How is recall measured in LTM retention?
Recall requires subjects to generate information on their own. The two types of recall are cued recall (fill-in-the-blank) and free recall (no cue).

The given example of a wave in the football game is of 'transverse wave'. The transverse wave is a wave in which points oscillates along the path at the right angle to the direction of travelling wave.

Change in velocity, \(\Delta v\), is the initial velocity subtracted from the final velocity. \(\Delta v = v_{2} - v_{1} = 35 - 20 = 15kmh^{-1}\)

The change in the velocity of Maria's car is equal to 15 Km/h.

Velocity can be defined as a vector measurement of the rate of motion and direction of an object. The velocity of an object can be defined as the rate of change in the position of the body with respect to time.

Velocity can be defined as a vector parameter as it exhibits both magnitude and direction. A mathematical formula to calculate the velocity of the object can be represented as:

v = d / t

Where v is the velocity of the object, d is the distance, and t is the time taken by the object.

Although the S.I. unit for the measurement of the velocity is m/s and can also expressed in the terms of miles per hour (mph), and kilometers per hour(kph).

Given, the initial velocity of Maria's car, v₁ = 20 m/s

The final velocity of Maria's car, v₂ = 35 m/s

The change in the velocity of Maria's car, Δv = v₂ - v

Δv = = 35 - 20

Δv = 15 km/h

Therefore, the change in velocity is 15 Km/h.

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Answer:

Troposphere, stratosphere, mesosphere and thermosphere. The next region is the exosphere, but that region is 500+ km from the Earth's surface.

\(\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}\)

Actually Welcome to the Concept of the Atmosphere.

The four layers of Atmosphere starting from bottom are as follows:

1.) Troposphere - The troposphere is the lowest layer of our atmosphere. Starting at ground level, it extends upward to about 10 km (6.2 miles or about 33,000 feet) above sea level.

2.) Stratosphere - The next layer up is called the stratosphere. The stratosphere extends from the top of the troposphere to about 50 km (31 miles) above the ground.

3.) Mesosphere - Above the stratosphere is the mesosphere. It extends upward to a height of about 85 km (53 miles) above our planet. Most meteors burn up in the mesosphere.

4.) Thermosphere - The layer of very rare air above the mesosphere is called the thermosphere. High-energy X-rays and UV radiation from the Sun are absorbed in the thermosphere, raising its temperature to hundreds or at times thousands of degrees.

The x and y component of the ball's final velocity are respectively 7.35 m/s and  4.90 m/s.

The rate at which a body's displacement changes in relation to time is known as its velocity. Velocity is a vector quantity with both magnitude and direction. SI unit of velocity is meter/second.

Given that:

Mass of the ball: M = 6.35 kg.

Initial velocity of ball: U = 8.49 m/s.

Mass of the pin at rest: m = 1.59 kg.

Final velocity of pin: v = 20.1 m/s at a -77.0° angle.

Let the x and y component of the ball's final velocity are respectively V₁ m/s and  V₂ m/s.

Appling conservation of momentum along x axis:

MU + m.0 = MV₁ + mvcos(-77.0°)

⇒ V₁ = u - (m/M) v cos(-77.0°)

After putting the values we get:

V₁ = 7.35 m/s.

Appling conservation of momentum along y-axis:

M.0 + m.0 = MV₂ + mvsin(-77.0°)

⇒ V₂ = - (m/M) vsin(-77.0°)

After putting the values we get:

V₂ = 4.90 m/s.

Hence, the x and y component of the ball's final velocity are respectively 7.35 m/s and  4.90 m/s.

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Answer:

Because it is the standard operational range for measuring the body temperature of a living individual.

Explanation:

When making a measuring equipment like a thermometer, you attempt to keep the reading range (Span) as small as possible so that the reading is as accurate as feasible. Because sensors are often rated in percent of span (in this example, span = 45 C–32 C = 13 degrees Celsius), One percent of 13 equals 0.13 degrees Celsius, thus if your sensor has a usual accuracy of 1%, your temperature reading is accurate to roughly 0.1 degrees Celsius.

Your precision would be 0.2 degrees Celsius if you doubled the span. The reading accuracy is now becoming unsatisfactory. Readability is another advantage of analog and mercury clinical thermometers. The markings for 0.1 deg C would be ten times closer together if a mercury thermometer had ten times the range (span). There is a limit to the device's readability.

Answer:

4.44 hr

Explanation:

distance= rate*time

400km/(90 km/hr)=time

time=4.44 hr

1) The force acting on the object during the time interval 0-3 seconds is 1/3 N.

2) The force acting on the object during the time interval 6-10 seconds is -0.5 N.

3) There is no time interval in which no force acts on the object.

(i) Force acting on the object in time interval 0-3 seconds. Force acting on the object is equal to the product of its mass and acceleration, i.e.,F = ma.

In the given velocity-time graph, the acceleration of the object can be determined by determining the slope of the velocity-time graph from 0 to 3 seconds.

Slope = (change in velocity) / (change in time)= (20-0) / (3-0) = 20/3 m/s^2

Acceleration, a = slope= 20/3 m/s^2

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × 20/3= 1/3 N.

Therefore, the force acting on the object during the time interval 0-3 seconds is 1/3 N.

(ii) Force acting on the object in time interval 6-10 seconds. Similar to the first question, the force acting on the object in time interval 6-10 seconds can be determined by determining the acceleration of the object during this time interval.

The slope of the velocity-time graph from 6 seconds to 10 seconds can be determined as follows:

Slope = (change in velocity) / (change in time)= (-20-20) / (10-6) = -40/4= -10 m/s^2 (negative sign indicates that the object is decelerating)

Mass of the object, m = 50 g = 0.05 kg

∴ Force acting on the object, F = ma= 0.05 × (-10)= -0.5 N.

Therefore, the force acting on the object during the time interval 6-10 seconds is -0.5 N.

(iii) Time interval in which no force acts on the object. There is no time interval in which no force acts on the object. This is because, as per Newton's first law of motion, an object will continue to remain in a state of rest or uniform motion along a straight line unless acted upon by an external unbalanced force.In other words, if the object is moving with a constant velocity, there must be a force acting on the object to maintain its motion.

Therefore, there is no time interval in which no force acts on the object.

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Answer:

Approximately \(0.293\; {\rm kg \cdot m^{2}}\).

Explanation:

Since the wheel started from rest, initial angular velocity will be \(\omega_{0} = 0\; {\rm rad \cdot s^{-1}}\). It is given that the angular velocity \(\omega_{1}\) is \(14.9\; {\rm rev \cdot s^{-1}}\) after \(t = 3.41\; {\rm s}\). Apply unit conversion and ensure that all angular velocity are measured in radians-per-second:

\(\begin{aligned} \omega_{1} &= 14.9\; {\rm rev \cdot s^{-1}} \times \frac{2\, \pi \; {\rm rad}}{1\; {\rm rev}} \\ &\approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}\).

Change in angular velocity:

\(\begin{aligned} \Delta \omega = \omega_{1} - \omega_{0} \approx 93.620\; {\rm rad \cdot s^{-1}}\end{aligned}\).

Since the tangential force is constant and there is no friction on the wheel, the angular acceleration \(\alpha\) of this wheel will be constant. Since the change in velocity \(\Delta \omega \approx 93.620\; {\rm rad \cdot s^{-1}}\) was achieved within \(t = 3.41\; {\rm s}\), the average angular acceleration will be:

\(\begin{aligned} \alpha &= \frac{\Delta \omega}{t} \\ &\approx \frac{93.620\; {\rm rad \cdot s^{-1}}}{3.41\; {\rm s}} \\ &\approx 27.45\; {\rm rad \cdot s^{-2}}\end{aligned}\).

At a distance of \(r = 0.110\; {\rm m}\) from the axis of rotation, the tangential force \(F = 73.0\; {\rm N}\) will exert on the wheel a torque \(\tau\) of magnitude:

\(\begin{aligned} \tau &= F\, r \\ &= (73.0\; {\rm N})\, (0.110\; {\rm m}) \\ &\approx 8.030\; {\rm N \cdot m}\end{aligned}\).

Let \(I\) denote the moment of inertia of this wheel. The equation \(\alpha = (\tau / I)\) relates angular acceleration \(\alpha\) to moment of inertia \(I\!\) and net torque \(\tau\). Rearrange this equation to find the moment of inertia:

\(\begin{aligned}I &= \frac{\tau}{\alpha} \\ &\approx \frac{8.030\; {\rm N\cdot m}}{27.45\; {\rm rad \cdot s^{-2}}} \\ &\approx 0.293\; {\rm N \cdot m \cdot s^{2}} \\ &= 0.293 \; {\rm kg \cdot m^{2}}\end{aligned}\).

Note that the unit "radians" is typically ignored. Additionally, \(1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}\).

Hence, the moment of inertia of this wheel is approximately \(0.293\; {\rm kg \cdot m^{2}}\).

Answer:the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.

Explanation:

The Enron scandal drew attention to accounting and corporate fraud as its shareholders lost $74 billion in the four years leading up to its bankruptcy, and its employees lost billions in pension benefits.

Enron's Board of Directors failed to fulfil its fiduciary duties towards the corporation's shareholders. Secondly, the top executives of Enron were greedy and acted in their own self-interest.

Enron's downfall was attributed to its reckless use of derivatives and special purpose entities. By hedging its risks with special purpose entities which it owned, Enron retained the risks associated with the transactions. This arrangement had Enron implementing hedges with itself.

Enron raised fundamental issues about corporate fraud, accounting transparency, and investor protection.

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The size of the angle θ of a point object moving from point A to point B along a circular path is 2πR / L.

To understand this, consider a simple example. Suppose that a point object that moves from point A to point B along a circular path with a radius of 1 meter. The distance between points A and B is also 1 meter. Therefore, the size of the angle θ is equal to 2π × 1 / 1 = 2π radians.

In general, the size of the angle θ = ratio of the circumference of the circle to the distance between points A and B. The circumference of the circle is equal to 2πR, where R = radius of the circle. Therefore, the size of the angle θ is equal to 2πR / L.

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Complete question:

A point object moves from point A to point B along a circular path with a radius R. What is the size of the angle θ?

a) The total surface area of tin G in terms of π is 170π cm².

b) The mass of tin H is 336 g.

To solve the given problem, we need to determine the total surface area of tin G in terms of π and the mass of tin H. Since the mass of an empty cylindrical tin is proportional to its surface area, we can use the given information to find the solutions.

a) Total surface area of tin G in terms of π:

The surface area of a cylinder consists of two circular bases and the lateral surface area. The formula for the lateral surface area of a cylinder is given by:

Lateral surface area = 2πrh

where r is the radius of the base and h is the height of the cylinder.

In the case of tin G, the given dimensions are a radius of 5 cm and a height of 12 cm. Substituting these values into the formula, we can calculate the lateral surface area:

Lateral surface area = 2π(5 cm)(12 cm)

Lateral surface area = 120π cm²

Since the total surface area of the cylinder includes the two circular bases as well, we need to add their areas. The area of a circle is given by:

Area of a circle = πr²

The radius of the circular base of tin G is 5 cm, so the area of each circular base is:

Area of each circular base = π(5 cm)²

Area of each circular base = 25π cm²

To find the total surface area of tin G, we sum the lateral surface area and the areas of the two circular bases:

Total surface area of tin G = Lateral surface area + 2 × Area of each circular base

Total surface area of tin G = 120π cm² + 2 × 25π cm²

Total surface area of tin G = 120π cm² + 50π cm²

Total surface area of tin G = 170π cm²

Therefore, the total surface area of tin G in terms of π is 170π cm².

b) Mass of tin H:

We are given that the surface area of tin H is 792π cm². We can assume that the same proportionality factor applies as in tin G, so we can set up the following proportion:

(surface area of tin G) / (mass of tin G) = (surface area of tin H) / (mass of tin H)

Using the given values, we have:

(170π cm²) / (72 g) = (792π cm²) / (mass of tin H)

Cross-multiplying and solving for the mass of tin H, we get:

(170π cm²) × (mass of tin H) = (72 g) × (792π cm²)

mass of tin H = (72 g) × (792π cm²) / (170π cm²)

mass of tin H = 336 g

Therefore, the mass of tin H is 336 g.

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Answer:

blanket:thermal conductivity

as it conduct heat.

aquarium glass:transparency

as light can be transfered from one side to

another.

Answer:

When a teen driver drives with a lot of his peers as passengers they may lead to distraction which may later end up in accident as the driver was distracted

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

When a teen driver drives with a lot of his peers as passengers they may direct to distraction which may later end up in casualty as the driver was distracted.

Several studies have indicated that passengers substantially increase the chance of crashes for young, novice drivers. This improved risk may result from distractions that young passengers complete for drivers.Teens driving with a teen or young adult passengers existence of teen or young adult passengers raises the crash risk of unsupervised teen drivers. This risk grows with each additional teen or a young adult passenger.

Crash risk is two- to six times more significant for those who utilize a cellphone while driving resembled for drivers who are not distracted. Using a phone delays reaction time increases lane deviations, and forces drivers to look away from the road for extended times.

Overconfidence, lack of focus, and phone while driving are the factors  contribute to accidents when a teen driver controls other teens as passengers,

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The resultant of two displacement vectors having opposite directions is the difference of the two displacements, having the same direction as the smaller vector.

When two displacement vectors have opposite directions, it means they are pointing in opposite ways. In other words, one vector is in the opposite direction of the other. To find the resultant of these vectors, we need to subtract one vector from the other.

If we consider two displacement vectors, let's say vector A and vector B, and they have opposite directions, we can represent them as A and -B.

To find the resultant, we subtract vector B from vector A: A - (-B) or A + B.

The resultant will have the same direction as the smaller vector. This is because when we subtract a larger vector from a smaller vector, the resultant will have the direction of the smaller vector.

Therefore, the correct option is: "The resultant is the difference of the two displacements, having the same direction as the smaller vector."

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Answer:

E = 4000 V / m

U = 1.92*10^-18 J

C' = 4.71 pF

1.2 times greater with di-electric

Explanation:

Given:-

- The potential difference between plates, V = 12 V

- The area of each plate, A = 7.6 cm^2

- The separation between plates, d = 0.3 cm

- The charge of the proton. q = 1.6*10^-19 C

- The initial velocity of proton, vi = 0 m/s

Solution:-

- The electric field ( E ) between the parallel plates of the air-filled capacitor is determined from the applied potential difference by the battery on the two ends of the plates.

- The separation ( d ) between the two plates allows the charge to be stored and the Electric field between two charged plates would be:

                          E = V / d

                          E = 12 / 0.003

                          E = 4,000 V/m ... Answer

- The amount of electrostatic potential energy stored between the two plates is ( U ) defined by:

                         U = q*E*d

                         U = (1.6 x10^-19)*(4000)*(0.003)

                         U = 1.92*10^-18 J  ... Answer

- The electrostatic energy stored between plates is ( U ) when the proton moves from the positively charges plate to negative charged plate the energy is stored within the proton.

- A slab of di-electric material ( Teflon ) is placed between the two plates with thickness equal to the separation ( d ) and Area similar to the area of the plate ( A ).

- The capacitance of the charged plates would be ( C ):

                        C = k*ε*A / d

Where,

            k: the di-electric constant of material = 2.1

            ε: permittivity of free space = 8.85 × 10^-12

- The new capacitance ( C' ) is:

                      C' = 2.1*(8.85 × 10^-12) *( 7.6 / 100^2 ) / 0.003

                      C' = 4.71 pF

- The new total energy stored in the capacitor is defined as follows:

                     U' = 0.5*C'*V^2

                     U' = 0.5*(4.71*10^-12)*(12)^2

                     U' = 3.391 * 10^-10 J

- The increase in potential energy stored is by the amount of increase in capacitance due to di-electric material ( Teflon ). The di-electric constant "k" causes an increase in the potential energy stored before and after the insertion.

- Hence, the new potential energy ( U' ) is " k = 2.1 " times the potential energy stored in a capacitor without the di-electric.

We can see that Shakespeare's structure when viewed through a formalist lens is to: C. Introduce a new motif that focuses on the inevitability of death.

Who is William Shakespeare?

William Shakespeare was known as an English actor, poet and also a playwright. Many regard Shakespeare as the greatest writer in the English language. He is often seen as the world's greatest dramatist.

William Shakespeare was known to have written several plays like:

The reader aims to comprehend how words interact with one another and how different parts of the text relate to one another.

Therefore, William Shakespeare employ this technique in his study and is likely to be interested in how the various components of the text interact.

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It would require 28618.56 J of heat to raise the temperature of 90 g of water from 12°C to 88°C.

Given that:

Mass of water m = 90g

Initial temperature = 12°C

Final temperature = 88°C,

The amount of heat required to raise the temperature of a substance is given by:

Q = mCΔT

Where Q is the amount of heat, m is the mass of the substance, C is the specific heat capacity of the substance, and ΔT is the change in temperature.

Note that: the specific heat capacity of water is 4.184 J/g°C.

So, for raising the temperature of 90 g of water from 12°C to 88°C, we have:

Q = mCΔT

Q = (90 g) × (4.184 J/g°C) × (88°C - 12°C)

Q = (90 g) × (4.184 J/g°C) × (76°C)

Q = 28618.56 J

Therefore, the required heat is 28618.56 Joules.

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Answer:

b Shiny

Explanation:

Trust me it's right

(a) For the density function, the value of B is 2.5 g/cm³   and the value of C is 1.3 g/cm⁴

(b) The total mass of the rod is 1470 g.

The density of the rod is a function of distance and it is given as;

ρ = B + Cx

where;

ρ = 2.5 g/cm³  +  (19.5 g/cm³ ) / ( 15 cm ) x

ρ = 2.5 g/cm³  +  1.3 g/cm⁴  x

The total mass of the rod is calculated by integrating the function;

dm = ( B + Cx)(8 cm² ) dx

m = 8B + 8Cx

m = 8Bx  +  8Cx² / 2

m = ( 8 x 2.5 x 15 )  +  ( 8 x 1.3 x 15² ) / 2

m = 1470 g

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If the universe were to suddenly begin shrinking rather than continuing to expand, it would have a significant effect on the cosmic microwave background radiation (CMB).

The CMB is the afterglow of the Big Bang and is observed as a nearly uniform background radiation in all directions. It is thought to have been emitted when the universe was about 380,000 years old and had cooled enough for neutral atoms to form.

If the universe were to suddenly begin shrinking, the photons in the CMB would lose energy as they travel through the contracting space. This would cause the CMB radiation to shift to shorter wavelengths, which is known as blue-shifting.

Therefore, the correct answer is B. It would blue-shift.

The future consequence of using windmills for wind energy that is most closely related to the given options is: A) It can harm birds and species nearby. Option A

One of the potential consequences of using windmills for wind energy is the risk of harm to birds and other species. Wind turbines can pose a threat to birds, especially large raptors and migratory birds, as they can collide with the spinning turbine blades.

The fast-moving blades can cause injury or death to birds that come into contact with them. Additionally, the construction and operation of wind farms can disrupt wildlife habitats and migration patterns, impacting local ecosystems.

While weather can certainly affect the quality and consistency of wind energy generation (option B), it is not specifically a consequence of using windmills. Weather patterns and variations in wind speed and direction can influence the efficiency and reliability of wind turbines, but this is an inherent characteristic of wind energy rather than a consequence.

Option C states that wind energy produces less noise than other energy sources. This is a positive attribute of wind energy, as wind turbines generally generate less noise compared to other forms of power generation, such as fossil fuel power plants. However, it is not a future consequence but rather a benefit of wind energy.

Option D refers to wind cells being used in isolated locations. This statement is not related to the consequences of using windmills for wind energy but rather describes the potential use of wind cells (small-scale wind energy systems) in remote or isolated areas.

In summary, the most appropriate answer is A) It can harm birds and species nearby, as the impact on wildlife is a significant consideration in the development and operation of wind energy projects.

Option A.

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The time taken for the cannonball to move upward before stopping, given that is was fired directly upward at 300 m/s is 30.6 seconds

Velocity and time are related according to the following equation of motion

v = u + gt

Where

From the question given above, the following data were obtained:

The time taken for the cannonball to move upward before stopping can be obtained as illustrated below:

v = u - gt (since the ball is going against gravity)

0 = 300 - (9.8 × t)

0 = 300 - 9.8t

Collect like terms

-9.8t = 0 - 300

-9.8t = -300

Divide both sides by -9.8

t = -300 / -9.8

t = 30.6 seconds

Thus, the time taken is 30.6 seconds

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The current flowing in the 2Ω and 1Ω is 1.14 A and the current flowing in the 3Ω and 4Ω is 0.286 A.

The value of the current in each resistor is calculated by applying Kirchoff voltage law as follows;

The total voltage in loop 1 is calculated as;

2 + 4 - I₁R₁ - (I₁ - I₂)R₂ - I₁R₃ = 0

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

The current flowing in loop 2 is calculated as;

I = V/R

I₂ = ( 6 V - 4 V ) / (3 + 4)

I₂ = 0.286 A

The value of the current flowing in loop 1 is calculated as;

6 - 2I₁ - 3(I₁ - I₂) - 1₁ = 0

6 - 2I₁ - 3(I₁ - 0.286) - 1₁ = 0

6 - 3I₁ - 3₁ + 0.858 = 0

-6I₁ = -6.858

I₁ = 6.858 / 6

I₁ = 1.14 A

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Answer:

1. A fossil that is widespread geographically but only occurs in one layer or a small number of layers of rock

2. Sediment deposited into water will spread in a horizontal and continuous sheet

3. Any undisturbed sequence of layered rocks has the oldest rock on the bottom and newest rock on the top

4. Study of fossils

5. Study of rock layers and the process that form them

Sorry if any of these are wrong

Answer:

the speed of the nerve impulse in miles per hour is 201.59 mi/hr

Explanation:

Given;

the speed of the nerve impulse, v = 90.1 m/s

To convert this speed in meters per second to miles per hour, we use the following method;

1,609 meter = 1 mile

3,600 s = 1 hour

\(v(mi/h) = 90.1 \ \frac{m}{s} \times \frac{1 \ mile}{1,609 \ m} \times \frac{3,600 \ s}{1 \ hour} = (\frac{90.1 \times 3,600}{1,609} )\frac{mi}{hr} = 201.59 \ mi/hr\)

Therefore, the speed of the nerve impulse in miles per hour is 201.59 mi/hr

\(\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}\)

                           \(\large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}\)

                   \(\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}\)

                         \(\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}\)

                         \(\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}\)

                         \(\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}\)

                         \(\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}\)

                  \(\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}\)

                          \(\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}\)  

                          \(\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}\)

                          \(\large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}\)

The electric field at point B, located at the midpoint between the upper left and right corners of the square, can be approximated as 3.244 x \(10^6\) N/C in the upward direction.

To find the electric field at point B, we can consider the contributions from the two charges placed at the lower corners of the square. Since the charges are the same and the distance to point B is the same for both charges, the magnitudes of the electric fields produced by each charge will be equal.

First, let's calculate the magnitude of the electric field produced by one of the charges at point B using Coulomb's Law:

Electric field due to a point charge (E) = (k * q) / \(r^2\)

Where:

- k is the electrostatic constant, approximately equal to 8.99 x 10^9 \(10^9\)N \(m^2/C^2\)

- q is the charge of the object, 7.25 μC (7.25 x \(10^-^6\) C)

- r is the distance from the charge to point B, which is half the length of the square's side, 0.201 m / 2 = 0.1005 m

Plugging in the values, we have:

E = (8.99 x \(10^9 N m^2/C^2\) * 7.25 x \(10^-^6\) C) / (0.1005 \(m)^2\)

E ≈ 1.622 x \(10^6\) N/C

Since the electric fields from the two charges at the lower corners have equal magnitudes and point in the same direction (upward), the total electric field at point B is twice the magnitude of the individual electric field:

Electric field at point B = 2 * E ≈ 2 * 1.622 x \(10^6\) N/C

Electric field at point B ≈ 3.244 x \(10^6\) N/C

Therefore, the electric field at point B, midway between the upper left and right corners of the square, is approximately 3.244 x \(10^6\)N/C upward.

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