Answer: 60 Joules (J)
Explanation:
To compute the work done by a force, you can use the formula:
Work = Force × Distance × cosθ
In this case, the force applied is 300 N, and the distance is 20 cm (which can be converted to meters by dividing by 100, giving us 0.20 m).
The angle θ represents the angle between the direction of the force and the direction of displacement. Since the force is used to compress the spring, the displacement is in the same direction as the force, and the angle θ is 0 degrees. The cosine of 0 degrees is 1, so we can omit the cosine term.
Now we can calculate the work done:
Work = 300 N × 0.20 m
Work = 60 Joules (J)
Therefore, the work done by the force in compressing the spring is 60 Joules (J).
In an experiment similar to the one pictured below, an electron is projected horizontally at a speed vi into a uniform electric field pointing up. The magnitude of the total vertical deflection, ye, of the electron is measured to be 1 mm. The same experiment is repeated with a proton (whose mass is 1840 times that of the electron) that is also projected horizontally at a speed vi into the same uniform electric field. What is the magnitude of the total vertical deflection, yp, for the proton
I think you need Graph to figure it out
Using Newton's second law and kinematic projectile motion we can find the proton deflection y = 5.43 10⁻⁷ m, in the opposite direction to the electron deflection.
given parameters
The deflection of the electorn y₁ = 1 mm = 0.001 m The initial velocity of the electron and proton v_i The mass of the proton m_p = 1840 meto find
deflection of the protonFor this exercise we will use Newton's second law where the force is electric
F = ma
F = q E
where F is the force, q the charge, E the electric field, m the mass and the acceleration of the particle
q E = m a
a = q / m E
This acceleration is the direction of the electric field that is perpendicular to the initial velocity (v_i)
Having the acceleration we can use the kinematics relations
If we make the direction of the initial velocity coincide with the x-axis
v_i = cte
v_i = x / t
t = x/ v_i
on the y-axis is in the direction of the electric field
y = v_{iy} t + ½ a t²
on this axis the initial velocity is zero
y = [tex]\frac{1}{2} (\frac{q}{m} E) \ t^2[/tex]
subtitute
y = (1)
Electron motion.
Let us propose the expression for the electron situation, the length of the displacement must be the same for electron and proton, suppose that it is x = L
In this case the charge q = -e and the mass m = m_e
its substitute in equation 1
y₁ = [tex]\frac{1}{2} \ ( \frac{-e}{m_e} E) \ \frac{x^2}{v_i^2}[/tex]
where y₁, is the lectron deflection.
Proton motion
Between the proton and the electron we have some relationships
q_p = -e
m_ = 1840 m_e
we substitute in the equation 1
y₂ = ½ e / 1840 me E x² / vi²
y₂ =
y₂ = - y₁ / 1840
y₂ = - 0.001 / 1840
y₂ = - 5.43 10⁻⁷ m
The negative sign indicates that the deflection of the proton is in the opposite direction to the deflection of the electron.
In conclusion they use Newton's second law and kinematics we can find the proton deflection is y = 5.43 10⁻⁷ m
learn more about electric charge movement here: https://brainly.com/question/19315467
A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move away from the object?
Answer:
v = 7.95 m/s
Explanation:
Given that,
Wavelength of a wave, [tex]\lambda=53\ cm=0.53\ m[/tex]
Frequency of a wave, f = 15 Hz
We need to find the speed of the wave. The speed of a wave is given by :
[tex]v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s[/tex]
So, the wave move with a speed of 7.95 m/s.
In picture 1, heat is flowing from the ____ to the _____ In picture 2, heat is flowing from the _______ to the ____
Answer: In picture 1, heat is flowing from the liquid to the air. In picture 2, heat is flowing from the air to the liquid
Explanation:
I don't know if I answered correctly, if not I can provide another answer
Fill in the question
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
please help me I'm begging you Define and give examples of elements and compounds the structure of atoms (electrons, neutrons, and protons)
In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position
Answer:
Explanation:
The relation between electric field and potential difference is as follows
E = - dV / dr
That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .
Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means
dV / dr = constant
E = constant
Electric field is constant .
So the option which is correct is
" points east and does not vary with position " .
PLEASE HELP ASAP! WILL GIVE BRAINLIEST TO CORRECT ANSWER! HELP!! HELP!!
The diagram shows the structure of an animal cell.
The image of an animal cell is shown with some organelles labeled numerically from 1 to 6. The outer double layer boundary of the cell is labeled 1. A stacked disc like structure is labeled 2. A broad rod shaped structure with an irregular shape inside it is labeled 3. The entire plain section that forms the background of the cell and is within the outer boundary is labeled 4. A small circular shape within the large circular shape is labeled 5. The large central circular shape is labeled 6.
Which number label represents the cell membrane?
1
2
4
6
(this is middle school science)
Answer:
1. cell membrane
2. golgi body
3. mitochondrion
4. cytoplasm
5. nucleolus
6. nucleus
Explanation:
The correct answer to this question is Option A; 6.
Why?
In a plant cell, the nucleus surrounds the nucleolous, which would be number 5. Therefore, number 6 would be your correct answer.
~Thank you~
An ordinary ruler is used to measure the area and its error of a rectangle. It is found that their sides are 5.0 cm long and 2.0 cm width. The error in area (in cm) is
Answer:
You need to know the accuracy to which you can read the ruler:
Suppose that you can read the read the ruler to the nearest milimeter
A = L * W your calculated area of the rectangle
A + ΔA = (L + ΔL) * (W + ΔW) = L W + L ΔW + W * ΔL + ΔL ΔA
Or ΔA = L ΔW + W ΔL
Where we have subtracted A = L * W and the term ΔL * ΔA is very small
So (5 + .1) * (2 + .1) - 5 * 2 = .1 * 2 + .1 * 5 = .7 cm^2
Then you report A = 10 cm^2 +- .7 cm^2 including the - sign for completeness
[BWS.02]If the same experiment is repeated in different parts of the world by different scientists,
the results will be the same
the results will become invalid
the outcome of the experiment will be non testable
the outcome of the experiment will be non observable
Answer:
the results will be the same.it may be
If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.
What is scientific experiment?An experiment is a procedure that is carried out to support or refute a hypothesis, or to determine the efficacy or likelihood of something that has never been tried before. Experiments shed light on cause-and-effect relationships by demonstrating what happens when a specific factor is changed.
Controls are typically included in experiments to minimise the effects of variables other than the single independent variable. This improves the reliability of the results, often by comparing control measurements to the other measurements. Scientific controls are an essential component of the scientific method.
Hence, If the same experiment is repeated in different parts of the world by different scientists, the results will be the same.
Learn more about experiment here:
https://brainly.com/question/11256472
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What are regular and irregular reflection of light? plz help its
urgent..
Explanation:
Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.
A toy car can go 5 mph. How long would it take to go 12 miles?
Earth's magnetic field is approximately 1/2 gauss, that is 50 micro-tesla because the SI field unit of a tesla is 10,000 gauss. Earth's north geographic pole is close to its south magnetic pole, and magnetic field is directed from the north to the south poles of a magnetic dipole so it goes from Earth's south geographic pole towards its north. Suppose you have wire carrying a large DC current from the south wall of a building to its north wall and that it is horizontal, on the floor. If Earth's field is parallel to the ground and does not dip, what force if any would the wire experience
Answer:
F = 0
Explanation:
The magnetic force is described by two expressions
for a moving charge
F = q v x B
for a wire with a current
F = I L xB
bold indicates vectors
let's write this equation in module form
F = I L B sin θ
where the angle is between the direction of the current and the direction of the magnetic field
In this case they indicate that the cable goes from the South wall to the North wall, so this is the direction of the current
The magnetic field of the Earth goes from the south to the north and in this part it is horizontal
Therefore the current and the magnetic field are parallel, the angle between them is zero
sin 0 = 0
consequently the magnetic force is zero
F = 0
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 60-m-diameter blades at that location. Take the air density to be 1.25 kg/m3. Cengel, Yunus; Cengel, Yunus. Thermodynamics: An Engineering Approach (p. 98). McGraw-Hill Higher Education. Kindle Edition.
Answer:
1767Kw
Explanation:
Velocity of wind = 10 m/s
diameter of the blades= 60m
ρ= air density = 1.25 kg/m3
Acceleration due to gravity= 9.81 m/s^2
Mechanical energy of the wind can be calculated using the expression below
Energy= (e*m)
= ρ V A e............eqn(1)
Where A= area
ρ= air density
e= wind energy per unit mass of air
e= (v^2)/2..........eqn(2)
If we substitute the values into eqn (2) we have
e= [(10)^2]/2
=50J/Kg
But Area=A= (πd^2)/4
Area= ( π× 60^2)/4
Area=2827.8m^2
If we input substitute the values into eqn (1) we have
Energy= 1.25 ×10 × 50×2827.8
=1767145.7W
We can convert to kilo watt
=1767145.7W/ 1000
= 1767Kw
Hence, the mechanical energy of air per unit mass and the power generation potential of a wind turbine is 1767Kw
As every amusement park fan knows, a Ferris wheel is a ride consisting of seats mounted on a tall ring that rotates around a horizontal axis. When you ride in a Ferris wheel at constant speed, what are the directions of your acceleration and the normal force on you (from the always upright seat) as you pass through (a) the highest point and (b) the lowest point of the ride
Answer:
Answer is explained in the explanation section below.
Explanation:
In this question, we are asked to find out the direction of acceleration and direction of the normal force acting upon us from the always upright seat.
a) You pass through the highest point:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the highest point, then the direction of of our acceleration will be towards the center or it will be towards downward direction.
And at the highest point on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards.
b) You pass through the lowest point of the ride:
When we sit in the Ferris wheel at the any amusement park, and when it starts rotating and the time when we reach the lowest point, then the direction of of our acceleration will be towards the center or it will be towards upward direction.
And at the lowest on the Ferris Wheel, the direction of the normal force F acting upon us will be upwards again.
a 2,400 kg car drives north towad a 60kg shopping cartthat has a velocity of zero the two objects collide giving the car a final velocity 4.33m/s north and the shopping cart 8.88m/s north what is the in itial velocity of the car
Answer:
4.552m/s
Explanation:
[tex]V=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{1} } =\frac{2400*4.33+60*8.88}{2400}=4.552m/s[/tex]
what element is produced when a gold nucleus loses a proton?
The string will break if the tension in
it exceeds 0.180 N. What is the
smallest possible value of d (in cm)
before the string breaks?
Answer:
define d first?
you need to list more variables
Answer:
list more valuable unit
A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its surface. Find the electric field for the following points: (a) for all points outside the spherical shell E = keq2/r2 E = q/4πr2 none of these E = keq/r2 E = 0 (b) for a point inside the shell a distance r from the center E = keq2/r2 E = keq/r2 E = 0 E = q/4πr2 none of these
Answer:
a) E = 0
b) [tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
From which we have;
[tex]E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0[/tex]
E = 0/A = 0
E = 0
b) [tex]\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}[/tex]
[tex]E \cdot A = \dfrac{+q }{\varepsilon _{0}}[/tex]
[tex]E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}[/tex]
By Gauss theorem, we have;
[tex]E\oint dS = \dfrac{q}{\varepsilon _{0}}[/tex]
Therefore, we get;
[tex]E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}[/tex]
The electrical field outside the spherical shell
[tex]E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }[/tex]
[tex]k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }[/tex]
Therefore, we have;
[tex]E = \dfrac{k_e \cdot q}{ r^2 }[/tex]
The voltage v(t) = 141.4 cos (ωt) is applied to a load consisting of a 10Ω resistor in parallel with an inductive reactance XL=ωL = 3.77Ω. Calculate the instantaneous power absorbed by the resistor and by the inductor. Also calculate the real and reactive power absorbed by the load, and the power factor. Draw all the voltage, current and power waveforms, also the draw the circuit and phasor diagrams.
Answer:
A) P(t) = 2651.25 [ 1 - cos2wt ] W
B) Real power = 999.79 watts
Reactive power = 2652.86 VA
c) power factor = 0.3526
Explanation:
Given data:
V(t) = 141.4 cos (ωt)
R(t) = 10 Ω
Inductive reactance XL = ωL = 3.77 Ω
Ir(t) = V(t) / R(t) = 14.14
A) Calculate the instantaneous power absorbed by the resistor and by inductor
By resistor :
Pr(t) = V(t) * Ir(t) = 141.4 * 14.14 [tex]cos^{2} wt[/tex] = 1999.396 [tex]cos^{2} wt[/tex]
hence Pr = 999.698 (cos2ωt + 1) w
By Inductor :
Pl(t) = V(t) I'L(t) = 141.4 cosωt * 37.5 cos(ωt - 90)
= 5302.5 [tex]sin^2 wt[/tex]
Hence Pl(t) = 5302.5 [tex]sin^2 wt[/tex] w = 2651.25 [ 1 - cos2wt ] W
B) calculate the real and reactive power
First we have to determine the power factor
Given that : V(t) = 141.4 cosωt v , Ir(t) = 14.14 cosωt A
IL(t) = 37.5 cos (ωt - 90° )
The phasor representation of the above is :
V = [tex]\frac{141.4}{\sqrt{2} } <0^{0} v[/tex] = 141.4 ∠0° , Ir = 10 ∠ 0° , IL = 37. 5 ∠ -90°
Total load current = Ir + IL = 28.35 ∠ -69.35°
power factor = cos -69.35° = 0.3526
Next we will determine the Real and reactive power using the relation below
S = VI = 100 ∠ 0° * 28.35 ∠ -69.35°
= 2835 ∠ 69.35°
S = P + jQ = 999.79 + 2652.85 j
Real power = 999.79 watts
Reactive power = 2652.85 VA
The carts are moving on a level, frictionless track. After the collision all three carts stick together. Find the speed of the combined carts after the collision.
Answer:
0.13 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law say about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.
Answer:
A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.
(Hope this helps! Btw, I am the first to answer.)
How do objects with the same charger interact
The interaction between two like-charged objects is repulsive. ... Positively charged objects and neutral objects attract each other; and negatively charged objects and neutral objects attract each other.
Answer:
they repel with each other. object of like charges repel while object of opposite charges attracts with each other.
A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 volts, how many volts will the secondary coil have?
242 volts
288
353
467
Answer:
467 volts
Explanation:
Vs/Vp = Ns/Np
Vs = Ns/Np × Vp
Vs = 350/150 × 200 = 7/3 × 200
Vs = 467 volts
What is the acceleration of a .3 kg mass when there is a net force of 25.9 N on it?
Answer:
86.33m/s^2
Explanation:
Acceleration = Force/Mass
= 25.9/0.3
= 86.33
A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:
Complete question is;
A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:
A) 10 feet
B) 30 feet
Answer:
A) 0.728 ft/s
B) 1.8 ft/s
Explanation:
Let the the position of the worker in ft be denoted by s.
Since he begins to walk away at a constant rate of 3 ft/s, then;
ds/dt = 3 ft/s
Now, the rope will form a triangle, with width "s" and the height 40. Since distance from the connection point to the ceiling = 40 ft
Using pythagoras theorem, we can find the length of the rope on this side of the pulley.
Hence, the length of rope on this side of the pulley = √(s² + 40²)
Meanwhile, on the other side the length will be;
(80) - √(s² + 40²)
Also, height of the weight will be;
h = 40 - ((80) - √(s² + 80²))
h = √(s² + 80²) - 40
Differentiating this, we have;
dh/dt = (ds/dt) × (s/√(s² + 40²))
From earlier, we saw that ds/dt = 3 ft/s
Thus;
dh/dt = 3s/√(s² + 40²)
A) when he has walked 10 ft, it means that s = 10. Thus;
dh/dt = (3 × 10)/√(10² + 40²)
dh/dt = 0.728 ft/s
B) when he has walked 30 ft, it means that s = 30. Thus;
dh/dt = (30 × 3)/√(30² + 40²)
dh/dt = 1.8 ft/s
Can someone please help, ty!!
Will mark brainliest.
Answer:
4. unbalanced and Accelerating
5. balance and rest
1. Clara stops for 10 minutes to catch up with a friend.
Answer:
Clara has speed of 80m/min
Explanation:
Clara was jogging at 600 m in 5 minutes. She stopped suddenly which reduced her velocity and then she waited for 10 minutes so that her friends comes near her. She stopped to catch her friend. During this 10 minutes the velocity of Clara is zero. She started to walk again at a slower speed of 80m/min.
. [30%] We first showed that The electric field for a point charge radiating in 3-dimensions has a distance dependence of 1/r 2 (see Equation 1). In Problem 1 you showed that the electric field for a point charge radiating in 2-dimensions has a distance dependence of 1/r . Consider again the 2-dimensional case described in Problem 1. What distance dependence do you expect for the electric potential
Answer:
Answer is explained in the explanation section below.
Explanation:
Note: This question is incomplete and lacks necessary data to solve. As it mentioned the reference of problem number 1, which is missing in this question. However, I have found that question on the internet and will be solving the question accordingly.
Solution:
The relation between electric field and the electric potential is:
E = [tex]\frac{dV}{dr}[/tex]
So, making dV the subject, we have:
dV = E x dr
Integrating the above equation, we get.
V = [tex]\int\limits^_ {} \,[/tex]E x dr Equation 1
Now, in 2-D
E is inversely proportional to the radius r.
E ∝ 1/r
So, we can write: replacing E ∝ 1/r in the equation 1
V ∝ [tex]\int\limits^_ {} \,[/tex][tex]\frac{1}{r}[/tex] x dr
Which implies that,
V ∝ log (r)
Hence, distance dependence expected for the electric potential = ln (r)
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa