Please help! quick and easy chart.

Answer:

the moment of inertia is 4.5 × 10⁻⁵ kg.m²  

Explanation:

Given that;

point mass m = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg

perpendicular distance r = 3m

We know that a point mass doesn't have a moment of inertia around its own axis but, but using the parallel axis theorem, a moment of inertia around a distant axis of rotation can be determined using;

\(I_{}\) = mr²

so we substitute

\(I_{}\) = (5 × 10⁻⁶ kg) × (3 m)²

\(I_{}\) = (5 × 10⁻⁶ kg) × 9 m²

\(I_{}\)  = 4.5 × 10⁻⁵ kg.m²  

Therefore; the moment of inertia is 4.5 × 10⁻⁵ kg.m²  

The moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm² at a perpendicular distance of 3 m.

The moment of inertia of given point mass can be determined by,

\(I = mr^2\)

Where,

\(I\)- moment of inertia

\(m\)- mass = 0.005 g = ( 0.005 / 1000 ) = 5 × 10⁻⁶ kg

\(r\) - perpendicular distance = 3 m

Put the values in the formula,

\(I = (5 \times 10^{-6}{\rm \ kg}) \times (3 {\rm \ m})^2\\\\I = 5 \times 10^{-6}{\rm \ kg} \times 9 {\rm \ m}\\\\I = 4.5 \times 10^{-5} kgm^2\)

Therefore; the moment of inertia of given point mass is 4.5 × 10⁻⁵ kgm².

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The magnitude of the resultant vector, representing the actual path of the boat, is approximately 1.42 m/s.

To find the magnitude of the resultant vector, we need to consider the boat's velocity and the velocity of the river. The boat's velocity is given as 0.75 m/s, and the river's velocity is given as 1.2 m/s.

Since the boat is moving in a river, we can think of the boat's velocity as a combination of two velocities: its own velocity and the velocity of the river. The resultant vector represents the actual path of the boat, considering both velocities.

To calculate the resultant vector, we can use vector addition. The magnitude of the resultant vector can be found by taking the square root of the sum of the squares of the boat's velocity and the river's velocity. Mathematically, we have:

Resultant magnitude = √(boat velocity^2 + river velocity^2)

Plugging in the given values, we have:

Resultant magnitude = √(0.75^2 + 1.2^2)

= √(0.5625 + 1.44)

= √2.0025

≈ 1.42 m/s

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Answer:

1. Letter A indicates the crest.

2. Letter E indicates the wavelength.

3. Letter C indicates the trough.

Explanation:

1. Determination of the the crest.

A crest is simply defined as a point on a wave where the displacement is maximum (i.e highest).

Considering the diagram given above, the point where the displacement is maximum is A. Therefore, A is the crest of the wave.

2. Determination of the wavelength.

The wavelength of a wave is simply defined as the distance between two successive crest or trough.

Obeserving the diagram above, we can see that E gives the distance between two successive crest. Therefore, E is the wavelength of the wave.

3. Determination of the trough.

A trough is simply defined as a point on a wave where the displacement is minimum (i.e lowest). Thus we can say that the trough is the opposite of the crest.

Considering the diagram given above, the point where the displacement is minimum is C. Therefore, C is the trough of the wave.

The electroscope will remain charged and continue to show the presence of an electric charge if the ground is withdrawn from the knob before the charging rod is removed.

An early scientific tool used to find electrical charge on a body is called an electroscope. The movement of a test object caused by the Coulomb electrostatic force is used to detect charge. Voltage and charge on an object are inversely correlated.

There are three traditional electroscope types: the needle electroscope, the gold-leaf electroscope, and the pith-ball electroscope (third).

electroscopes are utilized to determine whether a body has an electric charge. The movement of a test object caused by the Coulomb electrostatic force is used to detect charge. Voltage and charge on an object are inversely correlated.

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Answer:

which language is this

Explanation:

After pushing the pumpkin hard, you find yourself reversing direction at a speed of 4.0 m/s. 3.0 seconds after being pushed, the pumpkin will slide 12 m. Assume you weigh 60.0 kg.

We can use the conservation of momentum to solve this problem. After the push, the momentum of the system is given by:

p = (449 kg + 60 kg) * v

where v is the speed of the pumpkin and you after the push. Since you end up sliding backward at 4.0 m/s, we have:

v = -4.0 m/s

Substituting this into the expression for momentum, we find:

p = (449 kg + 60 kg) * (-4.0 m/s) = -2036 kg·m/s

The negative sign indicates that the momentum of the system is in the opposite direction of your motion.

During the sliding motion, the net force on the system is given by:

Fnet = (449 kg + 60 kg) * g * sin(θ)

where g is the acceleration due to gravity (9.81 m/s^2) and θ is the angle of the ramp. Since the ramp is smooth and horizontal, θ = 0 and Fnet = 0. Therefore, there is no net force to change the momentum of the system.

Using the equation for motion with constant acceleration, we can find the distance the pumpkin slides in 3.0 seconds:

x = x0 + v0t + (1/2)at²

Since the initial speed of the pumpkin is -4.0 m/s and there is no net force acting on it, its speed remains constant during the slide. Therefore, v0 = -4.0 m/s and a = 0. Substituting these values, we find:

x = x0 + v0t = (-4.0 m/s) * (3.0 s) = -12 m

The negative sign indicates that the pumpkin slides in the opposite direction to your motion. Therefore, the pumpkin slides 12 meters backward (i.e., towards you) in 3.0 seconds after the push.

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Answer:

Different colour lights (RBG) uses additive properties to bring light of a certain color to our eyes.

Explanation:

:))

The given scenario describes an experiment involving an electroscope, a glass bottle, a cardboard lid, a bolt or nail, a light wire, and an aluminum foil strip.

The purpose of the experiment is to demonstrate the behavior of electrons and their effect on the electroscope. Initially, when the electroscope is neutral, it means that there are equal numbers of protons and electrons on the foil leaves, causing them to hang down straight. The glass bottle and the cardboard lid act as insulators, preventing the rapid escape of electrons and maintaining equilibrium.

When the head of the bolt is touched with a plastic ruler that has been rubbed with fur or wool, the ruler gains excess electrons due to the process of friction. These excess electrons are transferred to the bolt since metal is a good conductor. The electrons then move down the bolt and accumulate on the foil leaves.

As the foil receives the additional electrons, the repulsive force between the like charges (electrons) causes the foil leaves to separate or fly apart from one another. This is a result of the electrostatic repulsion between the negatively charged leaves.

The purpose of using the glass bottle is to provide a protective barrier against air currents that could interfere with the experiment. It also allows observation of the behavior of the foil leaves as they move apart, indicating the presence of excess electrons.

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(1) The tension in cable 2 is 34.83 N.

(2) The coefficient of friction between the block and the floor is 0.45.

The tension in the cable 2 is determined by resolving the forces into various component as shown below.

The vertical component of the forces on the picture is given as;

∑Fy = T₁ sin (65) + T₂ sin (32) - 20 N = 0

T₁ sin (65) + T₂ sin (32) - 20 N = 0

1.7 x sin (65) + 0.53T₂ - 20 = 0

1.54 + 0.53T₂ = 20

0.53T₂ = 18.46

T₂ = 18.46 / 0.53

T₂ = 34.83 N

The value of coefficient of friction is calculated by applying Newton's second law of motion as follows;

F (net) = ma

(F₁ + F₂) - Ff = ma

where;

(F₁ + F₂) - μmg = ma

where;

When block moves at a constant velocity, the acceleration a = 0

(F₁ + F₂) - μmg = 0

μmg  = (F₁ + F₂)

μ = (F₁ + F₂) / mg

μ = ( 195 + 222 ) / ( 95 x 9.8 )

μ = 0.45

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

a) to the right

b) up

c) down

d) there will be no force on the proton

e) there will be no force on the proton

Explanation:

The complete question is

If the velocity of a proton is straight up(thumb pointing up) then RHR2 (right hand rule) shows that the force points to the left. What would the direction of the force be if the velocity were a)down, b)to the right, c)to the left, d)into the page, and e)out of the page?

The right hand rule for a positive charge states that...

Hold the thumb of the right hand at right angle to the rest of the fingers, and the rest of the fingers parallel to one another, and pointing away from the body. If the thumb shows the velocity of a positive charge in a magnetic field, and the fingers all point in the direction of the magnetic field, then the palm will push in the direction of the force on the positive charge (proton).

From this, we can deduce from the original statement about this proton that the direction of the field is into the screen of this computer. with that field direction held constant, we can work out that

a) if the thumb point down, the force will be to the right

b) if the thumb points to the right, the force will be upwards

c)  if the thumb points to the left, the force is downwards

d)  if the thumb points into the page, then there will be no force on the proton since the proton must travel perpendicularly to the magnetic field for a force to be induced on it

e) explanation is the same as foe option d.

Explanation:

c willl fall fast then a and b

Answer:

Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.

The positive charge is released from a point such that it will move along an uniform electric field to the position of lower potential and lower potential energy. Therefore, option (A) is correct,

When a positive charge (say +Q) is released from a point (say A) and moves in an uniform electric field to reach the point (say B), then some work is done on the charge. This work done is given as,

\(W=+Q(V_{A}-V_{B})\)

Here, \(V_{A}\) and \(V_{B}\) are the potential differences between the points A and B respectively..

This means the charge is moving from higher potential to lower potential. And since it is moving along the uniform electric field, therefore the electric potential energy of charged system is decreased.

Thus, we conclude that on releasing the positive charge from a point, it starts moving along the electric field towards the direction of lower electric potential and lower electric potential energy. Hence, option (A) is correct.

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Answer:

10,920 meters

Explanation:

distance = speed × time

which is, d = 21 × 520

Answer:

10920

Explanation:

distance traveled= 21 (speed) ×520 (time)

=10920

\(Question\)

How can scientist best confirm and validate the result of an experiment so they can publish their findings?

Answer:

Hi, there!

D. By creating a replicable experiment Is The correct Answer!

Hope this Helps!!

-xXxAnimexXx- ♚♛♕♔ッ✨♚

Answer:. By creating a replicable experiment

Explanation:

Answer:

A. –3.53 meters/second

Explanation:

Answer:

Your answer is wavelength

As per the given scenario, in this case, the coefficient of friction () is half and the coefficient of restitution (e) is zero.

Identify the body's starting velocity:

We may use the equation of motion to get the body's initial velocity (u)

\(v^2 = u^2 + 2as\)

\(0 = u^2 + 2(-9.8)m/s^2 * h\)

\(u^2 = 19.6h\)

u = √(19.6h)

The body's initial velocity (u) and initial relative velocity (u_rel) are the same.

The body's horizontal velocity immediately following the collision, which is zero, is the final relative velocity (v_rel).

\(e = v_{rel }/ u_{rel}\)

e = 0 / u_rel = 0 / u

Now, one can investigate the forces affecting the body: When a body is on an inclined plane.

There are two main forces at work on it: the frictional force that prevents the body from moving and the gravitational force that pulls it downward (mg).

The gravitational force has two components that act perpendicular to and parallel to the inclined plane, respectively: m*g*cos(45°) and m*g*sin(45°).

Determine the conditions for the body to stop:

μ * N = m * g * sin(45°)

μ * (m * g * cos(45°)) = m * g * sin(45°)

μ * cos(45°) = sin(45°)

(1/2) * cos(45°) = sin(45°)

Simplifying further, we have:

√2 / 4 = √2 / 2

Thus, the body will come to rest following the collision if the equation is valid.

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Answer:

1. Gravity

2. Increases

Explanation:

Gravity is the force that attracts all objects toward each other. This is the force that keeps us grounded on Earth instead of floating in the air.

Although people have gravitational forces, the attraction force is a negligible amount. However, as the mass of the objects are increased, they will exert  far greater gravitational powers to other objects or individuals such as humans.  

According to the given statement the band has been stretched far  is  0.25 m.

The energy that is stored when a force is used to bend an elastic object is known as elastic potential energy. Until the force is released as well as the object springs returns to its original form, doing work in the process, the energy is retained. The object may be squeezed, stretched, or bent during the deformation.

For example, if you pull a spring, it will return to its initial form when you release it (energy input equals energy output.) Elastic potential energy is rendered possible by this.

U= 1.94 J

k  = 58.7  N/m

U = 1/2 kΔ x²

1.94 = 1/2 kΔ x²

3.88/58.7 = Δ x²

x = √(3.88 /58.7)

x = 0.25 m

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(a) When m₂ is released, its acceleration will be approximately -3.27 m/s².

(b) The tension in the string is approximately -13.08 N.

To determine the acceleration of m₂ when it is released and the tension in the string, we need to consider the forces acting on the system.

(a) Acceleration of m₂:

Since the pulley is assumed to be frictionless, the tension in the string is the same on both sides of the pulley. We can consider the system consisting of m₁ and m₂ as one body. The net force acting on this system is the difference between the weight of m₁ and the weight of m₂:

Net force = m₁g - m₂g

Applying Newton's second law, F = ma, where F is the net force and a is the acceleration, we have:

m₁g - m₂g = (m₁ + m₂)a

Rearranging the equation to solve for the acceleration, we get:

a = (m₁g - m₂g) / (m₁ + m₂)

Substituting the given values, m₁ = 2.00 kg and m₂ = 4.00 kg, and the acceleration due to gravity, g = 9.8 m/s², we can calculate the acceleration:

a = ((2.00 kg)(9.8 m/s²) - (4.00 kg)(9.8 m/s²)) / (2.00 kg + 4.00 kg)

a = (19.6 N - 39.2 N) / 6.00 kg

a = -19.6 N / 6.00 kg

a = -3.27 m/s²

Therefore, when m₂ is released, its acceleration will be approximately -3.27 m/s². The negative sign indicates that the acceleration is in the opposite direction of the gravitational force.

(b) Tension in the string:

The tension in the string can be determined by considering the forces acting on m₂. The net force on m₂ is equal to its mass multiplied by its acceleration:

Net force = m₂a

Substituting the given values, m₂ = 4.00 kg and a = -3.27 m/s², we can calculate the tension:

Tension = (4.00 kg)(-3.27 m/s²)

Tension = -13.08 N

Therefore, the tension in the string is approximately -13.08 N. The negative sign indicates that the tension acts in the opposite direction of the weight of m₂.

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Answer:

B

Explanation:

Is the equilibrium stable or unstable for the third charge

Answer:

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In the circuit shown in Figure, initially K

1

is closed and K

2

is open. What are the charges on each capacitor. Then K

1

was opened and K

2

was closed (order is important), What will be the charge on each capacitor now? [C=1μF]

1796057

expand

Medium

Solution

verified

Verified by Toppr

When K

2

is open and K

1

is closed the capacitors C

1

and C

2

will charge and potential develops across them i.e., V

1

and V

2

respectively which will be equal to the potential of battery 9V

∴V

1

+V

2

=9……..I

∵V=

C

q

=orVα

C

1

Or

V

2

V

1

=

C

1

C

2

V

2

V

1

=

6C

3C

3V

2

=6V

1

V

2

=2V

1

…II

From Eqns. I and II

V

1

+2V

1

=9

3V

1

=9

V

1

=3

Volt

V

2

=2×3Volt=6Volt

∴q

1

=C

1

V

1

=6C×3=18C[(from II C=1μF)

=18×1μF=18μc

q

2

=C

2

V

2

=3C×6

=3×1μF×6=18μC

So, charges on each capacitor i.e., q

1

=q

2

=18μc

When k

1

is open and k

2

is closed then charge q

2

will be distributed among

C

2

and C

3

. Let it be q

2

and q

3

∴q

2

=q

2

+q

3

As C

2

and C

3

are now in parallel combination so their potentials

remain same (V)

∴q

2

=C

2

V+C

3

V

18μC=3×1μFxV+3×1μF×V

18=6V

V=3Volt

So potential on C

2

and C

3

capacitors are 3 Volt each

q

2

′

=C

2

V−3×1μF×3Volt=9μC

q

3

=C

3

V=3×1μF×3Volt=9μC

Answer:

True

Explanation:

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

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A Vector is a quantity which has both magnitude as well as direction.

vectors follow parallelogram law of addition and uses cosine of the angle between the vectors.

The addition of two vectors, i.e. c, is given by the formula

c=\(\sqrt{a^2+b^2+2abcos\alpha }\)

where \(\alpha\) is the angle between the vectors

now we know that maximum value of cos\(\alpha\) can be is 1

and minimum value of cos \(\alpha\) can be-1

now the maximum value cos\(\alpha\) is 1 at the angle of π/2

and the minimum value of cosα is -1 is at the angle of -π

Therefore, A and B should have 0 degree between them and in the same direction to have the maximum value

there fore they should be in same direction and 0 degree meaning they should be on same line

A and B will have minimum value when the angle is 180 degree which means they should be on same line and be inn opposite direction to each other.

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Based on the mass of the liquid and the mass of the cylinder, the reading shown on the balance is 1.593 kg.

To calculate the reading shown on the digital balance, we need to first find the mass of the liquid in the cylinder, and then add it to the weight of the empty cylinder.

The volume of the liquid can be found using the formula for the volume of a cylinder:

V = πr²h

where;

r is the radius of the cylinder, and h is the depth of the liquid.

Substituting the given values, we get:

V = π x (3.5 cm)² x 12.0 cm

= 1486.96 cm³

We need to convert the volume from cubic centimeters to cubic meters to use the density of the liquid in kilograms per cubic meter:

V = 1.48696 x 10⁻³ m^3

The mass of the liquid can be found using the formula:

mass = density x volume

Substituting the given values, we get:

mass = 900 kg/m³ x 1.48696 x 10⁻³ m³

= 1.33826 kg

Therefore, the reading shown on the digital balance is:

mass of empty cylinder + mass of liquid

mass of empty cylinder = 2.5/9.81

mass of empty cylinder = 0.2548 kg

reading shown on the digital balance =  0.2548 kg + 1.33826 kg

reading shown on the digital balance = 1.593 kg

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Answer:

a) 103.176 m / min

b) 1751.28 meters

Explanation:

Given:-

- Emma's and Lily's velocities ( E(t) and L(t) ) are given as functions respectively:

                         \(E(t) = \frac{7510}{t^2-7t + 80.22} \\\\L ( t ) = 12t^3*e^-^0^.^5^t\)

- Where, E ( t ) and L ( t ) are given in m / min

- Both run for a total time of 15 minutes in the same direction along the straight track defined by the absolute interval:

                         ( 0 ≤ t ≤ 15 ) mins                  

- It is known that Emma is 10 meters ahead of Lily at time t = 0.

Find:-

a) Find the value of \(\frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\) using correct units, interpret the meaning of

b) What is the maximum distance between Emma and Lily over the time interval 0 ≤ t ≤ 15?  

Solution:-

- The average value of a function f ( x ) over an interval [ a , b ] is determined using calculus via the following relation:

                          \(f_a_v_g = \frac{1}{b-a}\int\limits^a_b {f(x)} \, dx\)

- The first part of the question is asking us to determine the average velocity of Emma over the time interval of ( 2 , 8 ). Therefore, ( E_avg ) can be determined using the above relation:

                         \(E_a_v_g = \frac{1}{8 - 2}*\int\limits^8_2 {E(t)} \, dt\\\\E_a_v_g = \frac{1}{6}*\int\limits^8_2 {E(t)} \, dt\\\)

- We will evaluate the integral formulation above to determine Emma's average velocity over the 2 ≤ t ≤ 8 minute range:

                         \(E_a_v_g = \frac{1}{6}*\int\limits^8_2 {\frac{7510}{t^2 - 7t + 80.22} } \, dt\\\\E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{50t^2 - 350t + 4011} } \, dt\\\\\)

- Complete the square in the denominator:

                          \(E_a_v_g = \frac{1}{6}*37550\int\limits^8_2 {\frac{1}{(5\sqrt{2}*t - \frac{35}{\sqrt{2} })^2 + \frac{6797}{2} } } \, dt\\\\\)

- Use the following substitution:

                          \(u = \frac{5*(2t - 7 )}{\sqrt{6797} } \\\\\frac{du}{dt} = \frac{10}{\sqrt{6797} } \\\\dt = \frac{\sqrt{6797}}{10}.du\)

- Substitute the relations for (u) and (dt) in the above E_avg expression.

                          \(E_a_v_g = \frac{1}{6}*37550\int {\frac{\sqrt{6797} }{5*(6797u^2 + 67997) } } \, du\\\\E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}} \int {\frac{1 }{(u^2 + 1) } } \, du\)

- Use the following standard integral:

                          \(arctan(u) = \int {\frac{1}{u^2 + 1} } \, du\)

- Evaluate:

                         \(E_a_v_g = \frac{1}{6}*37550*\frac{1}{5\sqrt{6797}}* arctan ( u ) |\)

- Apply back substitution for ( u ):

                        \(E_a_v_g = \frac{1}{6}*[\frac{75100* arctan ( \frac{5*(16 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } - \frac{75100* arctan ( \frac{5*(4 - 7 )}{\sqrt{6797} } )}{\sqrt{6797} } ]\\\\\)

- Plug in the limits and find Emma's average velocity:

                        \(E_a_v_g = 151.82037*[arctan (0.54582 ) - arctan ( -0.18194 ) ]\\\\E_a_v_g = 103.176 \frac{m}{min}\)

Answer: Emma's average speed over the interval ( 2 ≤ t ≤ 8 ) is 103.179 meters per minute.

- The displacement S ( E ) of Emma from time t = 0 till time ( t ) over the absolute interval of  0≤t≤15 is given by the relation:

                    \(S (E) = S_o + \int\limits^t_0 {E(t)} \, dt\\\\S ( E ) = 10 + \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } |_0^t\\\\S ( E ) = 10 + [ \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - \frac{75100*arctan( \frac{5*(0 - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } ]\\\\S ( E ) = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } + 375.71098\\\)

- The displacement S ( L ) of Lily from time t = 0 till time ( t ) over the absolute interval of  0 ≤ t ≤ 15 is given by the relation:

                    \(S (L) = \int\limits^t_0 {L(t)} \, dt\\\\S (L) = \int\limits^t_0 ({12t^3 *e^-^0^.^5^t } )\, .dt\\\)

 Apply integration by parts:

  \(S ( L ) = 24t^3*e^-^0^.^5^t - 64*\int\limits^t_0 ({e^-^0^.^5^t*t^2} \,) dt\\\)

 Re-apply integration by parts 2 more times:

 \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2\int\limits^t_0 ({e^-^0^.^5^t*t} \,) dt ]\\\)             \(S ( L ) = -24t^3*e^-^0^.^5^t + 64*[ -2t^2*e^-^0^.^5^t - 2*( -2t*e^-^0^.^5^t - (4e^-^0^.^5^t - 4 ) ]\\\\\)    

\(S ( L ) = e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) - 512 \\\)

- The distance between Emma and Lily over the time interval 0 < t < 15 mins can be determined by subtracting S ( L )  from S ( E ):

                    \(S = S ( E ) - S ( L )\\\\S = \frac{75100*arctan( \frac{5*(2t - 7 )}{\sqrt{6797} }) }{\sqrt{6797} } - e^-^0^.^5^t* ( -24t^3 -128t^2+ 256t + 512) + 887.71098\\\)

- The maximum distance ( S ) between Emma and Lily is governed by the critical value of S ( t ) for which function takes either a minima or maxima.

- To determine the critical values of the function S ( t ) we will take the first derivative of the function S with respect to t and set it to zero:

              \(\frac{dS}{dt} = \frac{d [ S(E) - S(L)]}{dt} \\\\\frac{dS}{dt} = E(t) - L(t) \\\\\frac{dS}{dt} = \frac{7510}{t^2 - 7t+80.22} - 12t^3*e^-^0^.^5^t = 0\\\\( 12t^5 - 84t^4 + 962.64t^3) *e^-^0^.^5^t - 7510 = 0\\\\t = 4.233 , 11.671\)

- We will plug in each value of t and evaluate the displacement function S(t) for each critical value: