A force in the x direction acts on a particle moving also in the x direction, producing a potential energy U(x)=Ax4 where A=0.630J/m4. Find the magnitude and direction of this force, when the particle is at x=-0.800m?

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

Answer:

The work done by both Joel and Jerry is equal to 0 J.

Explanation:

The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,

W = F.d

where,

W = Work Done on the Body

F = Force Applied on the Body

d = displacement covered by the body

In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,

W = F(0)

W = 0 J (For both Joel and Jerry)

Answer:

Yes, work has been done on the mud.

Explanation:

Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.

Answer:

a) E = 0

b) E = 2.697 MN/C

Explanation:

Solution:-

- The Gauss Law makes life simpler by allowing us to determine the Electric Field strength ( E ) of symmetrically charged objects. By choosing an appropriate Gaussian surface and determine the flux ( Φ ) that passes through an imaginary closed surface.

- The Law states that the net flux ( Φ ) that passes through a Gaussian surface is proportional to the net charged ( Q ) stored within that surface. We can mathematically express the flux ( Φ ) as follows:

                              Φ  = Q / εo

Where,                   1 / εo : The proportionality constant

                              εo: The permittivity of free space = 8.85*10^-12

- The flux produced by a charged object is also given in form of a surface integral of Electric Field ( E ) over the entire surface area ( A ) of the Gaussian surface as follows:

                               Φ = [tex]_S\int\int [ E ] . dA[/tex]  

- We can combine the two relations as follows:

                              [tex]_S\int\int [ E ] . dA[/tex]  = Q / εo

- Now we will consider a charged metal sphere. The important part to note is that the charge on a conducting sphere ( Q ) uniformly distributed on the outside surface of the charged sphere.

- Lets consider a case, where we set up our Gaussian surface ( spherical ) with radius ( r ) < radius of the charged metal surface ( a ). We will use the combined relation and determine the Electric Field ( E ) within a charged metal sphere as follows:

                              [tex]E. ( 4\pi*r^2 ) = \frac{Q_e_n_c}{e_o} \\\\E = \frac{Q_e_n_c}{e_o4\pi*r^2}[/tex]

- However, the amount of charge enclosed in our Gaussian surface is null or zero. As all the charge is on the surface r = a. Hence (Q_enc = 0 ),

                             [tex]E = 0[/tex]                  ..... ( r < a )

- For the case when we set up our gaussian surface with radius ( r ) > radius of the charged metal surface ( a ). We placed a charge of Q = +3.0uC on the surface of the metal sphere. Therefore, the electric field strength at a distance ( r ) from the center of metal sphere is:

                            [tex]E = \frac{Q_e_n_c}{e_o*4*\pi*r^2 } = k\frac{Q_e_n_c}{r^2 }[/tex]    .... ( r > a )

- The above relation turns out to be the Electric Field strength ( E ) produced by a point charge at distance ( r ) from the center. Where, k = 8.99*10^9 is the Coulomb's constant.

a) The radius of the charged metal sphere is given to be a = 5.0 cm. The first point r = 1.0 cm lies within the metal sphere. We looked at the first case where, ( r < a ) the enclosed charge is zero. Hence, the magnitudue of Electric Field Strength ( E ) is zero. ( E = 0 )

b) The second point lies at 10 cm from the center. For this we will use the second case where, ( r > a ). The Electric Field Strength due to a point charge with an enclosed charge of Q = +3.0 uC is:

                            [tex]E = ( 8.99*10^9 ) * \frac{3.0*10^-^6}{0.1^2} \\\\E = 2697000 N / C[/tex]

Answer: The electric field strength at point 10 cm away from the center is 2.697 MN/C

Complete question:

Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast,  a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.

(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.

Answer:

The net force on the person as the air bad deploys is -6750 N backwards

Explanation:

Given;

mass of the passenger, m = 60 kg

velocity of the car at impact, u = 15 m/s

final velocity of the car after impact, v = 0

distance moved as the front of the car crumples, s = 1 m

First, calculate the acceleration of the car at impact;

v² = u² + 2as

0² = 15² + (2 x 1)a

0 = 225 + 2a

2a = -225

a = -225 / 2

a = -112.5 m/s²

The net force on the person;

F = ma

F = 60 (-112.5)

F = -6750 N backwards

Therefore, the net force on the person as the air bad deploys is -6750 N backwards

Answer:

The power used is 0.96 watts.

Explanation:

Recall the formula for electric power (P) as the product of the voltage applied  times the circulating current:

[tex]P=V\,\,I[/tex]

and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:

[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]

Then we can re-write the power expression as:

[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]

which in our case becomes:

[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]

Answer:

0.4757 mm

Explanation:

Given that:

Load P = 223,000 N

the length of the height of the aluminium column = 1.22 m

the diameter of the aluminum column = 10.2 cm = 0.102 m

The amount that the column has shrunk ΔL can be determined by using the formula:

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

where;

A = πr²

2r = D

r = D/2

r = 0.102/2

r = 0.051

A = π(0.051)²

A = 0.00817

Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:

[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]

[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]

[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]

ΔL = 4.757 × 10⁻⁴ m

ΔL =  0.4757 mm

Hence; the amount that the column has shrunk is 0.4757 mm

Answer:

vr = 142.90 m/s

the magnitude of its relative velocity is 142.90 m/s

Explanation:

Given;

A plane has an airspeed of 142 m/s (eastward)

vi = 142 m/s

16.0 m/s wind is blowing southward at the same time as the plane is flying

vb = 16.0m/s

Writing the relative velocity vector, we have;

Taking north and south as positive and negative y axis respectively, east and west as positive and negative x axis respectively.

v = 142i - 16j

The magnitude of the velocity is;

vr = √(vi^2 + vb^2)

vr = √(142^2 + 16^2)

vr = √(20420)

vr = 142.8985654231 m/s

vr = 142.90 m/s

the magnitude of its relative velocity is 142.90 m/s

Explanation:

We have

A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.

The frequency of a pendulum is equal to the no of oscillation per unit time. so,

[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]

Tim period is reciprocal of frequency. So,

[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]

The time period of a pendulum is given by :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

l is length of pendulum

[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]

So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.

Answer:

[tex]\boxed{\mathrm{1.50 \: E^4 \: N/C}}[/tex]

Explanation:

[tex]\displaystyle \mathrm{E=\frac{F_e}{q} }[/tex]

[tex]\displaystyle \mathrm{Electric \: field \: strength \: (N/C)=\frac{Electric \: force \: (N)}{Charge \: (C)} }[/tex]

[tex]\displaystyle \mathrm{E=\frac{0.751}{5.00\: E^{-5}} }[/tex]

[tex]\displaystyle \mathrm{E=\frac{0.751}{0.00005} }[/tex]

[tex]\displaystyle \mathrm{E=15020}[/tex]

The magnitude of the electric field at the location of the test charge is [tex]1.50E^{4}N/C[/tex]

Given that,

Based on the above information, the calculation is as follows:

[tex]= 0.751 \div 5.00E^{-5}\\\\= 0.751 \div 0.00005[/tex]

= 15020

Learn more: brainly.com/question/17429689

Answer:

Explanation:

F ×1 = 0.5×0.145×47×47

F = 160.15 N

Answer:

Explanation:

The magnetic field produced is expressed using the formula

[tex]B = \frac{\mu_0NI}{L}[/tex]

B is the magnetic field = 0.30T

I is the current produced in the coil = 4.5A

[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A

L is the length of the solenoid = 32 cm = 0.32 m

N is the number of turns in the solenoid.

Making N the subject of the formula from the equation above;

[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]

[tex]N = \frac{BL}{\mu_0I}[/tex]

Substituting the give values to get N;

[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]

The number of turns the solenoid must have is approximately 16,931 turns

Answer:

The distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

Explanation:

Given;

mass of the bigger child, M = 30 kg

mass of the smaller child, m = 20 kg

distance between the two children, d = 3 m

This information can be represented diagrammatically;

                                    3m

         |<------------------------------------------------>|

----------------------------------------------------------------------------

         ↓             x            Δ            3-x           ↓

         20kg                                                 30kg

x is the distance from the pivot point that the small child will sit in order to maintain the balance

Take moment about the pivot;

Clockwise moment = anticlockwise moment

30(3-x) = 20x

90 -30x = 20x

90 = 20x + 30x

90 = 50x

x = 90 / 50

x = 1.8 m

Therefore, the distance from the pivot point that the small child will sit in order to maintain the balance is 1.8 m

The distance from the pivot point which the small child must sit in order to maintain the balance is 1.8 meters.

Given the following data:

To determine what distance from the pivot point is the small child sitting in order to maintain the balance, we would take moment about a pivot:

Note: The distance of the child from the pivot is equal to [tex]3-n[/tex]

For moment:

Clockwise moment = anticlockwise moment

[tex]30(3-n) = 20n\\\\90-30n=20n\\\\90=20+30n\\\\90=50n\\\\n=\frac{90}{50}[/tex]

n = 1.8 meters

Read more: https://brainly.com/question/2400211

Answer:

The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion

Explanation:

I hope this will help you:)

Answer:

θ_c = 53.65°

Explanation:

The point after which the light ray had started reflecting internally will be when the reflecting angle is at 90°. The incident angle at this point is called the critical angle and this can be calculated through Snell's law as;

n1 sin θ_c = n2 sin 90

Where;

n1 is the refractive index of the medium through which the incident rays will pass through.

n2 is the Refractive index of the medium through which the refracted rays will pass through.

θ_c is the critical angle at which the incident ray will reflect totally internally.

Now, since sin 90 = 1

Thus;

n1 sin θ_c = n2

θ_c = sin^(-1) (n2/n1)

Now, we are told that the reflection travels in flint glass and reflected from water.

Thus, the first medium is flint glass and the second medium is water.

So, from tables,

Refractive index of flint glass; n1 = 1.655

Refractive index of water; n2 = 1.333

Thus;

θ_c = sin^(-1) (1.333/1.655)

θ_c = 53.65°

The complex, highly technical formula for capacitors is

Q = C V

Charge = (capacitance) (voltage)

Charge = (3 F) (24 V)

Charge = 72 Coulombs

The positive plate of the capacitor is missing 72 coulombs worth of electrons.  They were sucked into positive terminal of the battery stack.

The negative plate of the capacitor has 72 coulombs worth of extra electrons.  They came from the negative terminal of the battery stack.

You should be aware that this is a humongous amount of charge !  An average lightning bolt, where electrons flow between a cloud and the ground for a short time, is estimated to transfer around 15 coulombs of charge !

The scenario in the question involves a "supercapacitor".  3 F is is no ordinary component ... One distributor I checked lists one of these that's able to stand 24 volts on it, but that product costs $35 apiece, you have to order at least 100 of them at a time, and they take 2 weeks to get.  

Also, IF you can charge this animal to 24 volts, it will hold 864J of energy.  You'd probably have a hard time accomplishing this task with a bag of leftover AA batteries.

Answer:

Explanation:

We shall apply Bernoulli's equation of flow of liquid

1 / 2 ρ v² + ρ gh + P = constant

For calculating velocity in second floor

A₁ V₁ = A₂ V₂

1  x 2.8 = 2 x V₂

V₂ = 1.4 m /s

1 / 2 ρ v₁² + ρ gh₁ + P₁ = 1 / 2 ρ v₂² + ρ gh₂ + P₂

.5 x 10³ x 2.8² + 10³ x 9.8h₁ + 340 x 10³ = .5 x 10³ x 1.4² + 10³ x 9.8 x h₂ + P₂

P₂ = 3.92 x 10³ + 9.8  x 10³ ( h₁ - h₂ ) + 340 x 10³ - .98 x 10³  

= 3.92 x 10³ - 9.8  x 10³ x 4 + 340 x 10³ - .98 x 10³

= 303.74  x 10³ Pa

= 303.74 kPa .

Answer:

M1 Vx1 + M2 Vx2 + M3 Vx3 = 0     conservation of momentum in x direction

Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3

Vx3 = - 320 * 2 / 100 = -6.4 m/s      M2 has no x-component of momentum

Likewise:

Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3

Vy3 = - 355 * 1.5 / 100 = -5.33 m/s

tan theta = -5.33 / -6.4 = .833    where theta is in the third quadrant and measured from the negative x-axis

theta = 39.8 deg

180 + 39.8 = 219.8     from the positive x-axis

Answer:

I2 = 3.076 A

Explanation:

In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:

[tex]B=\frac{\mu_oI}{2R}[/tex]        (1)

I: current in the wire

R: radius of the wire

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:

[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex]         (2)

I1: current of the first ring = 8A

R1: radius of the first ring = 0.078m

I2: current of the second ring = ?

R2: radius of the first second = 0.03m

To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:

[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]

The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.

Answer:

The change in gravitational potential energy of the climber-Earth system is  [tex]\Delta PE = 396900 \ J[/tex]

Explanation:

From the question we are told that

    The mass of the hiker is  [tex]m = 75 \ kg[/tex]

    The time  taken is  [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]

    The  vertical elevation after time  T is  [tex]H = 540 \ m[/tex]

The  change  in gravitational potential is  mathematically represented as

         [tex]\Delta PE = mgH[/tex]

here g is the acceleration due to gravity with value  [tex]g = 9.8 \ m/s^2[/tex]  

     substituting values  

        [tex]\Delta PE = 75 * 9.8 * 540[/tex]

       [tex]\Delta PE = 396900 \ J[/tex]

Answer:

the answer is option A = photoelectric effect

Explanation:

If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.

-310J

The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e

ΔE = Q + W       ----------------------(i)

If heat is released by the system, Q is negative. Else it is positive.

If work is done on the system, W is positive. Else it is negative.

In this case, the system is the balloon and;

Q = -0.659kJ = -695J    [Q is negative because heat is removed from the system(balloon)]

W = +385J  [W is positive because work is done on the system (balloon)]

Substitute these values into equation (i) as follows;

ΔE = -695 + 385

ΔE = -310J

Therefore, the change in internal energy is -310J

PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.

Answer:

Explanation:

Hope this helped,

Kavitha

Answer:

180.1 g

Explanation:

Data provided in the question

Mass of lead sinker = 225 grams

Density = 11.3 g/cm^3

The Wooden block of mass = 25 grams

Density = 0.5/g cm^3

Based on the above information, the apparent weight is

Before that we need to do the following calculations

[tex]V_1 = \frac{m_1}{D_1}[/tex]

[tex]= \frac{225}{11.3}[/tex]

= 19.91 cm^3

[tex]V_2 = \frac{m_2}{D_2}[/tex]

[tex]= \frac{25}{0.5}[/tex]

= 50 cm^3

Now as we know that

V = V_1 + V_2

= 19.91 cm^3 + 50 cm^3

= 69.91 cm^3

Now the weight of dispacement of water is

[tex]m = VD_{water}[/tex]

[tex]= 69.91 cm^3 (1 \frac{g}{cm^3} )[/tex]

= 69.91 g

Therefore the apparent weight is

[tex]W = m_1 + m_2 - m[/tex]

= 225 + 25 - 69.91 g

= 180.1 g

Answer:

The additional force is  [tex]F_3 = 120 \ N[/tex]

Explanation:

From the question we are told that

      The horizontal force in one direction is  [tex]F_i = 480 \ N[/tex]

       The horizontal force in the opposite direction is  [tex]F_f = -600 \ N[/tex]

The negative sign shows that it is acting in the opposite direction

Generally for the box to remain stationary the net force on it must be equal to zero  that is  

        [tex]F_1 + F_2 +F_3 = 0[/tex]

Where [tex]F_3[/tex] is the additional force required  

    So

          [tex]F_3 = -F_1 - F_2[/tex]

substituting values

         [tex]F_3 = -480 - [-600][/tex]

        [tex]F_3 = -480 + 600[/tex]

       [tex]F_3 = 120 \ N[/tex]

Answer:

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]

I is the rotational inertia = 16kgm²

[tex]\theta = angular\ displacement[/tex]

[tex]\theta = 2 rev = 12.56 rad[/tex]

[tex]\alpha \ is \ the\ angular\ acceleration[/tex]

To get the angular acceleration, we will use the formula;

[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]

[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

=  257.23/2.0

= 128.61 Watts

Answer: It is done to prevent the necessary compound from solidifying along with the debasements. It expels any insoluble pollutions from the appropriate response (as opposed to separating the predetermined item). With since quite a while ago stemmed channels, the gems kick off inside the progression because the arrangement cools, obstructing the pipe. utilizing a stemless channel keeps this from occurring.

Explanation:

it is good to use the stemless funnel for hot gravity filtration experiment,  to prevent the necessary compound from solidifying, expels any insoluble pollutions from the appropriate response.

Recrystallization is the process of getting pure crystals from an impure compound in a solvent and Hot gravity filtration remove the impurities from a solution prior to recrystallization.

In this technique the filtration equipment and the sample are heated and the filtration is needed for recrystallization which requires a hot solution as it need to be supersaturated for crystals to form on cooling.

Hot solutions hold more solute in a suspension than a cold solution as the solubility of solids increases with a increase in temperature, that means saturated solution contain more dissolved solute.

When the hot solution cool down, it will be supersaturated  and hold more dissolved solute than its cold. The  main objective to  choose a solvent is that it dissolves the compound when heated, but that doesn’t dissolve the impurity at high temperatures.

For more details regarding gravity filtration, visit:

https://brainly.com/question/28503164

#SPJ2

Answer:

The value of A is 1.5m/s^2 and B is 0.5m/s^³

Explanation:

The mass of the rocket = 2540 kg.

Given velocity, v(t)=At + Bt^2

Given t =0  

a= 1.50 m/s^2

Now, velocity V(t) = A*t + B*t²

If,  V(0) = 0, V(1) = 2

a(t) = dV/dt = A+2B × t  

a(0) = 1.5m/s^²  

1.5m/s^²  =  A + 2B ×  0  

A = 1.5m/s^2

now,

V(1) = 2 = A× 1 + B× 1^²  

1.5× 1 +B× 1 = 2m/s

B = 2-1.5  

B = 0.5m/s^³

Now Check V(t) = A× t + B × t^²

So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² ×  1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s  

Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)

Answer:

The product of the object's weight and the horizontal distance between the two positions.

Explanation:

Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.

Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = [tex]q_{int}[/tex] / ε₀

where Ф the electric flow, qint is the charge inside the surface

To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

           R <r <R_a

for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field

          q_{int} = Q₁

The surface area is

          A = 4π r²

we substitute

          E 4π r² = Q₁ /ε₀

          E = 1 / 4πε₀ Q₁ / r²

          k = 1/4πε₀

          E = k Q₁ / r²