An 18-kg bicycle carrying a 62-kg girl is traveling at a speed of 7 m/s. What is the kinetic energy of the girl
and bicycle combined together?​

There are no reactions or targets given to drag the appropriate labels to their respective targets. However, I can provide you with some information that may help you identify the type of reaction (SN2, SN1, E1, or E2) for a reaction with an alkyl halide.

When an alkyl halide undergoes a substitution reaction with a nucleophile, there are two possible mechanisms: SN1 and SN2. Similarly, elimination reactions can occur by either an E1 or E2 mechanism.

SN1 mechanism:

The SN1 reaction is a two-step process in which the halide ion departs from the substrate in the first step. This step leads to the formation of a carbocation, which is an intermediate that is highly reactive and unstable. A nucleophile can then react with the carbocation to form the product. Because the rate-determining step only involves the alkyl halide, the reaction rate is proportional to the concentration of the alkyl halide. This reaction works best with tertiary substrates.

SN2 mechanism:

The SN2 reaction is a one-step process in which the nucleophile attacks the substrate at the same time as the halide ion departs. The reaction proceeds with inversion of configuration at the reaction center. This reaction works best with primary substrates.

E1 mechanism:

In an E1 reaction, the leaving group departs to form a carbocation, which is then deprotonated by a base to form an alkene. This reaction works best with tertiary substrates.

E2 mechanism:

The E2 reaction is a one-step process in which the leaving group departs at the same time as the base removes a proton from an adjacent carbon atom. The reaction proceeds with inversion of configuration at the reaction center. This reaction works best with primary substrates.

When reacting with an alkyl halide, there are two possible mechanisms for substitution reactions: SN1 and SN2. For elimination reactions, there are two possible mechanisms: E1 and E2. The type of reaction that occurs depends on the structure of the substrate, as well as the identity of the nucleophile or base. A main answer to this question is that in order to identify the type of reaction (SN2, SN1, E1, or E2) that occurs, one must consider the structure of the substrate and the reaction conditions.

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Answer:

 U = 25 J

Explanation:

The energy in a set of charges is given by

          U = \(k \sum \frac{q_i}{r_i}\)

in this case we have three charges of equal magnitude

          q = q₁ = q₂ = q₃

with the configuration of an equilateral triangle all distances are worth

          d = a

          U = k ( \(\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3}\) )

we substitute

           15 = k q² (3 / a)

            k q² /a = 5

For the second configuration a load is moved to the measured point of the other two

          d₁₃ = a

The distance to charge 2 that is at the midpoint of the other two is

          d₁₂ = d₂₃ = a / 2

           U = k (\frac{q_1q_2}{ r_1_2 } + \frac{q_1q_3}{r_1_3} + \frac{q_2q_3}{r_2_3})

            U = k q² (  \(\frac{2}{a} + \frac{1}{a} + \frac{2}{a}\) )

            U = (kq² /a) 5

substituting

            U = 5 5

            U = 25 J

Explanation:

1) left ⬅️

2) into the page

3) upward ⬆️

4) right ➡️

5) downward ⬇️

The current through the 20Ω resistor in Figure 1 when the potential difference is 300V is 15A (amperes).

In Figure 1, we see a circuit with a 20Ω resistor and a δv of 300 V. To determine the current through the resistor, we can use Ohm's Law, which states that the current flowing through a conductor is directly proportional to the voltage and inversely proportional to the resistance. Mathematically, this can be expressed as I = V/R, where I is the current, V is the voltage, and R is the resistance.

In this case, we know the resistance of the 20Ω resistor and the voltage across it, which is 300 V. Substituting these values into the equation gives us I = 300 V / 20Ω = 15 A. Therefore, the current flowing through the 20Ω resistor in Figure 1 is 15 A.

It's important to note that this calculation assumes that the circuit is ideal, meaning that there are no other resistances or sources of energy in the circuit. In reality, circuits are often more complex, with multiple resistances and energy sources, which can affect the current flowing through each component.

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The speed of the water leaving the end of the hose \({0.405 }\frac{ \mathrm{~m}}{ \mathrm{s}} $$\)

As per the given data in question:

Initial diameter of the hose \(D_1=2.80 \mathrm{~cm}$\)

Volume of the bucket \(\mathrm{V}=21.0 \mathrm{~L}=0.021 \mathrm{~m}^3$\)

Time = 1.30 minutes.

The cross-sectional area of the hose is

\(\mathrm{A}_1=\frac{\pi \mathrm{D}_1{ }^2}{4} \\{A}_1=\frac{\pi\left(2.80 \times 10^{-2} \mathrm{~m}\right)^2 }{4}\\{A}_1=6.16 \times 10^{-4} \mathrm{~m}^2$\)

The volume flow rate

\($\dot{V}=\frac{V}{t}\)

\(=\frac{0.021 \mathrm{~m}^3}{1.30 \times 60 \mathrm{~s}}\\=2.50 \times 10^{-4} \frac{\mathrm{~m}^3}{\mathrm{s}$}\)

Speed over water is only required to be used for collision avoidance and not necessarily for navigation. By using speed over ground, a navigator is more aware of the situation than otherwise. For example, if the ship is drifting towards a danger, navigator will know it better if he has speed over ground in radar.

Then, the speed of the water leaving the end of the hose is

\($$\mathrm{v}_1=\frac{\dot{V}}{\mathrm{A}_1} =\frac{\left(2.50 \times 10^{-4} \mathrm{~m}^3 / \mathrm{s}\right)}{\left(6.16 \times 10^{-4} \mathrm{~m}^2\right)} ={0.405 }\frac{ \mathrm{~m}}{ \mathrm{s}} $$\)

Therefore the value of speed of the water leaving the end of the hose is 0.405 m/s

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The correct option is (b) i.e. radio wave, electromagnetic wave can travel the farthest distances because it has a wavelength range of greater than 1x10-1 meters.

Radio wave have the longest wavelength range (greater than 1x10-1 meters) among all the electromagnetic waves, so they can travel the farthest distances.

Radio waves have a wavelength range that can be as long as several kilometers, which makes them suitable for long-distance communication. They can travel through the atmosphere, around the Earth's curvature, and even through solid objects like buildings, making them ideal for broadcast and communication purposes.

Gamma rays, X-rays, and microwaves, on the other hand, have much shorter wavelengths, making them less suitable for long-distance travel. Gamma rays have the shortest wavelength range and are highly energetic and penetrate through materials, making them useful for imaging and radiation therapy. X-rays have a slightly longer wavelength range and are used for medical imaging. Microwaves have a slightly longer wavelength range compared to X-rays and are used for communication, radar, and cooking.

Therefore, radio waves are the electromagnetic waves that can travel the farthest distances because they have a wavelength range that is greater than 1x10-1 meters.

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(a) The tension in the tow line is 500 N.

(b) The distance travelled by the car before coming to rest is 10 m.

The tension in the tow line is calculated by applying the following formula as shown below.

T = ma

where;

T = 1000 kg x 0.5 m/s²

T = 500 N

The distance travelled by the car before coming to rest is calculated as follows;

a = F / m

a = ( 5000 N ) / ( 1000 kg )

a = 5 m/s²

v² = u² - 2as

where;

2as = u²

s = ( u² ) / ( 2a )

s = ( 10² ) / ( 2 x 5 )

s = 10 m

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The object displacement over the 9 seconds interval will be 0 m. Displacement can be define as the shift in the position of an object with respect to the reference frame.

An object's location changes if it moves with regard to a reference frame, like when a passenger moves to the back of an airplane or a lecturer moves to the right with respect to a whiteboard. Displacement is the term used to explain this shift in position.

The distance an item travels from its starting point until it reaches a certain velocity can be calculated using the formulas t = (v vi)/g and y = gt2/2 + vit. The equation for the object's velocity at a particular distance from the starting point is the outcome of this.

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Answer:

#A

Acceleration=a

\(\\ \tt\hookrightarrow a=\dfrac{v-u}{t}=\dfrac{8}{4}=2m/s^2\)

#B

\(\\ \tt\hookrightarrow a=\dfrac{0-8}{4}=\dfrac{-8}{4}=-2m/s^2\)

Answer(s):

Solution for Practice-A:

We know that:

Solution:

Hence, the acceleration is 2 m/s for practice A.

Solution for Practice B:

We know that:

Solution:

Hence, the acceleration is -2 m/s for practice B.

Hoped this helped!

Answer:

Si un objeto se mueve en relación a un marco de referencia (por ejemplo, si una profesora se mueve a la derecha con respecto al pizarrón, o un pasajero se mueve hacia la parte trasera de un avión), entonces la posición del objeto cambia. A este cambio en la posición se le conoce como desplazamiento. La palabra desplazamiento implica que un objeto se movió, o se desplazó.

Explanation:

El desplazamiento se define como el cambio en la posición de un objeto. Se puede definir de manera matemática con la siguiente ecuación:

\text{desplazamiento}=\Delta x=x_f-x_0desplazamiento=Δx=x  

f

​  

−x  

0

​  

start text, d, e, s, p, l, a, z, a, m, i, e, n, t, o, end text, equals, delta, x, equals, x, start subscript, f, end subscript, minus, x, start subscript, 0, end subscript

x_fx  

f

​  

x, start subscript, f, end subscript se refiere al valor de la posición final.

x_0x  

0

​  

x, start subscript, 0, end subscript se refiere al valor de la posición inicial.

\Delta xΔxdelta, x es el símbolo que se usa para representar el desplazamiento.

Debemos ser cuidados al usar la palabra distancia, ya que hay dos maneras de usar el término en física. Podemos hablar acerca de la distancia entre dos puntos, o podemos hablar de la distancia recorrida por un objeto.

La distancia se define como la magnitud o el tamaño del desplazamiento entre dos posiciones. Observa que la distancia entre dos posiciones no es la misma que la distancia recorrida entre ellas.

Es importante darse cuenta que la distancia recorrida no tiene que ser igual a la magnitud del desplazamiento (es decir, la distancia entre dos puntos). De manera específica, si un objeto cambia de dirección en su trayecto, la distancia total recorrida será mayor que la magnitud del desplazamiento entre esos dos puntos. Ve los ejemplos resueltos a continuación.

Answer:I think it would start moving

Explanation:

The % error in the computing of the resistivity of wire of resistance 1.04 Ω estimated error 0.01, length = 53.4 cm estimated error 0.1 cm, diameter = 0.57 mm estimated error 0.01 mm is 4.7 %

ρ = π d² R / 4 l

ρ = Resistivity

R = Resistance

d = Diameter

l = Length

R = 1.04 Ω

ΔR = 0.01 Ω

l = 53.4 cm

Δl = 0.1 cm

d = 0.57 mm

Δd = 0.01 mm

Δρ / ρ = 2 Δd / d + ΔR / R + Δl / l

Δρ / ρ = 2 ( 0.01 / 0.57 ) + ( 0.01 / 1.04 ) + ( 0.1 / 53.4 )

Δρ / ρ = 0.035 + 0.01 + 0.002

Δρ / ρ = 0.047

Δρ / ρ = 4.7 %

Therefore, the % error in the computing of the resistivity is 4.7%

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This is a what do you think question.

The refractive index of the liquid is 1.51.

The diameter of the nth ring in a Newton's rings experiment is given by:

d_n = (2n - 1) * λ / 2 * μ,

where λ is the wavelength of the light used and μ is the refractive index of the medium between the lens and the plate.

If the diameter of the tenth ring changes from 1.48 cm to 1.28 cm, the difference is:

Δd = d_10' - d_10 = (2 * 10 - 1) * λ / 2 * μ' - (2 * 10 - 1) * λ / 2 * μ = λ / μ * (1 / μ' - 1 / μ) * (2 * 10 - 1),

where μ' is the refractive index of the liquid.

Solving for μ', we get:

μ' = μ * (1 - Δd / (λ * (2 * 10 - 1) * (1 / μ - 1))).

Substituting the given values, we get:

μ' = 1.51.

Therefore, the refractive index of the liquid is 1.51.

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Answer:(B) 25.0 N

Explanation:

This is what makes most since. You are wanting to look at the stress of the first wire. 25 *3 for the amount of balls handing of the string you get 75.0 N which is the max the top wire can hold without breaking! Hope this helps!

We can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field and when the cosmic ray proton enters our solar system, it experiences an interplanetary magnetic field that is perpendicular to its velocity.

(a) To find the magnetic field in the region of space where the cosmic-ray proton is executing a circular orbit, we can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field. The centripetal force is given by the equation:

F = qvB

Where:
F is the centripetal force,
q is the charge of the particle (in this case, the charge of a proton),
v is the velocity of the particle,
B is the magnetic field strength.

The centripetal force can also be expressed as:

F = mv²/r

Where:
m is the mass of the proton,
v is the velocity of the proton,
r is the radius of the circular orbit.

Equating the two expressions for the centripetal force, we have:

qvB = mv²/r

Rearranging the equation, we can solve for the magnetic field B:

B = mv/rq

Given the energy of the proton (13.0 MeV), we can use the relation between energy and velocity for a particle with rest mass m:

E = mc² = (γ - 1)mc²

Where:
E is the energy of the particle,
m is the rest mass of the particle,
c is the speed of light,
γ is the Lorentz factor.

The Lorentz factor can be expressed as:

γ = E/mc² + 1

Substituting the given energy and rest mass of the proton, we can calculate the Lorentz factor.

Now, we can substitute the values of m, v, and r into the equation for the magnetic field B and solve for B.

(b) When the cosmic ray proton enters our solar system, it experiences an interplanetary magnetic field that is perpendicular to its velocity. In this case, the proton will follow a helical path in the magnetic field.

To find the radius of the proton's circular orbit in this field, we can use the formula for the radius of a helical path in a magnetic field:

r = mv/|q|B

Where:
m is the mass of the proton,
v is the velocity of the proton,
|q| is the magnitude of the charge of the proton,
B is the magnitude of the magnetic field strength.

Substituting the given values of m, v, and B into the equation, we can calculate the radius of the proton's circular orbit in this field.

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A. the change in volume of the cube is approximately 0.120601 cm^3. B. the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm. C. the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

(a) To find the change in volume of the cube as x increases from 2.00 to 2.01 centimeters, we need to calculate the difference in volume between these two values.

The volume of a cube is given by V = x^3, where x is the edge length.

For x = 2.00 cm, the volume V1 = (2.00 cm)^3 = 8.00 cm^3.

For x = 2.01 cm, the volume V2 = (2.01 cm)^3 = 8.120601 cm^3.

The change in volume is ΔV = V2 - V1 = 8.120601 cm^3 - 8.00 cm^3 ≈ 0.120601 cm^3.

Therefore, the change in volume of the cube is approximately 0.120601 cm^3.

(b) The average rate of change of volume of the cube with respect to an edge length as x increases from 2.00 to 2.01 centimeters can be calculated by dividing the change in volume by the change in edge length.

ΔV = 0.120601 cm^3 (from part a)

Δx = 2.01 cm - 2.00 cm = 0.01 cm

The average rate of change of volume is ΔV/Δx = 0.120601 cm^3 / 0.01 cm ≈ 12.0601 cm^3/cm.

Therefore, the average rate of change of volume of the cube with respect to an edge length is approximately 12.0601 cm^3/cm.

(c) The instantaneous rate of change of volume of the cube with respect to an edge length at the instant when x = 2 centimeters can be found by taking the derivative of the volume function V = x^3 with respect to x and evaluating it at x = 2.

dV/dx = 3x^2

At x = 2 cm, the instantaneous rate of change of volume is dV/dx evaluated at x = 2:

dV/dx = 3(2 cm)^2 = 12 cm^2.

Therefore, the instantaneous rate of change of volume of the cube with respect to an edge length at x = 2 centimeters is 12 cm^2.

To find the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3, we need to determine the slope of the tangent line.

The derivative of f(x) = 1/x is given by f'(x) = -1/x^2.

At x = 3, the slope of the tangent line is f'(3) = -1/(3^2) = -1/9.

Since the line we want to find is parallel to the tangent line, it will have the same slope. So the slope of the line is -1/9.

Using the point-slope form of a linear equation, we can write the equation of the line as:

y - y1 = m(x - x1),

where (x1, y1) is the given point (1, 5) and m is the slope.

Substituting the values, we have:

y - 5 = (-1/9)(x - 1).

Expanding and rearranging the equation, we get:

y = (-1/9)x + 14/9.

Therefore, the equation of the line that passes through the point (1, 5) and is parallel to the tangent line to f(x) = 1/x at x = 3 is y = (-1/9)x + 14/9.

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Explanation:

The new volume of water = 25 ml

The old volume of water = 15 ml

The difference = 25 - 15 but what are the units?

Since the question asks for force, the units must start out as 10 mL

In water 1 mL has a mass of 1 gram, so the answer is 10 grams.

Grams are units of mass, not weight. You should convert this into newtons.

10 grams = 1/1000 = 0.01 kg

1 kg has a weight of 9.81 Newtons

0.01 kg has a weight 0.081 Newtons

If you have never seen a Newton before, then the answer is 10 grams

The average kinetic energy of the methane CH₄ molecule at 273K and 585K is  565 x 10⁻²¹ J and 1.21 x 10⁻²⁰ J. respectively.

The average kinetic energy of a gas molecule is given by the formula,

K = 3/2K'T

Where,

K is the average kinetic energy,

K' is the Boltzmann constant,

T is the temperature.

So, putting values, to find the average kinetic energy of CH₄ molecule, first for 272K

K = 3/2 x 1.38 x 10⁻²³ x 273

K = 565 x 10⁻²¹ J

Now, for 585 K,

K = 3/2 x 1.38 x 10⁻²³ x 585

K = 1.21 x 10⁻²⁰ J.

So, these are average kinetic energies of the CH₄ molecule at 273K and 585K.

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Answer:

Resistor A

Explanation:

Resistors A, B and C are connected in parallel to a battery of voltage represented by V.

Since they are in parallel, the same voltage, V, passes across the three of them.

And from Ohm's law, the voltage (V) through a resistor is the product of the current (I) flowing through it and the resistance (R) of the resistor. i.e

V = I x R

=> I = \(\frac{V}{R}\)

Therefore, assuming the values of the resistances of resistors A, B and C are A, B and C respectively,

(i) the current, \(I_{A}\) through A is

\(I_{A}\) = \(\frac{V}{A}\)

(ii) the current, \(I_{B}\) through B is

\(I_{B}\) = \(\frac{V}{B}\)

(iii) the current, \(I_{C}\) through C is

\(I_{C}\) = \(\frac{V}{C}\)

From the foregoing, it can be deduced that the current is inversely proportional to the resistance. Therefore, the higher the resistance, the lower the current. Consequential of this, the resistor that carries the highest current is the one with the smallest resistance, which is A

Resistor A

Given:

Resistors A, B and C are connected in parallel to a battery of voltage represented by V.

Since they are in parallel, the same voltage, V, passes across the three of them.

Ohm's law:

It states that voltage (V) through a resistor is the product of the current (I) flowing through it and the resistance (R) of the resistor. i.e.

V = I x R

⇒ I = V/R

Therefore, assuming the values of the resistances of resistors A, B and C are A, B and C respectively,

(i) the current,  through A is

\(I_A=\frac{V}{A}\)  

(ii) the current,  through B is

\(I_B=\frac{V}{B}\)

(iii) the current,  through C is

\(I_C=\frac{V}{C}\)  

From the Ohm's law we can say that the current is inversely proportional to the resistance.

Therefore, higher the resistance, the lower is the current.

It is given that resistance of A being the smallest, thus is will carries the highest current.

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Answer:

The change in momentum of the ball = 81 kg·m/s

Explanation:

Given:

m = 1.5 kg

V₁ = - 27 m/s

V₂ = + 27 m/s

____________

Δp - ?

The change in momentum of the ball:

Δp = m·ΔV

Δp = m· (V₂ - V₁) = 1.5·(27 - (-27)) = 1.5·(27 + 27) = 81 kg·m/s

If you add a vector with a magnitude of 1 to a vector of magnitude 2, the possible magnitudes for the vector sum range from 1 to 3.

This is because of the triangle inequality theorem.The triangle inequality theorem states that the magnitude of the sum of two vectors is less than or equal to the sum of their magnitudes. In other words, if we have two vectors A and B, then the magnitude of their sum C is given by:

|C| ≤ |A| + |B|

If we have a vector of magnitude 1, we can call it vector A. If we have a vector of magnitude 2, we can call it vector B.

Then the magnitude of their sum, vector C, is given by:|C| = |A| + |B|.

Since |A| = 1 and |B| = 2, we can substitute those values into the equation: |C| = 1 + 2

|C| = 3

Therefore, the maximum magnitude of the vector sum is 3.

However, the minimum magnitude of the vector sum is found when the vectors are pointing in opposite directions.

In this case, the magnitude of the sum would be:

|C| = |B| - |A|

|C| = 2 - 1

|C| = 1.

Therefore, the possible magnitudes for the vector sum range from 1 to 3.

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Also because field lines entering and leaving a closed surface are equal, the magnetic flux through the surface is zero.

→ds=qϵ0. As we all know, the amount of charge contained by a surface determines the quantity of electric flux through it rather than its size, shape, or area. Given that the enclosed charges in the figures are identical, the electric flux across each surface should be identical.

Gauss' law states that the electroluminescence through some kind of closed surface is commensurate to the net charge contained by the surface. Consequently, when the surface's net charge is contained, is 0, the closed surface's electric flow is also zero.

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First, let's calculate the total resistance in the circuit.

Since the two resistors are in series, the total resistance is the sum of each resistance:

Now, to calculate the current, we can divide the voltage by the total resistance:

Therefore the current is 2.083 A.

The Current throughout the circuit is 2.083 A.

To calculate the current, we will use the formula:

I = V / R

Where,

I = current,

V = Voltage of Battery.

R = Total Resistance.

Now, we will use the given values in the question,

V = 25 V

R = 7Ω + 5Ω

Now, we have to find total Resistance,

Total Resistance =  7Ω + 5Ω

= 12Ω.

Now, we will calculate the current using the above formula,

Putting all the values,

I = 25 / 12

I = 2.083 A.

Therefore, the Current running throughout the circuit is 2.083 A.

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Answer: The Answer is B. It is directed inward, toward the center of a circle.

Explanation:

The electric flux through the square is 2.14 × 10⁴ Nm²/C.

Given: A charge of 4 μC is 30 cm above the center of a square of side length 60 cm.

To find: The flux through the square.Solution:We know that the Electric flux through a closed surface S is given by;

Φ = ∫ E . dA,

where E is the electric field vector and dA is a small area element of the surface with direction given by the normal to the surface and magnitude equal to the area of the element.For a point charge, the electric field E at a distance r from it is given by;

E = kq/r²

Where q is the charge, k is the Coulomb constant, and r is the distance between the charge and the point of observation.So the electric field at the center of the square due to the given charge will be;

E = kq/r² = (9 × 10⁹) × (4 × 10⁻⁶) / (0.3)²= 5.93 × 10⁴ N/C

The total electric flux through the square will be given by;

Φ = Φ1 + Φ2

Where Φ1 is the flux through the top and bottom faces, and Φ2 is the flux through the four side faces.The top and bottom faces are parallel to the electric field, so the electric flux through each will be;

Φ1 = E × A = (5.93 × 10⁴) × (0.6 × 0.6) = 2.14 × 10⁴ Nm²/C

The side faces are perpendicular to the electric field, so the electric flux through each will be zero.Hence;Φ = Φ1 = 2.14 × 10⁴ Nm²/CAnswer:

Therefore, the electric flux through the square is 2.14 × 10⁴ Nm²/C.

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