Consider the f(x) = Acos(x) function shown in the figure in blue color. What is the value of amplitude A for this function?
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.

Answer:  It simply means that a freely falling object would increase its velocity by 9.8 m/s per second

Answer:

Hello!

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that

That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

The summary of the step you might use to carry out an investigation using scientific method include:

This is defined as a systematic approach which helps to predict how several elements in the universe works.

This usually starts with observing a phenomenon and making a hypothesis. This is usually followed by conducting experiment to ascertain its authenticity before a conclusion is drawn.

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Answer:

4 m/s

Explanation:

Conservation of momentum ,  mv

Before collision  80 kg * 6 m/s  = 480  kg-m/s  

   after the collision,

         the momentum is the same but the mass is 80+40 kg =120 kg

 mv = 480

120 * v = 480

v = 4 m/s

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

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Answer:

B Gravitational potential energy

Explanation:

The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity

=> 0.029kg × 200 = 3.429 × velocity of pendulum

=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s

1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²

=> 10 = 13.44× amplitude²

=> Amplitude = √(10/13.44)

= 0.86m

Angle = tan inverse (amplitude/length of pendulum)

= tan inverse (0.86/2.5)

= 20°

Maximum displacement along vertical direction= L - L×cos20°

= 2.5 - 2.5×cos20°

= 0.15 m

Maximum displacement along horizontal direction= Lsin20°

= 2.5 × sin 20°

= 0.85m

Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

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4. a poor insulator

If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.

Answer:

4 conductors and insulators are opposites of each other.

Explanation:

The value of parameter B (angular frequency) for this function is 2.7.

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(Bx)

where;

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue color; at y = -0.9, x = -7.5 cm

-0.9 = cos(-7.5B)

7.5B = arc cos (0.9)

7.5B = 2.7

B = 7.5/2.7

B = 2.7

Thus, the value of parameter B (angular frequency) for this function is 2.7.

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The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

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Answer:

Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].

Explanation:

Assume that the radius of this orbit is [tex]r[/tex].

Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].

The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].

In other words:

[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].

The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:

[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].

Substitute this value into the expression for [tex]r[/tex] and evaluate:

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].

(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)

Answer:

the earth receives the suns energy as radiation which is able to travel without the need of matter from point A to point B and this radiation is also in the form of heat.

Explanation:

Sun's energy fall on the earth in form of radiation and heats up the atmosphere and produces unequal heating due to its spherical nature and impacts climate of different areas.

The Sun is the only source of energy for the Earth's climate system. Fundamental to atmospheric composition is the warming of the atmosphere caused by solar radiation, which also causes global wind patterns and aids in the development of clouds, storms, and rainfall.

When the sun is at its maximum (peak in sunspots, prominences, and flares), it produces more ultraviolet (UV) radiation. More ozone is produced as a result of UV light, warming the stratosphere. The lower atmosphere and Earth's surface warm slightly as a result of the increased solar radiation.

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The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

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6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.

The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.

The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms,  [tex]x_{c}[/tex] and  [tex]x_{o}[/tex] their distances relative to C atom.

Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )

[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]

[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m

Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.

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A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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Given the following data:

First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.

The electric field due to B₁ to B₃ is given by:

E₁ = kq₁/r₁²

E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)

E₁ = 153 × 10³/0.0125

Electric field, E₁ = 12.24 × 10⁶ N/C.

The electric field due to B₂ to B₃ is given by:

E₂ = kq₂/r₂²

E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)

E₂ = 45 × 10³/0.01

Electric field, E₂ = 4.5 × 10⁶ N/C.

Also, the magnitude of the angle formed is given by tan trigonometry:

Tanθ = 5/10

Tanθ = 0.5

θ = tan⁻¹(0.5)

θ = 26.56°.

Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁x = E₁cosθ - E₂

E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶

E₁x = 10.95 × 10⁶ - 4.5 × 10⁶

E₁x = 6.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁y = -E₁sinθ

E₁y = -12.24 × 10⁶ × sin(26.56)

E₁y = -12.24 × 10⁶ × 0.4471

E₁y = -5.5 × 10⁶ N/C.

Also, the magnitude of the electric field is given by:

Exy = √(E₁x² + E₁y²)

Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)

Exy = √4.225 × 10¹³ + 3.025 × 10¹³)

Exy = √(7.25 × 10¹³)

Exy = 8.5 × 10⁶ N/C.

We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:

E₂x = -E₂sinθ

E₂x = -4.5 × 10⁶ × sin(90)

E₂x = -4.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₂y = E₂cosθ

E₂y = -4.5 × 10⁶ × cos(90)

E₂y = 0 N/C.

The magnitude of the net electric field at B₃ is given by:

Eₙ = √(E₁x + E₂x)² + E₁y²)

Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)

Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)

Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)

Eₙ = √(3.65 × 10¹³)

Eₙ = 6.04 × 10⁶ N/C.

For the direction, we have:

Tanθ = x/y

Tanθ = 6.5/-5.5

Tanθ = -1.1818

θ = tan⁻¹(-1.1818)

θ = 49.76°.

The magnitude of the force on B₃ due to this net electric field is given by:

F = B₃ × Eₙ

F = 26 × 10⁻⁶ × 6.04 × 10⁶

F = 157.04 Newton.

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Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

Answer: By dividing airflow by duct cross section.

Explanation:

In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.

In thicker materials the particles can move more easily, therefore the resistance has to decrease.

Hence Option (b) is correct.

Resistivity, which is sometimes represented by the Greek letter rho, is quantitatively equal to a specimen's resistance R times its cross-sectional area A times its length.

ρ = R × A/ l

R = ρ × l/ A

It means the Resistance of the material is inversely proportional to the area of the material.

As the area of the material increases, the resistance decreases and vice versa.

Hence, In thicker particles, there is a greater area available and that's why resistance has to decrease so that particles can move more easily.

Hence, Option (b) is correct.

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Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.

The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.

Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.

What is work if the displacement is not in the direction of force ?

The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.

Also, the work done will be zero, if the displacement is zero.

In the question above, the displacement is completely perpendicular to the direction of the applied force.

Therefore, the work done on the bag is zero.

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The force required to pull the two hemispheres is 46622.72N

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

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The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Mass of the astronaut, [tex]m_a[/tex] = 126 kg

Speed he acquires, [tex]v_{a}[/tex]  = 2.70 m/s

Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

[tex]P_f = m_av_a + m_cv_c[/tex]

[tex]P_I[/tex] = 0

[tex]m_av_a + m_cv_c = 0[/tex]

[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]

   [tex]= \frac{126* 2.70}{1800}[/tex]

   [tex]= - 0.189[/tex] m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     [tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     = [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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The correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

Energy balance refers to the way in balance is achieved when intake of energy is equal to energy expended.

Energy refers to the impetus behind all motion and all activity. If is also the capacity to do work. Energy is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent.

So therefore, the correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

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(a) Walking 73.2 m at 1.22 m/s would take

[tex]\dfrac{73.2\,\rm m}{1.22 \frac{\rm m}{\rm s}} = 60 \,\rm s[/tex]

and running 13.2 m at 3.02 m/s would take

[tex]\dfrac{13.2\,\rm m}{3.02\frac{\rm m}{\rm s}} \approx 4.37\,\rm s[/tex]

You've undergone a total displacement of 73.2 + 13.2 = 86.4 m in a matter of approximatly 64.37 s, so your average velocity is

[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{86.4\,\mathrm m}{64.37\,\rm s} \approx \boxed{1.34\dfrac{\rm m}{\rm s}}[/tex]

(b) In the first 1.00 min = 60 s, you undergo a displacement of

[tex](60\,\mathrm s) \left(1.22 \dfrac{\rm m}{\rm s}\right) = 73.2 \,\rm m[/tex]

and in the second minute, you undergo a displacement of

[tex](60\,\mathrm s) \left(3.05\dfrac{\rm m}{\rm s}\right) = 183 \,\rm m[/tex]

Your total displacement is then 73.2 + 183 = 256.2 m in a matter of 2.00 min = 120 s, so your average velocity is

[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{256.2\,\mathrm m}{120\,\rm s} \approx \boxed{2.14\dfrac{\rm m}{\rm s}}[/tex]

(c) For part (a), your displacement [tex]x(t)[/tex] (in m) at time [tex]t[/tex] (in s) is given by

[tex]x(t) = \begin{cases}1.22t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.02 (t-60) & \text{for } t > 60\end{cases}[/tex]

and for part (b), your displacement is given by the very similar

[tex]x(t) = \begin{cases}1.22 t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.05(t-60) & \text{for } t > 60 \end{cases}[/tex]

See the attached plots. The average velocity for the given situation is the slope of the dotted line.

As the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.

Gravitational force is a force that attracts any two objects with mass.

According to Newton's law of universal gravitation, the force of attraction between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.

F = Gm₁m₂/R²

where;

Thus, as the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.

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Answer: 1.348 meters

Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point  c . Adding the four moments together, we get:

(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0

6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0

11.5c=15.50

c=  1.348 metres

The center of mass is on the y axis at  y  = 1.348 metres.

The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.

An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.

High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.

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Answer:

The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2[tex]\pi \sqrt{0.5*9.8*0.3}[/tex]

T=1.2042s

The period of the oscillations.T = 1.2042s

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From the first circuit (see attached), it is correct to state that I₄ = 40mA. See the explanation below.

A path that is designed to transport electric current is that is referred to as a circuit.

I = 200mA

= 20mA + 140mA + I₄

I₄ = 200 - 160

I₄ = 40mA

From the above, we can rightly posit:

R₄ = 3/(40x 10⁻³)

= 75 Ω

Hence

V₆ =  V₄ - V₅

= 3v - 2v

= 1v

R₆ therefore, =

V₆/I₆

= 1/ (40 x 10 ⁻³)

= 14.28 Ω

I₃ = I₄ + I₆

= 40 + 140

= 180mA

R₃ = 9/(180 x 10⁻³)

= 50Ω

R₂ = 12/20 x (10⁻³)

= 60 Ω

Then,

Vbattery =

200 x 10⁻³ x90

= 18 volts

The above indices is used to complete the image. See Circuit Image II.

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The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

The term report refers to any formal presentation. It is usually made as part of an academic course or as part of professional development. It could be given while a person is hoping for promotion at a job that he or she is doing.

We have to bear in mind that the audience to whom the report is being presented is a familiar audience hence the presentation can be quite informal and devoid of the formality of presentation of reports as we know it to be.

Now, the focus is to make the presentation to look informal. The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

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