"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and the Sun."

Answer: 0.0617

Explanation:

Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15

Let X be a binomial variable that represents the number of days having wet weather.

Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.

Here, n= 7 ( 1 week = 7 days)

The probability that it will take a week for it three wet weather on 3 separate days:

[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]

Hence, the required probability =0.0617

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω ,  R₃ = 1.72 10³ Ω ,  R₄ = 5.4 10³ Ω    1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

   V = i R

let the voltage

1st resistance

         V = i R

         R₁ = V / i

         R₁ = 14.7 / 1 10⁻³

         R₁ = 14.7 10³ Ω

power is

        P = V i

        P = 14.7 1 10⁻³

        P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

          R₂ = 8.18 / 1 10⁻³

          R₂ = 8.18 10³ Ω

Power

          P = 8.18 1 10⁻³

          P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

          V₁ + V₂ + V₃ = 24.6

          V₃ = 24.6 - V₁ -V₂

          V₃ = 24.6 - 14.7 - 8.18

          V₃ = 1.72 V

          R₃ = 1.72 / 1 10⁻³

          R₃ = 1.72 10³ Ω

power

          P = Vi

          P = 1.72 10⁻³

          P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

           30 = V₁ + V₂ + V₃ + V₄

           V₄ = 30 - V₁ -V₂ -V₃

           V₄ = 30 -14.7 - 8.18 - 1.72

           V₄ = 5.4 V

          R₄ = 5.4 / 1 10⁻³

          R₄ = 5.4 10³ Ω

Power

         P = V i

         P = 5.4 10⁻³

         P = 0.0054 W

⅛ W resistance

The values ​​of these resistance are commercially

Let's check the consumption of the circuit

  R_total = R₁ + R₂ + R₃ + R₄

  R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

   R_total = 30 10³

the current circulating in the circuit is

     i = V / R_total

     i = 30/30 10³

     i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

Answer:

The net force is directed downwards.

Explanation:

Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.

Answer:

Electric field is zero at point 4.73 m

Explanation:

Given:

Charge place = 50 cm  = 0.50 m

change q1 = 5 µC

change q2 = 4 µC

Computation:

electric field zero calculated by:

[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]

Where electric field is zero,

First distance = x

Second distance = (x-0.50)

So,

E1 = E2

[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]

[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]

x = 0.263 or x = 4.73

So,

Electric field is zero at point 4.73 m

Answer:

The electric field strength needed is 4 x 10⁵ N/C

Explanation:

Given;

magnitude of magnetic field, B = 0.1 T

velocity of the charge, v = 4 x 10⁶ m/s

The velocity of the charge when there is a balance in the magnetic and electric force is given by;

[tex]v = \frac{E}{B}[/tex]

where;

v is the velocity of the charge

E is the electric field strength

B is the magnetic field strength

The electric field strength needed is calculated as;

E = vB

E = 4 x 10⁶ x 0.1

E = 4 x 10⁵ N/C

Therefore, the electric field strength needed is 4 x 10⁵ N/C

Answer:

d' = 75.1 cm

Explanation:

It is given that,

The actual depth of a shallow pool is, d = 1 m

We need to find the apparent depth of the water in the pool. Let it is equal to d'.

We know that the refractive index is also defined as the ratio of real depth to the apparent depth. Let the refractive index of water is 1.33. So,

[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{1\ m}{1.33}\\\\d'=0.751\ m[/tex]

or

d' = 75.1 cm

So, the apparent depth is 75.1 cm.

Answer:

in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

Answer:

10.8rev

Explanation:

Using

Wf²-wf = 2 alpha x theta

0²- 56.36x56.36/ 2(-20.13) x theta

Theta = 68.09 rad

But 68.09/2π

>= 10.8 revolutions

Explanation:

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

Answer:

cost_cost = $ 96

Explanation:

In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system

         performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)

         perfomance= 8,927 km / l

now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km

          #_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters

Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos

            cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638

             cost_gasoline = #_gasoline   cost_litro

             cost_gasoline = 225.72   0.638

             cost_gasoline = $ 144

This cost is for the one way trip, the total round trip cost is

             cost_total = 2 cost_gasoline

             cost_total = $ 288

Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is

                cost_person = total_cost / #_people

                cost_person = 288/3

                cost_cost = $ 96

Answer:

The projectile strikes the tower at a height of 354.824 meters.

Explanation:

The projectile experiments a parabolic motion, which consist of a horizontal motion at constant speed and a vertical uniformly accelerated motion due to gravity. The equations of motion are, respectively:

Horizontal motion

[tex]x = x_{o}+v_{o}\cdot t \cdot \cos \theta[/tex]

Vertical motion

[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta +\frac{1}{2} \cdot g \cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]x[/tex] - Initial and current horizontal position, measured in meters.

[tex]y_{o}[/tex], [tex]y[/tex] - Initial and current vertical position, measured in meters.

[tex]v_{o}[/tex] - Initial speed, measured in meters per second.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

[tex]t[/tex] - Time, measured in seconds.

The time spent for the projectile to strike the tower is obtained from first equation:

[tex]t = \frac{x-x_{o}}{v_{o}\cdot \cos \theta}[/tex]

If [tex]x = 600\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]\theta = \frac{\pi}{4}[/tex], then:

[tex]t = \frac{600\,m-0\,m}{\left(120\,\frac{m}{s} \right)\cdot \cos \frac{\pi}{4} }[/tex]

[tex]t \approx 7.071\,s[/tex]

Now, the height at which the projectile strikes the tower is: ([tex]y_{o} = 0\,m[/tex], [tex]t \approx 7.071\,s[/tex], [tex]v_{o} = 120\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])

[tex]y = 0\,m + \left(120\,\frac{m}{s} \right)\cdot (7.071\,s)\cdot \sin \frac{\pi}{4}+\frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (7.071\,s)^{2}[/tex]

[tex]y \approx 354.824\,m[/tex]

The projectile strikes the tower at a height of 354.824 meters.

Answer:

The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:

1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.

2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.

3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.

Explanation:

Answer:

1,860 g

Explanation:

In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

final. After the crash

      [tex]p_{f}[/tex] = m [tex]v_{1f}[/tex] + m v_{2f}

Recall that velocities are a vector so it has x and y components

       p₀ = p_{f}

we write this equation for each axis

X axis

       m v₁₀ = m v_{1fx} + m v_{2fx}

Y Axis  

       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

     v_{1f}  cos θ = -0.7738

we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

Answer:

The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

Given that,

A beam of light from a laser illuminates a glass.

Suppose, the length of piece is [tex]L=25.21\times10^{-2}\ m[/tex]

Index of refraction is 2.83.

We need to calculate the speed of light pulse in glass

Using formula of speed

[tex]v=\dfrac{c}{\mu}[/tex]

Put the value into the formula

[tex]v=\dfrac{3\times10^{8}}{2.83}[/tex]

[tex]v=1.06\times10^{8}\ m/s[/tex]

We need to calculate the time of short pulse of light beam

Using formula of velocity

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

Put the value into the formula

[tex]t=\dfrac{25.21\times10^{-2}}{1.06\times10^{8}}[/tex]

[tex]t=2.37\times10^{-9}\ sec[/tex]

Hence, The time of short pulse of light beam is [tex]2.37\times10^{-9}\ sec[/tex]

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

Answer:

The  angular acceleration is  [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Explanation:

From the question we are told that

    The moment of inertia is  [tex]I = 0.034\ kg \cdot m^2[/tex]

     The  net torque is  [tex]\tau = 0.11\ N \cdot m[/tex]

Generally the net torque is mathematically represented as

           [tex]\tau = I * \alpha[/tex]

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

        [tex]\alpha = \frac{\tau }{I}[/tex]

substituting values

         [tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]

        [tex]\alpha = 3.235 \ rad/s ^2[/tex]

Answer:

364days

Explanation:

Pls see attached file

Explanation:

The moon will take 112.7 days to make one revolution around the planet.

The period of the satellite around any planet only depends upon the distance between the planet's center and satellite and also depends upon the planet's mass.

Given, the distance from the moon's center to the planet's surface,

h = 2.165 × 10⁵ km,

The radius of the planet, r = 4175 km  

The mass of the planet = 6.70 × 10²² kg

The total distance between the moon's center to the planet's center:

a = r +h = 2.165 × 10⁵ + 4175

a = 216500 + 4175

a = 220675

a = 2.26750 × 10⁸ m

The period of the planet can be calculated as:

[tex]T =2\pi \sqrt{\frac{a^3}{Gm} }[/tex]

[tex]T =2\3\times 3.14 \sqrt{\frac{(2.20675 \times 10^8)^3}{(6.67\times 10^{-11}).(6.70\times 10^{22})} }[/tex]

T = 9738253.26 s

T = 112.7 days

Learn more about Kepler's law, here:

https://brainly.com/question/1608361

#SPJ5

Answer:

0.273m/s

Explanation:

first find out the meaning of 0.90×10−4m3/s

literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s

1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?

cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²

so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²

they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s

Answer:

8 seconds

160 meters

Explanation:

Given in the y direction:

Δy = 0 m

v₀ = 40 m/s

a = -10 m/s²

Find: t

Δy = v₀ t + ½ at²

0 m = (40 m/s) t + ½ (-10 m/s²) t²

0 = 40t − 5t²

0 = 5t (8 − t)

t = 0 or 8

Given in the x direction:

v₀ = 20 m/s

a = 0 m/s²

t = 8 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²

Δx = 160 m

Answer:

At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.

Explanation:

The number of particles (N)  counts are inversely proportional to the distance between the source and the detector.  

By using the below formula we can find the number of counts.

[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]

The number of count per second, when the distance is 3 meters.

[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]

Number of count per second when the distance is 10 meters.

[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]

Answer:

Explanation:

When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .

Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .

hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

Answer:

Explanation:

Formula for calculating the relationship between  the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:

[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]  where;

Vs and Vp are the emf in the secondary and primary coil respectively

Ns and Np are the number if turns in the secondary and primary coil respectively

Ip and Is are the currents in the secondary and primary coil respectively

Since the are all equal to each other, then we can equate any teo of the expression as shown;

[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

Given parameters

Np = 500-turns

Ns = 2000-turns

Is = 3.0Amp

Required

Current in the primary coil (Ip)

Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]

[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]

[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]

Hence the current in the primary coil is 12Amp

Answer:

The maximum force that can be lifted by this system is  51,478.4 kg

Explanation:

Given;

maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²

diameter of the piston, d = 65 cm = 0.65 m

The maximum gauge pressure of the piston is given as;

[tex]Hp = \frac{F}{A}[/tex]

Where;

F is the maximum force of the piston

A is the area of the piston

[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]

F = Hp x A

F = 1.52 x 10⁶N/m² x 0.3319m²

F = 504488 N

Force is given as;

F = mg

m = F/g

m = 504488/9.8

m = 51,478.4 kg

Therefore, the maximum force that can be lifted by this system is  51,478.4 kg

Answer:

69.58 m tall

Explanation:

Pls see attached file

Answer:

2 m/s

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

[tex]E = Blv[/tex]

where [tex]E[/tex] is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

[tex]l[/tex] is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

[tex]v[/tex] is the velocity of the fluid through the field = ?

[tex]B[/tex] is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x [tex]v[/tex]

2.88 x 10^-3 = 1.44 x 10^-3 x [tex]v[/tex]

[tex]v[/tex] = 2.88/1.44 = 2 m/s

Answer:

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.

Explanation:

When a container is rigid, the process is supposed to be isochoric, that is, at constant volume. Then, the equation of state for ideal gases can be simplified into the following expression:

[tex]\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}[/tex]

Where:

[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Initial and final pressures, measured in pascals.

[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Initial and final temperatures, measured in Kelvins.

In addtion, the specific heat at constant volume for monoatomic ideal gases, measured in joules per mole-Kelvin is given by:

[tex]\bar c_{v} = \frac{3}{2}\cdot R_{u}[/tex]

Where:

[tex]R_{u}[/tex] - Ideal gas constant, measured by pascal-cubic meters per mole-Kelvin.

If [tex]R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}[/tex], then:

[tex]\bar c_{v} = \frac{3}{2}\cdot \left(8.314\,\frac{Pa\cdot m^{2}}{mol\cdot K} \right)[/tex]

[tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex]

And change in heat energy ([tex]Q[/tex]), measured by joules, by:

[tex]Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})[/tex]

Where:

[tex]n[/tex] - Molar quantity, measured in moles.

The final temperature of the monoatomic ideal gas is now cleared:

[tex]T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}[/tex]

Given that [tex]T_{1} = 300\,K[/tex], [tex]Q = 6000\,J[/tex], [tex]n = 4\,mol[/tex] and [tex]\bar c_{v} = 12.471\,\frac{J}{mol\cdot K}[/tex], the final temperature is:

[tex]T_{2} = 300\,K + \frac{6000\,J}{(4\,mol)\cdot \left(12.471\,\frac{J}{mol\cdot K} \right)}[/tex]

[tex]T_{2} = 420.279\,K[/tex]

The final pressure of the system is calculated by the following relationship:

[tex]P_{2} = \left(\frac{T_{2}}{T_{1}}\right) \cdot P_{1}[/tex]

If [tex]T_{1} = 300\,K[/tex], [tex]T_{2} = 420.279\,K[/tex] and [tex]P_{1} = 6.00\times 10^{4}\,Pa[/tex], the final pressure is:

[tex]P_{2} = \left(\frac{420.279\,K}{300\,K} \right)\cdot (6.00\times 10^{4}\,Pa)[/tex]

[tex]P_{2} = 8.406\times 10^{4}\,Pa[/tex]

The final pressure of the monoatomic ideal gas is 8.406 × 10⁶ pascals.