frac{60.8}{10}[/tex]

[tex]10 \mathrm{kg} \text { block will slide on } 40 \mathrm{kg} \text { slab with }=6.08 \mathrm{m} / \mathrm{s}^{2}[/tex]

[tex]\text { Frictional force on 40 kg slab by 10 kg block, normal reaction \times \mu_{kinetic } }[/tex]

Frictional force on 40 kg slab by 10 kg block = 98 × 0.4  

Frictional force on 40 kg slab by 10 kg block = 39.2 N  

[tex]40 \mathrm{kg} \text { slab will move with } \frac{\text { frictional force }}{\text { mass }}[/tex]

[tex]40 \mathrm{kg} \text { slab will move with }=\frac{39.2}{40}[/tex]

40 kg slab will move with = [tex]0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

[tex]\text { The "resulting action" on the slab is } 0.98 \mathrm{m} / \mathrm{s}^{2}[/tex]

Answer:

a) 1.13 10-8 T.  b) +y direction

Explanation:

a)

For an electromagnetic wave propagating in a vacuum, the wave speed is c = 3. 108 m/s.

At a long distance from the source, the components of the wave (electric and magnetic fields) can be considered as plane waves, so the equations for them can be written as follows:

E(z,t) = Emax cos (kz-ωt-φ) +x

B(z,t) = Bmax cos (kz-ωt-φ) +y

In an electromagnetic wave, the magnetic field and the electric field, at any time, and at any point in space, as the perturbation is propagating at a speed equal to c (light speed in vacuum), are related by this expression:

Bmax = Emax/c

So, solving for Bmax:

Bmax = 3.4 V/m / 3 108 m/s = 1.13 10-8 T.

b) As we have already said, in an electromagnetic wave, the electric field and the magnetic field are perpendicular each other and to the propagation direction, so in this case, the magnetic field propagates in the +y direction.

To take place the process of nuclear fusion basically seeks to reach heavy nuclei through light nuclei. Reaching this process implies a release of energy that is what makes this process attractive because it is possible to obtain significant volumes of energy. The procedure to arrive at this process also implies a high cost concerning high temperatures and exorbitant pressures as it is necessary to be able to overcome the barrier of electrostatic repulsion.

This process does not generate any type of radioactive waste like other processes, therefore it is not as dangerous as nuclear fission. For this reason the correct answer is A. Fusion products are generally not radioactive.

Answer:

a) v = 7,207 m / s

, b) a = 42.3 m / s²

Explanation:

We will solve this exercise using the concept of mechanical energy, We will write it in two points before the car touches the springs and in point of maximum compression

Initial

    Em₀ = K = ½ m v²

Final  

    [tex]Em_{f}[/tex] = 2 Ke = ½ k x²

The two is placed because each barred has two springs and each does not exert the same force

    Emo = [tex]Em_{f}[/tex]

     ½ m v² = 2 ½ k x²

     v = √(2k/m) x

     v = √ (2 3,134 10⁵/4550) 0.614

     v = 7,207 m / s

Let's take this speed to km / h

     v = 5,096 m / s (1km / 1000m) (3600s / 1h)

     v = 25.9 km / h

This speed is common in school zones

Let's use kinematics to calculate the average acceleration

      vf² = v₀² - 2 a x

       0 = v₀² - 2 a x

       a = v₀² / 2 x

       a = 7,207²/2 0.614

       a = 42.3 m / s²

We buy this acceleration with the acceleration of gravity

       a / g = 42.3 / 9.8

       a / g = 4.3

This acceleration is well below the maximum allowed

When the separation between the plates of a parallel plate capacitor is increased, the amount of charge on the plates decreases due to the decrease in capacitance (option e), with the voltage remaining constant.

When parallel plate capacitor plates are pulled away from each other while connected to a battery maintaining a constant potential difference, the capacitance decreases. This is because the capacitance is inversely proportional to the distance between the plates. As the capacitance decreases, the charge on the plates also decreases since the voltage (V) remains constant, and the relation between charge (Q), capacitance (C), and voltage (V) is given by Q = CV. Therefore, the amount of charge on the plates decreases because the capacitance decreases (option e).

In order to solve this problem it is necessary to apply the concepts related to intensity and specifically described in Malus's law.

Malus's law warns that

[tex]I = I_0 cos^2\theta[/tex]

Where,

[tex]\theta=[/tex] Angle between the analyzer axis and the polarization axis

[tex]I_0 =[/tex]Intensity of the light before passing through the polarizer

The intensity of the beam from the first polarizer is equal to the half of the initial intensity

[tex]I = \frac{I_0}{2}[/tex]

Replacing with our the numerical values we get

[tex]I = \frac{46}{2}[/tex]

[tex]I = 23W/m^2[/tex]

Therefore the  intensity of the light that emerges from the filter is [tex]23W/m^2[/tex]

Final answer:

The rate at which the concrete loses thermal energy by conduction through the air layer can be calculated using Fourier's Law of Heat Conduction. The formula involves the thermal conductivity, area, temperature difference, and thickness of the air layer. However, without the thermal conductivity value for air, the calculation cannot be completed.

Explanation:

To calculate the rate at which the concrete slab loses thermal energy by conduction through the surrounding air layer at sunset, we can apply Fourier's Law of Heat Conduction. This law states that the heat transfer rate (Q) through a material is directly proportional to the temperature difference across the material (ΔT), the area through which heat is being transferred (A), and the thermal conductivity (k), and inversely proportional to the thickness of the material (L).

The formula to calculate the rate of heat loss is given by Q = k*A*(ΔT/L), where ΔT is the temperature difference between the two sides of the material, A is the contact area, k is the thermal conductivity of the material, and L is the thickness of the material.

Unfortunately, without the thermal conductivity value for air in the provided data, we cannot calculate the exact rate of heat loss for this specific scenario. Thermal conductivity is required to solve this problem, and it's typically obtained from tables in textbooks or scientific references.

Answer:

[tex]27.57713\ MPa\sqrt{m}[/tex]

Explanation:

Y = Fracture parameter

a = Crack length

[tex]\sigma[/tex] = Stress in part

Plane strain fracture toughness is given by

[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow Y=\frac{K_I}{\sigma\sqrt{\pi a}}\\\Rightarrow Y=\frac{39}{270\times \sqrt{\pi 0.00282}}\\\Rightarrow Y=1.53462[/tex]

When a = 1.41 mm

[tex]K_I=Y\sigma\sqrt{\pi a}\\\Rightarrow K_i=1.53462\times 270\sqrt{\pi 0.00141}\\\Rightarrow K_I=27.57713\ MPa\sqrt{m}[/tex]

The value of plane strain fracture toughness is [tex]27.57713\ MPa\sqrt{m}[/tex]

Answer:

The gauge pressure is 1511.11 psi.

Explanation:

Given that,

Flow rate = 94 ft³/min

Diameter d₁=3.3 inch

Diameter d₂ = 5.2 inch

Pressure P₁= 15 psi

We need to calculate the pressure on other side

Using Bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2[/tex]

We know that,

[tex]V=Av[/tex]

[tex]v=\dfrac{V}{A}[/tex]

Where, V = volume

v = velocity

A = area

Put the value of v into the formula

[tex]P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2[/tex]

Put the value into the formula

[tex]15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2[/tex]

[tex]P_{2}=1525.8\ psi[/tex]

We need to calculate the gauge pressure

Using formula of gauge pressure

[tex]P_{g}=P_{ab}-P_{atm}[/tex]

Put the value into the formula

[tex]P_{g}=1525.8-14.69[/tex]

[tex]P_{g}=1511.11\ psi[/tex]

Hence, The gauge pressure is 1511.11 psi.

Answer:

449.37412 N

Explanation:

G = Gravitational constant = 6.67259 × 10⁻¹¹ m³/kgs²

m = Mass of satellite = 438 kg

M = Mass of planet

T = Time period of the satellite = 24 h

r = Radius of planet = [tex]1.94\times 10^8\ m[/tex]

The time period of the satellite is given by

[tex]T=2\pi\sqrt{\frac{r^3}{GM}}\\\Rightarrow M=4\pi^2\frac{r^3}{T^2G}\\\Rightarrow M=4\pi^2\times \frac{(1.94\times 10^8)^3}{(24\times 3600)^2\times 6.67259\times 10^{-11}}\\\Rightarrow M=5.78686\times 10^{26}\ kg[/tex]

The gravitational force is given by

[tex]F=G\frac{Mm}{r^2}\\\Rightarrow F=6.67259\times 10^{-11}\times \frac{5.78686\times 10^{26}\times 438}{(1.94\times 10^8)^2}\\\Rightarrow F=449.37412\ N[/tex]

The force acting on this satellite is 449.37412 N

Gravity, is often known as gravitation. The gravitational force between the planet and the satellite can be written as 0.4332 N.

Gravity, often known as gravitation, is the universal force of attraction that acts between all matter in mechanics. It is the weakest known force in nature, and so has no bearing on the interior properties of ordinary matter.

[tex]F =G\dfrac{m_1m_2}{r^2}[/tex]

As it is given that the value of the gravitational constant is G = 6.67259 × 10−11 N m² /kg², while the radius of the planet is 1.94×10⁵. And the time taken by the satellite to revolve around the planet is 24 hours. therefore, the Mass of the planet can be written as,

The Time period of the satellite is given as:

[tex]T = 2\pi \sqrt{\dfrac{r^3}{GM}}[/tex]

Substitute the values in the formula,

[tex]24 = 2\pi \sqrt{\dfrac{(1.94 \times 10^5)^3}{6.67259 \times 10^{-11}M}}\\\\M = 5.78686 \times 10^{17}\rm\ kg[/tex]

Thus, the mass of the planet can be written as 5.5786×10¹⁷ kg.

Now, the gravitational force can be written as,

[tex]F = G\dfrac{Mm}{r^2}\\\\F = 6.67259\times 10^{-11} \times \dfrac{5.5786 \times 10^{17} \times 438}{1.94 \times 10^5}\\\\ F = 0.4332\rm\ N[/tex]

Hence, the gravitational force between the planet and the satellite can be written as 0.4332 N.

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Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.

The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.

Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.

Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.

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The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.

To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.

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