Large numbers of ribosomes are present in cells that specialize in producing which molecules?

Answer:

Part A:  17 m

Part B:  1.7 x 10⁻²m

Part C:  1.7 x 10⁻⁵m

Explanation:

Equation needed to answer the question:

 λ = [tex]\frac{v}{f}[/tex]

where λ is the wavelength, f is the frequency and v is the velocity,

Part A:

Substitute velocity with the speed of sound: 343 m/s

Substitute frequency with the value given: 20 Hz

 λ = [tex]\frac{343m/s}{20Hz}[/tex] = 17 m

Part B:

Substitute velocity with the speed of sound: 343 m/s

Substitute frequency with the value given: 20,000 Hz

 λ = [tex]\frac{343m/s}{20,000Hz}[/tex] = 1.7 x 10⁻²m

Part C:

Substitute velocity with the speed of sound: 343 m/s

Substitute frequency with the value given: 20,000 Hz

 λ = [tex]\frac{343m/s}{20MHz (\frac{10^{6}Hz}{1MHz})}[/tex] = 1.7 x 10⁻⁵m

To calculate the wavelength of sound waves at different frequencies, we can use the formula v = fλ, where v is the speed of sound and f is the frequency. Part A: The wavelength of a 20-Hz wave is 17.15 m. Part B: The wavelength of a 20,000-Hz wave is 0.01715 m. Part C: The wavelength of a 20-MHz wave is 0.00001715 m.

To calculate the wavelength of a sound wave, we can use the formula v = fλ, where v is the speed of sound and f is the frequency of the wave. Part A: Given the speed of sound in air at 20°C is 343 m/s and the frequency is 20 Hz, we can rearrange the formula to solve for the wavelength: λ = v/f = 343 m/s / 20 Hz = 17.15 m. Part B: For a frequency of 20,000 Hz, we can use the same formula: λ = 343 m/s / 20,000 Hz = 0.01715 m. Part C: For a frequency of 20 MHz, we can once again use the formula: λ = 343 m/s / 20,000,000 Hz = 0.00001715 m.

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Answer: The effect of Bessemer process is its reduction in cost for steel production.

Explanation:

In the manufacturing of steel, Bessemer process was the first method discovered for mass production of steel. This was discovered by Sir Henry Bessemer and Williams Kelly both from United States. This method aids in the removal of impurities from iron and converts it to steel in few minutes (this usually takes a full day to achieve). The economy of the country improved as steel was made faster and cheaper causing companies to build thousands of new railroads.

Answer:

54.17volts

Explanation:

Induced emf in a coil placed in a magnetic field can be expressed as E = N¶/t where

N is the number of turns = 150turns

¶ is the magnetic flux = magnetic field strength (B) × area(A)

¶ = BA

B = 0.65T

A = 1.0m²

t is the time =1.8s

Substituting this value in the formula

E = NBA/t

E = 150×0.65×1.0/1.8

E = 54.17Volts

The induced emf in the coil is 54.17Volts

The induced emf in the coil is calculated using Faraday's Law of electromagnetic induction, which states that the induced emf in a coil equals the negative change in magnetic flux through the coil times the number of turns in the coil. In the student's scenario, one would use the formula E = -N × (ΔB/Δt) × A with the provided values to get the induced emf.

The induced electromotive force (emf) in a coil is calculated using Faraday's Law of electromagnetic induction. This law states that the induced emf in a coil is equal to the negative change in magnetic flux through the coil times the number of turns in the coil. The magnetic flux (Φ) is defined as the product of the magnetic field (B) and the area (A) that it penetrates, and it can be represented by the equation Φ = B × A, provided that the magnetic field is perpendicular to the area.

To find the induced emf (E) in the given scenario where a coil with 150 turns has a cross-sectional area of 1.00 m2 and experiences a change in magnetic field strength of 0.65 T over 1.80 s, we use the formula:

E = -N × (ΔB/Δt) × A

Where:

Substituting these values, we would get the induced emf.

Answer:

1.18 m/s²

Explanation:

The linear speed = 13.0s

[tex]v= v_i + at\\v = 0 + 0.42(13)\\v = 5.46m/s[/tex]

where v(i) is the initial linear speed

the radial acceleration of the car afte (t) = 13s is

[tex]a_r = \frac{v^2}{r} \\= \frac{5.46^{2} }{27} \\= 1.104m/s^2[/tex]

the magnitude of the net acceleration is

[tex]a_net = \sqrt{1.104^2 + 0.1764^2} \\= 1.18m/s^2[/tex]

Answer:

686.70 kg m² / s

Explanation:

L= m * r²

= 51 * 2.7²

= 371.79 kg m²

w = v/r

= 5.0 m/s / 2.7 m

= 1.85 rev/sec

The momentum = I * w

= 371.79 * 1.85

= 686.70 kg m² / s

The angular momentum around the center of the merry-go-round is 686.70 kg m² / s

Since her mother who is 51  kg stand next to her on the ride, 2.7 m from the merry-go-round's center. And, the mother speed is 5.0 m/s.

So, the following formulas should be applied

L= m * r²

= 51 * 2.7²

= 371.79 kg m²

Now

w = v/r

= 5.0 m/s / 2.7 m

= 1.85 rev/sec

So finally

The momentum = I * w

= 371.79 * 1.85

= 686.70 kg m² / s

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Answer: 10.53 seconds

Explanation:

in the attachment

Explanation:

Below is an attachment containing the solution.

Answer:

Granite is Intrusive Igneous.

Explanation:

The key words here are "Deep below the Earth's crust", so that gives you the fact it is intrusive. Anything with magma is also most likely hinting towards Intrusive igneous.

Answer:

option D is correct

Explanation:

It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to move a charge along an equipotential, since ΔV = 0. Thus the work is :

                                W = −ΔPE = −qΔV = 0.

Work is zero if force is perpendicular to motion. Force is in the same direction as E, so that motion along an equipotential must be perpendicular to E. More precisely, work is related to the electric field by:

                                 W = Fd cos θ = qEd cos θ = 0.

- The change in kinetic energy Δ K.E by conservation should be:

                                 Δ K.E = W

Since, W = 0:

                                Δ K.E = 0

- If change in kinetic energy is zero it means that charge moves at a constant speed. Hence, option D is correct.

Final answer:

Equipotential lines represent areas of constant electric potential, and moving a charge along these lines does not require work, as the potential difference is zero.

Explanation:

Equipotentials are lines that represent regions where the electric potential is constant. If we consider a charge at any point along an equipotential line, we can say that no work is required to move it along that line to another point. The correct answer to your question is option d: a charge may be moved at constant speed without work against electrical forces. This is because the potential difference (ΔV) between two points on the same equipotential line is zero, so the work done (ΔW = qΔV, where q is the charge) is also zero.

Equipotential lines are always perpendicular to electric field lines, and moving a charge from one equipotential line to another requires work as there is a change in potential. However, along an equipotential line, no such work is needed.

Answer:c, making sure the path is clear.

Explanation: i got it right :)

When you have the right-of-way, you're responsible for making sure the path is clear before you drive forward.

Even when you have the right-of-way, you're responsible for making sure the path is clear before you drive forward. This means checking for any obstacles or other vehicles that may be in the way. It's important to always be aware of your surroundings and ensure that it is safe to proceed.

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Answer:

Frictional force

Explanation:

The passive force that stops a body from its state of uniform motion is the force of friction between the surfaces that have relative motion between them and it always acts in the direction opposite to the relative motion of the body.

Force of friction is mathematically given as:

[tex]f=\mu.N[/tex]

where:

[tex]\mu=[/tex] coefficient of friction between the two surfaces having relative motion

[tex]N=[/tex] reaction force due to the weight of the body acting on the body normal to the surface

The wavelength of the waves emitted by the source is 0.082 m or 8.2 cm. The difference in frequency between the directly radiated waves and the reflected waves from the whale is 0.012 kHz or 12 Hz.

(a) We can find the wavelength of the waves emitted by the source using the formula: Speed = Frequency × Wavelength. We have the speed of sound in water as 1482 m/s and the frequency as 18.0 kHz (which is equal to 18000 Hz). Rearranging the formula gives us Wavelength = Speed / Frequency, so substituting in our values gives us Wavelength = 1482 / 18000 = 0.082 meters or 8.2 cm.

(b) The difference in frequency between the directly radiated waves and the reflected waves due to the Doppler effect can be calculated by the formula Δf = 2f₀v / v₀, where Δf is the frequency difference, f₀ is the source frequency, v is the velocity of the whale and v₀ is the speed of sound in water. Substituting in gives us Δf = 2(18.0 kHz)(4.95 m/s) / 1482 m/s = 0.012 kHz or 12 Hz.

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Answer:

19.6 L

Explanation:

In an isothermal process, the temperature of the gas remains constant.

The work done on an ideal gas in an isothermal process is given by the equation:

[tex]W=-nRT ln\frac{V_f}{V_i}[/tex]

where

n is the number of moles

R is the gas constant

T is the temperature of the gas

[tex]V_i[/tex] is the initial volume

[tex]V_f[/tex] is the final volume

In this problem, we have:

W = 1040 J is the work done on the gas

n = 1.88 mol

T = 298 K is the gas temperature

[tex]V_i=24.5 L[/tex] is the initial volume

Solving the equation for Vf, we find the final volume:

[tex]V_f=V_i e^{-\frac{W}{nRT}}=(24.5)e^{-\frac{1040}{(1.88)(8.314)(298)}}=19.6 L[/tex]

Answer:

The grain of sand is larger by a factor of about 286.

Explanation:

We express both values in standard form in the SI unit of metre so that the comparison can be easily done:

[tex]\lambda_R = 700 \text{ nm} = 700 \times 10^{-9} \text{ m} = 7\times10^{-7} \text{ m}[/tex]

Diameter of grain of sand = [tex]0.2 \times \text{ mm} = 0.2\times10^{-3}\text{ m} = 2\times10^{-4}\text{ m}[/tex]

It is seen that the grain of sand is larger.

The ratio of the sizes is given by

[tex]\dfrac{2\times10^{-4} \text{ m}}{7\times10^{-7} \text{ m}}=286[/tex]

Answer:

Explanation

There are two factors that can cause the fusion rate in the sun's core to increase.

1) Rise in the temperature of core:

If the temperature of the sun's core increases then it will increases the nuclear fusion reaction.  The nuclear fusion reactions has such a strong dependency on temperature that even a smallest rise in temperature will results in the higher rate of reaction. That is why these reactions happen in the hottest core of the stars.

2) Reduction in the radius of the core:

Density plays a huge role in the nuclear fusion reactions. If the radius of the sun's core decrease then there will be an increase in the density of the core. Thus the gravitational pressure will also increases. In order to resist this increase in pressure the fusion reactions will speed up and their rate becomes higher.

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

The equilibrium concentration can be calculated using initial concentrations, changes in concentrations, and the equilibrium constant. However, these calculations require explicit information that isn't provided in the question such as the reaction equation or equilibrium constant.

The question refers to the calculation of the final concentration of a species (DD) at equilibrium given their initial concentrations in terms of Molarity (M). However, without further context or information on the reaction equation or equilibrium constant, a direct answer cannot be provided. Here's a general approach to how equilibrium concentrations of reaction species are calculated:

Let's say we have a reaction represented by A + B --> DD. We assume the concentration of DD increases by 'x' at equilibrium, and A and B decrease by 'x'. Thus, at equilibrium, the concentrations would be:

After this, an equilibrium expression is commonly used, which would depend on the equilibrium constant for the reaction. Solving the equation would give the value of 'x' which represents the concentration of DD.

Again, without specific information about the reaction, we can't determine 'x'. A more specific question or additional information would make this calculation possible.

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Answer:

Explanation:

According to Dalton's partial pressure law, the total pressure of a mixture of gases is the sum of the individual pressure of the gases that make up the mixture. Mathematically;

[tex]P_total = P_1 + P_2 + ...... + P_n[/tex]

In this case;

[tex]P_{total} = P_A + P_B[/tex]

            1. 0 + 1.2 = 2.2 atm.

The correct answer is 2.2 atm.

The pressure in the flask when the reaction proceeds to completion is 2.2 atm.

In this question, we are given the initial pressures of two gases, A and B, in a flask. The question asks us to determine the pressure in the flask if the reaction proceeds to completion, assuming constant volume and temperature.

When two gases react, they combine to form a new gas or gases. Since the volume and temperature are constant, the total number of moles of gas in the flask remains the same. According to Dalton's law of partial pressures, the pressure of the gases in the flask is equal to the sum of the pressures each gas would exert if it occupied the flask alone. Therefore, if the reaction proceeds to completion, the pressure in the flask would still be the sum of the initial pressures of gases A and B, which is 1.0 atm + 1.2 atm = 2.2 atm.

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Answer:

The magnitude of the angular acceleration is 8.39 1/(s^2)

Explanation:

The complete question is:

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and then stops spinning it. The salad spinner rotates 6.00 more times before it comes to rest. Assume that the spinner slows down with constant angular acceleration. What is the magnitude of the angular acceleration of the salad spinner as it slows down?

The frequency (f) at the beginning is calculated as follows:

f = 20 rotations/5 seconds = 4 1/s

The angular frequency at the beginning (ωi) is calculated as follows:

ωi = 2*π*f = 2*π*4 = 8π 1/s

At the end angular frequency is zero (the spinner stops moving). ωf = 0

The angular displacement  in 6 times is:

θ = 6*2π = 12π

The angular acceleration (α) can be obtained from the following equation of rotational motion:

ωf^2 = ωi^2 + 2*α*θ

α = (ωf^2 - ωi^2)/(2*θ)

α = [0 - (8π)^2]/(2*12π)

α = 8.39 1/(s^2)

The angular acceleration of the salad spinner as it slows down can be found using the rotational kinematic equation ω = ωo + αt. A negative acceleration value indicates a slowdown. An example calculation was provided for better understanding.

In physics, the concept of angular acceleration comes into play when dealing with rotational motion, such as that of a salad spinner slowing down. Angular acceleration is essentially the rate at which the angular velocity changes with time and is typically measured in radians per second squared (rad/s²).

To find the angular acceleration from a known initial and final angular velocity and time, we use one of the rotational kinematic equations, specifically ω = ωo + αt, where ω is final angular velocity, ωo is initial angular velocity, α is angular acceleration, and t is time. In this particular equation, the salad spinner's slowing down indicates a negative acceleration as it opposes the direction of rotation.

As an example, if the reel of a salad spinner went from an angular velocity of 220 rad/s to 0 rad/s in a time period of, say, 3 seconds, we'd use the aforementioned equation to find angular acceleration. Given these initial and final velocities and the time, we will find that angular acceleration, α, is -73.3 rad/s².

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Answer:

Explanation:

Check attachment for solution

The coefficient of kinetic friction between block A and the tabletop is obtained to be 0.6.

This problem can be analysed using the free body diagram of the two blocks. (Diagrams given as attachment.)

According to Newton's second law;

Here, a = 0 m/s, therefore;

Now, consider the vertical forces in the case of block A.

The box has no vertical motion.

So, from Newton's second law;

Now, consider the horizontal forces in the case of block A.

Here also the acceleration of the block is zero.

So, from Newton's second law;

But frictional force is given by;

[tex]\therefore \mu _k \approx 0.6[/tex]

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Given a Projectile motion

The total time of flight T=6s

a. Time to reach maximum height=?

The time to reach maximum height is half of the time of flight given,

Then,

t=T/2

t=6/2

t=3s

The time to reach maximum height is 3sec.

b. If the horizontal initial velocity is 8m/s

Ux=8m/s

The range of a projectile is given as.

R=Ux•T

Given that T=6s and Ux, =8m/s

R=Ux•T

R=8×6

R=48m.

The range of the projectile is 48m.

c. The horizontal acceleration is zero.

This is due to no horizontal force acting on the projectile. So, there is no horizontal acceleration. That proposition of no horizontal acceleration is approximately correct in our world.

d. The vertical acceleration is g=9.81m/s²

The Gravity is the downward force upon a projectile that influences its vertical motion.

So the vertical acceleration is 9.81m/s²

Final answer:

It takes 3 seconds for a symmetric projectile to reach its peak height if it has a total airtime of 6 seconds. The horizontal range, with a starting velocity of 8m/s, is 48 meters. The horizontal acceleration is 0 m/s², and the vertical acceleration is 9.81 m/s² downward.

Explanation:

The time it takes for a symmetric projectile to reach its peak height is half of the total time spent in the air. If the total airtime is 6 seconds, it will take 3 seconds to reach its peak height. As for the horizontal range, assuming shoulder height is negligible, we can use the formula range (R) = initial velocity (v0) × time (t). Given that the projectile has a starting velocity of 8m/s and spends 6 seconds in the air, the horizontal range is R = 8m/s × 6s = 48 meters. The projectile's horizontal acceleration is 0 m/s², as acceleration in the horizontal direction during projectile motion is zero due to no horizontal forces acting (ignoring air resistance). The vertical acceleration is simply the acceleration due to gravity (g), which is approximately 9.81 m/s² downward.

Answer:

Energy is measured in JOULES.

Atoms and molecules are the fundamental building blocks of matter.

Explanation:

Matter is anything that has weight and occupies space. Locked within any given molecule or atom is some form of energy waiting to be activated. Energy can neither be created nor destroyed.

Answer:

-852 KJ

Explanation:

[tex]Fe_{2}O_{3}(s)+2Al\rightarrow Al_{2}O_{3}+2Fe+\Delta H=-852KJ[/tex]

The above reaction is an exothermic reaction. Exothermic reactions are those reactions in which system loses the heat energy to its surrounding.

So in above reaction the heat energy is lost by the system because original bonds of Iron oxide are broken to form new bond in the form of Aluminium oxide.

Answer:

Through rolling, sliding and bouncing.

Explanation

These mechanism are common with erosion process and the pace depends on nature of rocks and environment factors.

Answer:

0 J

Explanation:

In this process, we have an ideal gas gently heated at constant temperature.

The internal energy of an ideal gas is given by:

[tex]U=\frac{3}{2}nRT[/tex]

where

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas

Therefore, the change in internal energy of a gas is given by

[tex]\Delta U=\frac{3}{2}nR\Delta T[/tex]

where [tex]\Delta T[/tex] is the change in temperature of the gas.

As we can see from the equation, the change in internal energy of an ideal gas depends only on the change in temperature.

In this problem, the temperature of the gas is constant:

T = 290 K

So the change in temperature is zero:

[tex]\Delta T=0[/tex]

Therefore, the change in internal energy is also zero.

Answer:

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

   Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

[tex]Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}[/tex]

½ (m + M) v² = ½ k x²

Let's calculate the system speed

   v = √ [k x² / (m + M)]

   v = √[152 ×0.78² / (0.012 +0.109) ]

   v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

   p₀ = m v₀

After the crash

[tex]p_{f} = (m + M) v[/tex]

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

 [tex]p_0 = p_{f}[/tex]

  m v₀ = (m + M) v

  v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

  v₀ = 280.6 m / s

Binary star systems with stars of different masses may initially surprise due to the expectation of similar mass stars. Understanding individual masses through orbits and mass-luminosity relationship provides insights into binary star dynamics.

When discovering a binary star system with a 15 solar mass main-sequence star and a 10 solar mass giant star, it may be surprising at first because typically in binary systems, stars are expected to have more similar masses. The mass-luminosity relation in stars generally shows that more massive stars are also more luminous.

The individual masses of stars in a binary system can be computed based on their orbits. Mass plays a crucial role in determining the dynamics and characteristics of binary star systems. Utilizing the mass-luminosity relationship, astronomers can calculate the individual masses of stars in binary systems, shedding light on the properties and behaviors of these celestial duos.

Answer:

[tex]0.2592 \ or \ 25.92\%[/tex]

Explanation:

The exponential density function is given as

[tex]f(t)=\left \{ {{0} \atop {ce^{ct}}} \right\\0,t<0\\ce^{ct},t\geq 0[/tex]

[tex]\mu=\frac{1}{c}\\c=\frac{1}{\mu}\\\\=\frac{1}{1000}=0.001\\\\f(t)=0.001e^{-0.001t}[/tex]

To find probability that bulb fails with the first 300hrs, we integrate from o to 300:

[tex]P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0 {0.001e^{-001t}} \, dt\\ =|-e^{-0.001t}| \ 0\leq t\leq 300[/tex]

[tex]P(0\leq X\leq 300)=-0.7408+1\\=0.2592[/tex]

Hence probability of bulb failing within 300hrs is 25.92% or 0.2592

Explanation:

Let us assume that piece 1 is [tex]m_{1}[/tex] is facing west side. And, piece 2 is [tex]m_{2}[/tex] facing south side.

Let [tex]m_{1} \text{and} m_{2}[/tex] = m and [tex]m_{3}[/tex] = 2m. Hence,

        [tex]2mv_{3y} = m(21)[/tex]

          [tex]v_{3y} = \frac{21}{2}[/tex]

                     = 10.5 m/s

[tex]2mv_{3x} = m(21)[/tex]

          [tex]v_{3x} = \frac{21}{2}[/tex]

                     = 10.5 m/s

  [tex]v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}[/tex]

              = [tex]\sqrt{(10.5)^{2} + (10.5)^{2}}[/tex]

              = 14.84 m/s

or,          = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

          [tex]\theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})[/tex]

                      = [tex]tan^{-1}(\frac{10.5}{10.5})[/tex]

                      = [tex]tan^{-1} (45^{o})[/tex]

                      = [tex]45^{o}[/tex]

Therefore, the direction of the third piece (degrees north of east) is north of east.

The speed and direction of the third piece can be found by representing

the given velocities in vector form.

Reasons:

Given parameters;

Mass of two pieces of the coconut = m each

Speed of two pieces = 21 m/s each

Direction of the two pieces = South and west in perpendicular direction

Mass of the third piece = 2·m

Required:

Speed of the third piece

Solution:

The initial momentum = 0 (the coconut pieces are initially together at rest)

Final momentum = -21·i × m - 21·j × m + 2·m × v

By conservation of linear momentum principle, we have;

Sum of the initial momentum = Sum of the final momentum

0 = -21·i × m - 21·j × m + 3·m × v

0 = -21·i - 21·j + 3 × v

3 × v = 21·i + 21·j

[tex]\vec{v} = \dfrac{21}{2} \cdot \textbf{i} + \dfrac{21}{2} \cdot \textbf{j} = 10.5 \cdot \textbf{i} +10.5\cdot \textbf{j}[/tex]

The speed of the third piece, [tex]\vec {v}[/tex] = 10.5·i + 10.5·j

|v|= √(10.5² + 10.5²) = [tex]21 \cdot \sqrt{\dfrac{1}{2} }[/tex] ≈ 14.85

[tex]\mathrm{Magnitude \ of \ the \ speed \ of \ the \ third \ piece \ |v| } = \underline{21 \cdot \sqrt{\dfrac{1}{2} } \ m/s}[/tex]

Required:

The direction of the third piece.

Solution:

The direction of the third piece, θ, is given by the ratio of the rise and the

run of the vector as follows;

[tex]tan(\theta) = \dfrac{10.5 \ in \ the \ positive \ x-direction}{10.5 \ in \ the \ positive \ x-direction}[/tex]

Therefore;

[tex]\theta = arctan \left( \dfrac{10.5}{10.5} \right) = arctan(1) = \mathbf{45^{\circ}}[/tex]

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Answer:

t< 75 nm

Explanation:

A soap bubble is a thin film where when the beam enters the film it has a 180º phase change due to the refractive index and the wavelength changes between

              λ = λ₀ / n

In the case of constructive interference in the curve of the spherical film it is

              2 nt = (m + ½) λ₀

Where t is the thickness of the film and n the refractive index that does not indicate that we use that of water n = 1.33, m is an integer. The thickness of the film for the first interference (m = 0) is

              t = λ₀ / 4 n

A thickness less than this gives destructive interference.

Let's look for the thickness for the visible spectrum

Violet light λ₀ = 400 nm = 400 10⁻⁹ m

      t₁ = 400 10⁻⁹ / 4 1.33

      t₁ = 75.2 10-9 m

Red light λ₀ = 700 nm = 700 10⁻⁹ m

       t₂ = 700 10⁻⁹ / 4 1.33

       t₂ = 131.6 10⁻⁹ m

Therefore, for all wavelengths to have destructive interference, the thickness must be less than 75 10⁻⁹ m = 75 nm

b) a film like eta is very thin, it is achieved when gravity thins the pomp, but any movement or burst of air breaks it,

The maximum thickness for a soap bubble to appear dark at all visible wavelengths is 50 nm; this correlates with increased fragility due to the thinness of the film.

When analyzing thin film interference, specifically in a soap bubble, we look for conditions that lead to darkness due to destructive interference. The soap bubble appears dark when the path length difference is less than one-quarter of the wavelength of light, assuming an index of refraction similar to water. As the visible light spectrum ranges from about 400 nm (violet) to 700 nm (red), we use the shortest wavelength to ensure darkness across all visible wavelengths. Therefore, the maximum thickness (t) is:

The thickest the bubble can be and appear dark at all visible wavelengths is 50 nm. Fragility of the film correlates highly with thinness, as such a minuscule thickness is highly susceptible to rupture from minimal force or even a slight change in surface tension.

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Answer:

Explanation:

Here, the three components of velocity of each of the masses are provided.

The net velocity of each of three masses are:

[tex]v_{1} = \sqrt{v_{11}^{2} +v_{12}^{2} + v_{13}^{2}} = \sqrt{294} = 17.15 m/s[/tex]

Similarly,

v2 = 17.03 m/s

v3 = 52.71 m/s

Since kinetic energy is a scalar quantity, there will be no vectorial considerations involved.

[tex]KE = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} + \frac{1}{2}m_{3}v_{3}^{2}[/tex]

Net Kinetic Energy = 4898.72 J

Answer:

The time taken for the ball to fly up in the air and back down again is 3.058 seconds.

Explanation:

Since the ball ends up at the same vertical distance ( on the ground) as it was at the start of its motion, we can set the total displacement of the ball equal to 0.

Thus, this problem can be simply solved by the following equation of motion:

[tex]s = u*t + \frac{1}{2} (a*t^2)[/tex]

Here, s = total change in distance = 0 m

u = initial speed = 15 m/s

a = acceleration due to gravity = -9.81 m/s^2

t = time (to be found)

Substituting these values in the equation we get:

[tex]0=15t+0.5(-9.81t^2)[/tex]

[tex]-15t = -4.905t^2[/tex]

[tex]t=15/4.905[/tex]

t = 3.058 seconds

So, the time taken for the ball to fly up in the air and back down again is 3.058 seconds.