The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.35 rad/s2. It accelerates for 28.3 s, then maintains a constant angular velocity. Calculate the total angle through which the wheel has turned 58.9 s after it begins rotating.

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.

Answer:

Water molecules attracting other water molecules is called cohesive attraction.

Explanation:

They are basically two forces in liquids that determine their wetting characteristics, they are cohesive and adhesive forces.

Cohesion is the attraction between molecules of same liquid example water and water, while adhesion is attraction between molecules of different liquids example alcohol and water.

Therefore, Water molecules attracting other water molecules is called cohesive attraction.

Answer:

a) 10.7° ≈ 11°

b) 0.19

Explanation:

If the road is banked at an angle, without seeking the help of friction, (i.e. frictionless road), the forces acting on the car are shown in the attached free body diagram to the question

In the y - direction

mg = N cos θ (eqn 1)

mg = weight of the car.

N = normal reaction of the plane on the car

And in the direction parallel to the inclined plane,

(mv²/r) = N sin θ (eqn 2)

(mv²/r) = force keeping the car in circular motion

Divide (eqn 2) by (eqn 1)

(v²/gr) = Tan θ

v = velocity of car = 60 km/h = 16.667 m/s

g = acceleration due to gravity

r = 150 m

(16.667²/(9.8×150)) = Tan θ

θ = Tan⁻¹ (0.18896)

θ = 10.7° ≈ 11°

b) In the absence of banking, the frictional force on the road has to balance the force keeping the car in circular motion

That is,

Fr = (mv²/r)

Fr = μN = μ mg

μ mg = mv²/r

μ = (v²/gr) = (16.667²/(9.8×150)) = 0.19

Hope this Helps!!!

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

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: 65mph, 75mph

Explanation:

Let us assume x to be the speed of the slower train, in mph (miles per hour).

Then the speed of the other train is (x+10) mph, according to the question.

We then would have an equation like this

1.6x + 1.6(x+10) = 224.

This is because, the first addend in the left side is the distance covered by the slower train.

The second addend in the left side is the distance covered by the faster train.

The sum is 224 miles, because they together covered all the distance to the moment when they meet each other.

1.6x + 1.6x + 16 = 224

3.2x + 16 = 224

3.2x = 224 - 16

3 2x = 208

x = 208/3.2

x = 65

Thus the speed of the slower train is 65mph, and that of the other train is 65 + 10 = 75mph

Final answer:

By setting up and solving a linear equation, the speed of the slower train is determined to be 65 mph, while the faster train's speed is 75 mph as one travels 10 mph faster than the other.

Explanation:

To solve the problem of the two trains travelling on parallel tracks towards each other, we must first define the variables for their speeds. Let's say one train travels at x mph and the other at (x + 10) mph. Since they are moving towards each other, you can add their speeds together to find how fast the distance between them is closing. The total closing speed is then x + (x + 10) mph, which simplifies to 2x + 10 mph.

The total distance to be covered by the two trains until they pass each other is 224 miles. We know they pass each other after 1.6 hours, so using the formula Distance = Speed x Time, we can set up the equation: 224 miles = (2x + 10 mph) x 1.6 hours. Solving for x gives us the speed of the slower train.

Performing the algebraic steps:

224 = (2x + 10) x 1.6

224 = 3.2x + 16

208 = 3.2x

x = 65 mph.

Therefore, the speed of the slower train is 65 mph and the speed of the faster train is 75 mph.

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:

Explanation:

Given

balloon is rising with a speed of [tex]u_y=6\ m/s[/tex]

Person throws a ball out of basket with a horizontal velocity of [tex]u_x=10\ m/s[/tex]

Considering upward direction to be positive

When ball is thrown it has two velocity i.e. in upward direction and in horizontal direction so net velocity is

[tex]v_{net}=\sqrt{(u_x)^2+(u_y)^2}[/tex]

[tex]v_{net}=\sqrt{(6)^2+(10)^2}[/tex]

[tex]v_{net}=\sqrt{36+100}[/tex]

[tex]v_{net}=\sqrt{136}[/tex]

[tex]v_{net}=11.66\ m/s[/tex]

Direction of velocity

[tex]\tan \theta =\dfrac{u_y}{u_x}[/tex]

[tex]\tan \theta =\dfrac{6}{10}[/tex]

[tex]\theta =30.96^{\circ}[/tex]

where [tex]\theta [/tex] is angle made by net velocity with horizontal .

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:

1.44 s

Explanation:

Since it is a projectile motion, we use the formula for the total time of flight,t

t = 2Usinθ/g where U = initial velocity of ball = 10 m/s, θ = 45 and g = 9.8 m/s²

t = 2Usinθ/g = 2 × 10sin45/9.8 = 1.44 s

So, the time needed for the ball to return to the ground is most nearly: 1.44 s

We have that for the Question it can be said that the time needed for the ball to return to the ground is most nearly

T=1.44

From the question we are told

A ball is thrown into the air with an initial velocity of 10 m/s at an angle of 45 degrees above the horizontal, as represented above. If air resistance is negligible the time needed for the ball to return to the ground is most nearly

Generally the Newton's equation for the vertical displacement  is mathematically given as

[tex]y=ut+1/2at^2\\\\Therefore\\\\T=\frac{2Usin\theta}{g}\\\\T=\frac{2*10*sin45}{9.8}\\\\[/tex]

T=1.44

Therefore

the time needed for the ball to return to the ground is most nearly

T=1.44

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Answer:

[tex]\boxed {3.43 x 10^{3}}[/tex]

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as [tex]v=\frac {d}{t}[/tex] where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

[tex]d=vt[/tex]

Substituting 343 m/s for d and 10 s for t then

[tex]d= 343\times10= 3430= 3.43 x 10^{3}[/tex]

Therefore, the distance between speaker and deter is [tex]\boxed {3.43 x 10^{3}}[/tex]

Answer:

The wire should be tightened because the present tension is 532.54 N where the required tension is 800 N and the higher the tension the more tightening is required.

Explanation:

To solve the question

v = [tex]\sqrt{\frac{F_t}{\mu} } = \sqrt{\frac{L*F_t}{m} }[/tex] where

v = velocity of the pulse in the string = 46.154 m/s

[tex]F_t[/tex] = Required tension force = 800 N

m = Mass of the wire = 15 kg

L = length of the wire to be tension-ed = 60 m

Since the pulse travels twice the distance of 60 m in 2.6 s the velocity is given by

v = 2×60/2.6 = 46.154 m/s

Therefore making [tex]F_t[/tex] the subject of the formula and substituting the values, we have

[tex]F_t[/tex] = [tex]\frac{v^2m}{L}[/tex] =[tex]\frac{46.154^{2*15} }{60}[/tex] = 532.54 N

This means that, as it is, the present tension in the wire is 532.54 N which is less than  the required 800 N, therefore the wire should be tightened

The wire should be tightened because the current tension on the wire is less than the required tension.

The given parameters;

The speed of the wave as the pulse traveled from one pole to the other two times, is calculated as follows;

[tex]v = \frac{2d}{t} \\\\v = \frac{2 \times 60}{2.6} \\\\v = 46.154 \ m/s[/tex]

The tension created on the wire during the pulse motion is calculated as follows;

[tex]v = \sqrt{\frac{T}{m/L} } \\\\v ^2 = \frac{TL}{m} \\\\T = \frac{v^2 m}{L} \\\\T = \frac{(46.154)^2 \times 15}{60} \\\\T = 532.55 \ N[/tex]

The current tension on the wire (532.55 N) is less than the required tension of 800 N. Thus, the wire should be tightened.

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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

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:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude [tex]g[/tex].

If we have the initial velocity [tex]v_o[/tex] and its angle [tex]\theta[/tex], we can obtain the vertical component of the velocity [tex]v_{oy}[/tex] using trigonometry:

[tex]v_{oy}=v_osin\theta[/tex]

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

[tex]v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }[/tex]

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

[tex]g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]

Finally, from the equation of horizontal motion with constant speed, we have that:

[tex]x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

[tex]v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m[/tex]

In words, the projectile travels 7.49m horizontally before it lands.

Answer:

Layer 3 switch.

Explanation:

A layer 3 switch carrys out both the function of a switch and a router. It acts as a switch that links all the devices that are on the same subnet or virtual LAN at lightning speeds and has IP routing intelligence built into it to carry out the function of a router. It can support routing protocols, check incoming packets, and can also carry out routing decisions based on the source and the destination addresses.

Answer:

Valence shell electron pair repulsion theory

Explanation:

It is also called the Gillespie-Nyholm theory after its two main discoverers, Ronald Gillespie and Ronald Nyholm.

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[tex]\red:\implies[/tex][tex]\underline{\underline{\textbf{\pink{Valence Shell Electron Pair Repulsion \: : -}}}}[/tex]

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:

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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Answer: 10.53 seconds

Explanation:

in the attachment

Explanation:

Below is an attachment containing the solution.

Answer: Nitrogen gas

Explanation:

Using ideal Gas's law

PV = nRT

where

Pressure of gas, P= 1atm

Volume of gas, V= 5.1L

no of moles of gas, n=

Ideal gas constant, R= 0.0821

Temperature of gas, T= 20°C = 20+273 = 293K

also, n= (mass/molar mass)

mass of the gas m = 5.9g

Molar mass of the gas = ?

So, PV = (mRT/M)

We're looking for molar mass M, then

M = mRT/PV

M = (5.9 * 0.0821 * 293)/(1 * 5.1)

M = 141.93/5.1

M = 27.8g/mol ~ 28g/mol

Since the gas is diatomic, then we say,

Atomic mass of gas = 1/2 * molar mass

Atomic mass = 1/2 * 28

Atomic mass = 14

Therefore, the gas is nitrogen.

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

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

The Potential energy stored in the system is 1 J

Explanation:

Given-

Mass, m = 4 kg

Spring constant, k = 800 N/m

Distance, x = 5cm = 0.05m

Potential energy, U = ?

We know,

Change in potential energy is equal to the work done.

So,

[tex]U = \frac{1}{2} k (x)^2\\\\[/tex]

By plugging in the values we get,

[tex]U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\[/tex]

Therefore, Potential energy stored in the system is 1 J

The maximum mass of block B, for which block A remains at rest on a 35-degree incline given a static friction coefficient of 0.40 and a 10kg mass of block A, is approximately 1.98 kg.

To solve this question, we will need to use the principles of static friction, inclined planes, and gravitational force. Static friction is what keeps block A from sliding down the incline. It has to overcome the downward force due to gravity on block A which is a component of the weight of block A acting downwards the inclined plane.

To determine the maximum mass of block B, we can equate the static frictional force to the net force acting downward on the inclined plane. The static frictional force is [tex]\mu N[/tex] where μ is the coefficient of static friction and N is the normal force. N = [tex]mAgcos\theta[/tex] where mA and g are the mass and acceleration due to gravity respectively and θ is the angle of the inclined plane. So, static frictional force = [tex]\mu mAgcos\theta[/tex]. The downward force is the sum of components of the weight of block A acting downwards and the weight of block B. So, [tex]mAgcos\theta (\mu)[/tex] = mAg*[tex]sin\theta[/tex] + mBg.

From this equation, you can solve for the mass of Block B: MB = [tex]mA\mu cos\theta[/tex] - mA*[tex]sin\theta[/tex].

Plugging in the given numbers, we get:

MB = 10(0.4*cos(35 degrees) - sin(35 degrees)).

This gives us approximately 1.98 kg as the maximum mass of block B before block A begins to slide.

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To find the largest mass for block MB, we first need to calculate the gravitational force and the static friction on block A, using the given values for mass, gravitational acceleration, coefficient of static friction and incline angle. The tension in the string, created by block MB, has to balance these forces. By setting the net forces equal, we can calculate the value of MB.

The problem revolves around understanding the concept of static friction and forces acting on an inclined plane. Firstly, let's calculate the force due to gravity acting on block A. This is simply F_gravity = mg sin θ where m=10 kg is the mass of the block, g=9.8 m/s² is the gravitational acceleration, and θ is 35°. Next, we need to calculate the frictional force that prevents the block from sliding. Since block A is at rest, static friction is at work here, and we can use the formula F_friction = μN, where μ=0.40 (the coefficient of static friction) and N is the normal force acting on the block, which equals mg cos θ.

To keep block A at rest, the tension T in the string due to the dangling mass MB must balance both gravity and friction. Therefore: T=F_gravity + F_friction. The weight of the dangling mass brings about the tension in the string, and hence, T=MBg. From these, we can calculate the largest mass of MB for which A remains at rest.

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THE ANSWER IS : THERE ARE ONLY ABOUT 100 DIFFERENT KINDS OF ATOMS THAT COMBINE TO FORM ALL SUBSTANCES

Explanation:

Answer:

The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Explanation:

The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.

The boat moves a distance of the change in centre of mass

Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.

For convention, we call the original position of the 59 kg person as x=0

This means,

59 kg person is at x = 0 m

88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.

78 kg person is at x = 2.90 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

For the initial setup,

X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)

X = (353.8/225)

X = 1.572 m

(Don't forget that this is 1.572 m from the end we designated x=0 m)

When the people exchange positions,

59 kg person is now at the other end of the boat with x = 2.90 m

88 kg of the boat still acts at the centre of the boat at x = 1.45 m

And 78 kg person is now at the end of the boat with x = 0 m

Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)

X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)

X = (298.7/225)

X = 1.328 m

(This is 1.328 m from the end we designated x=0 m from the start)

So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.

Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.

Hope this Helps!!!

Answer:

0.6m/s

Explanation:

Totally inelastic collision

m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)

Where v (v₁ and v₂) is the initial velocity of the objects

v(final) is the  final velocity of the objects stuck together

Toy car a , m₁ = 3kg, v₁ = 1m/s

Toy car b , m₂ = 2kg, v₂ = 0m/s

m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)

3(1) + 2(0) = (3 + 2) v(final)

3 = 5 v(final)

v(final) = 0.6m/s

Explanation:

Below is an attachment containing the solution.

Explanation:

The potential difference of 180 V is applied across resistance = 8.4 ohm

Thus the maximum current flowing I₀ = [tex]\frac{V}{R}[/tex] = [tex]\frac{180}{8.4}[/tex] = 21.4 A

The rms value of the current is = [tex]\frac{I_0}{\sqrt{2} }[/tex]  =  I₀ x 0.7 = 15 A

This value of current is greater than the 12.5 A . Thus the circuit will break .

Answer: non-shifting lines indicate non moving star.

Explanation: when a star is moving toward the detector, the wavelength will decrease - there will be a blue shift.

When it's moving away from the earth or detector, the wavelength will increase - there will be a red shift.

Identifiable patterns of absorption lines that appear shorter or longer wavelengths than normal indicate that the star is moving

In visualisation, the bottom spectrum shows the normal position of absorption line for a star that is not moving toward or away from the earth.

Non-shifting lines in the binary star system's spectrum are attributed to absorption by interstellar clouds, which unlike the stars, do not move relative to us. The narrowness of these lines indicates the low pressure of the absorbing gas.

When observing a binary star system and noting the Doppler Effect in the spectral lines, if certain lines do not shift, it indicates they originate from something that is not moving with respect to us. Most of the absorption lines shift due to the motion of the binary stars, which causes Doppler shifts as the stars move toward or away from us. This movement results in the spectral lines being blue-shifted when the star is approaching us, and red-shifted when it's receding. However, the lines that remain constant are likely due to the absorption by interstellar clouds located between Earth and the stars. The non-shifting lines are also much narrower, suggesting that the absorbing gas is at a very low pressure. This non-movement of specific lines helped in discovering the presence of interstellar materials, as their spectral lines do not participate in the Doppler shifts associated with the stars' orbits.