(refer to photo attached) Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C

If a charge of −4.72 µC is placed at this point, what are the magnitude and direction of the force on it? Magnitude _______N

Direction?

- toward the left
- upward
-downward
- toward the right

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.

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 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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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.

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) =  e^(-bx)

ln[-0.857 / cos(ω)] = -bx  

ln[-0.857 / cos(ω)] / b = - x  ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx  

ln[-0.57 / cos(2ω)] /2b = - x  ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) =  cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω  - cosω - 6 = 0

let cosω  = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω  = arc cos(-0.92)

ω  = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b  ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

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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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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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The electric filed in V/m is 1.58 * 10^6 V/m

We know that the electric field is obtained as the ratio of the voltage to the distance that separates the plates.

Thus;

E = V/d

E = electric field

V = voltage

d = distance of separation

E = 24 * 10^3 V/1.52 * 10^-2 m

E = 1.58 * 10^6 V/m

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(a) The electric field strength at a point 1.00 cm to the left of the middle is  2.0 x 10⁷ N/C.

(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)

E = +2.0 x 10⁷ N/C

F = Eq

F = 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = -94.4 N

Thus, the direction of the force will be towards the left.

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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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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

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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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 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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The fraction of the magnitude of the kinetic energy lost is [tex]\frac{(change) KE }{KE} = 1 - \frac{m}{m + M }[/tex]. = 0.955.

using the law of conservation of momentum,

[tex]mv=(m+M)V[/tex]

[tex]V= m ( \frac{1}{M+m} ) v\\[/tex]

kinetic energy lost,

Δ[tex]KE=KE_{i} -KE_{f}[/tex]

(see image )

now, for the other part

the fraction of the kinetic energy lost,

ΔKE/ KE = [tex]1- m (\frac{1}{m+M} )[/tex]

ΔKE/ KE = [tex]1 - 18 ( \frac{1}{18 + 380 } )[/tex]

ΔKE/ KE = 0.955.

what is kinetic energy ?

The energy that an object has as a result of motion is known as kinetic energy. It is stated as the effort needed to transfer a bulk body from rest to the given velocity. The body holds onto the kinetic energy it gained during its acceleration until its speed changes.

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(a) The amplitude of the magnetic field produced is 2.667 mV/m.

(b) The intensity of the laser is 2.832 W/m².

(c) The  power delivered by the laser is 2.22 x 10⁻⁶ W.

In electromagnetic waves, the amplitude of magnetic field is the maximum field strength of the magnetic fields.

B₀ = E₀/c

where;

B₀ = (0.8 MV/m) / (3 x 10⁸)

B₀ = (0.8  x 10⁶ V/m) / (3 x 10⁸)

B₀ = 2.667 x 10⁻³ V/m

B₀ = 2.667 mV/m

The intensity of the laser is calculated as follows;

I = ¹/₂ε₀E₀²

I = (0.5)(8.85 x 10⁻¹²)(0.8 x 10⁶)²

I = 2.832 W/m²

P = IA

where;

A = πd²/4

where;

A = π(1 x 10⁻³)²/4

A = 7.854 x 10⁻⁷ m²

Power = (2.832 W/m²) x (7.854 x 10⁻⁷ m²)

Power = 2.22 x 10⁻⁶ W

Thus, the amplitude of the magnetic field produced is 2.667 mV/m.

The intensity of the laser is 2.832 W/m².

The  power delivered by the laser is 2.22 x 10⁻⁶ W.

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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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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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The body moves at a velocity of 1.62m/s after the bullet emerges.

Mass of bullet, [tex]m_1[/tex] = 22g

                               = 0.022 kg

Mass of the block, [tex]m_2[/tex] = 1.9 kg

Velocity of bullet , [tex]v_1[/tex] = 265 m/s

[tex]v_2 = 0[/tex]

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

[tex]v^{'}_1 =125 m/s[/tex]

[tex]v^{'}_2=?[/tex]

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

   [tex]m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s[/tex]

Therefore, It moves with a velocity of 1.62 m/s.

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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.

(a) The speeds of the tips of both rotors; main rotor 178.3 m/s and  tail rotor 218.4 m/s.

(b) The speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.

v = ωr

where;

v(main rotor) = (444 rev/min x 2π rad x 1 min/60s) x (0.5 x 7.67 m)

v(main rotor) = 178.3 m/s

v(tail rotor) = (4,130 rev/min x 2π rad x 1 min/60s) x (0.5 x 1.01 m)

v(tail rotor) = 218.4 m/s

% speed (main motor) = 178.3/343 = 0.52 = 52 %

% speed (tail motor) = 218.4/343 = 0.64 = 64 %

Thus, the speed of the main rotor is 0.52 speed of sound, and the speed of the tail rotor is 0.64 speed of sound.

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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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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:

A i think...

Explanation:

Sorry if its wrong

Answer:

Explanation:

yes

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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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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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 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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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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