Answer:
375 is the answer.
Explanation:
Speed : Distance / Time taken
S: m/ s
s: 1500/4
375 m / s answer
Answer:
375m per minute
Explanation:
if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame
A battery is two or more individual cells connected together. Some large trucks utilize large 24 volt lead acid batteries. How many lead acid cells would be required to construct a battery with this voltage
Answer:
#_pile = 12 celdas
Explanation:
Lead acid sulfur batteries generate each cell a potential of 2 volts. By colonato to reach the voltage of 24 volts
#_pile = 24/2
#_pile = 12 cledas
serially connected
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
Both of these questions are the same but their answers in the answer key are different. Why?
A conductor is placed in a steady external electric field. Which of the following is FALSE?
a) All excess charge is distributed on the surface of the conductor
b) The electric field inside the conductor is the same as the external electric field
c) The electric field is zero inside the conductor
d) the electric field just outside the surface of the conductor is perpendicular to the surface
Answer:
a
Explanation:
because the electric field doesn't effect the conductor and its goes into storage for later
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
En la siguiente expresión matemáticas w=mg el peso w con relación a se relaciona con la masa m en una proporción
a) Directamente proporcional b) Inversamente proporcional c) Es constante
d) Ninguna de las anteriore
Answer:
a) Directamente proporcional
Explanation:
El peso se puede definir como la fuerza que actúa sobre un cuerpo o un objeto como resultado de la gravedad.
Matemáticamente, el peso de un objeto viene dado por la fórmula;
[tex] Peso = mg [/tex]
Donde;
m es la masa del objeto.
g es la aceleración debida a la gravedad.
De la expresión matemática, podemos deducir que el valor del peso de un objeto es directamente proporcional a la masa del objeto.
Por lo tanto, un aumento en la masa de un objeto provocaría un aumento en el peso del objeto y viceversa.
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?
Answer:
the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Explanation:
Given;
length of the conductor, L = 1 cm = 0.01 m
current carried by the solenoid, I₁ = 15 A
radius of the solenoid, r = 4 cm
number of turns per length of the solenoid, n = 800 turns/m
current carried by the solenoid, I₂ = 40 mA = 0.04 A
The magnetic field of the solenoid is calculated as;
B = μnI₂
where;
μ is the permeability of free space = 4π x 10⁻⁷ Tm/A
B = ( 4π x 10⁻⁷) x (800) x (0.04)
B = 4.022 x 10⁻⁵ T
The magnitude of the magnetic force on the wire segment is calculated as;
F = BI₁L sinθ
where
θ is the angle made by the wire segment against the solenoid = 90⁰
F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)
F = 6.03 x 10⁻⁶ N
Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N
Question 5 of 10
What must be the same for two resistors that are connected in parallel?
Answer:
in parallel combination : potential difference between two terminal of resistors are always constant. ... hence, potential difference ( voltage ) must be same across each resistor .
Explanation:
A converging lens is used to focus light from a small bulb onto a book. The lens has a focal length of 10.0 cm and is located 40.0 cm from the book. Determine the distance from the lens to the light bulb.
Answer:
[tex]u=13.3cm[/tex]
Explanation:
From the question we are told that:
Focal Length [tex]F=10.0cm[/tex]
Distance [tex]d=40cm[/tex]
Generally the equation for Focal length is mathematically given by
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]
[tex]\frac{1}{10}=\frac{1}{u}+\frac{1}{40}[/tex]
[tex]\frac{1}{u}=\frac{3}{40}[/tex]
[tex]u=13.3cm[/tex]
Focal length is the distance from the center of the lens to principle foci. The distance of the from the lens to the light bulb is 13.3 cm.
The distance can be determined by the formula,
[tex]\bold {\dfrac 1{f} = \dfrac 1{u} + \dfrac 1{v} }[/tex]
Where,
f - focal length = 10 cm
u - distance of object = ?
v = distance of image = 40 cm
Put the values in the equation,
[tex]\bold {\dfrac 1{10} = \dfrac 1{u} + \dfrac 1{40} }\\\\\bold {\dfrac 1{u} = \dfrac 3{40}}\\\\\bold {\dfrac 1{u} = 13.3 cm}[/tex]
Therefore, the distance of the from the lens to the light bulb is 13.3 cm.
To know more about the focal length,
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A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
Cuando el pistón tiene un volumen de 2x10^-4 m^3, el gas en el pistón está a una presión de 150 kPa. El área del pistón es 0.00133 m^2. Calcular la fuerza que el gas ejerce sobre el embolo del pistón.
Answer:
F = 1.128 10⁸ Pa
Explanation:
Pressure is defined by
P = F / A
If the gas is ideal for equal force eds on all the walls, so on the piston area we have
F = P A
We reduce the pressure to the SI system
P = 150 kpa (1000 Pa / 1kPa = 150 103 Pa
we calculate
F = 150 10³ / 0.00133
F = 1.128 10⁸ Pa
Two forces A and B act at a point. If their resultant is [given by] (B - A) in the direction of B, then
A. A and B are equal
B. A is greater than B
C. the angle between A and B is 0°
D. the angle between A and B is 90°
E. the angle between A and B is 180°
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
Please see this question related to Sum of Vectors for further details: https://brainly.com/question/11881720
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
Learn more about elastic collision:https://brainly.com/question/5719643
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A refrigerator has a coefficient of performance equal to 4.00. The refrigerator takes in 110 J of energy from a cold reservoir in each cycle. (a) Find the work required in each cycle. J (b) Find the energy expelled to the hot reservoir. J
Answer:
The correct answer is:
(a) 27.5 Joules
(b) 141.5 Joules
Explanation:
Given:
Energy,
[tex]Q_c = 110 \ J[/tex]
Coefficient of performance refrigerator,
[tex]Cop(refrig)=4[/tex]
(a)
As we know,
⇒ [tex]Cop(refrig) = \frac{Q_c}{Work}[/tex]
or,
⇒ [tex]Work=\frac{Q_c}{Cop(refrig)}[/tex]
[tex]=\frac{110}{4}[/tex]
[tex]=27.5 \ Joules[/tex]
(b)
⇒ [tex]Heat \ expelled = Heat \ removed +Work \ done[/tex]
or,
⇒ [tex]Q_h = Q_c+Work[/tex]
[tex]=114+27.5[/tex]
[tex]=141.5 \ Joules[/tex]
Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting at 2 m3 and ending at 1 m3. How much work do we need to do on the gas to perform this operation
Answer: [tex]-5013.65\ J[/tex]
Explanation:
Given
No of moles [tex]n=3[/tex]
Temperature [tex]T=290\ K[/tex]
Initial volume [tex]V_1=2\ m^3[/tex]
Final volume [tex]V_2=1\ m^3[/tex]
Work done in constant temperature process is
[tex]W=nRT\ln \left(\dfrac{V_2}{V_1}\right)[/tex]
Insert the values
[tex]\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J[/tex]
Which of the following is a form of mechanical energy?
A. Chemical energy
B. Gravitational potential energy
C. Thermal energy
D. Nuclear energy
Answer:
B
Explanation:
no reason for this answer
difference between wavefront and wavelets
Answer:
A wavefront is the locus of all the particles which are in phase. A wavelet is an oscilation that starts from zero, then the amplitude increases and later decreases to zero
A ball is thrown upward from the edge of a cliff with an initial velocity of 6 m/s. (a) How fast is it moving 0.5 s later? In what direction? (b) How fast is it moving 2 s later? In what direction?
Answer:
Explanation:
Kinematic equation
v = u + at
If UP is assumed to be the positive direction and we let gravity be 10 m/s² which will be in the downward direction so will be negative.
a) v = 6 + (-10)(0.5) = 1 m/s the result is positive, so upward
b) v = 6 + (-10)(2) = -14 m/s the result is negative, so downward
1. Estimate the buoyant force that air exerts on a man. (To do this, you can estimate his volume by knowing his weight and by assuming that his weight density is about equal to that of water. Assume his weight is 940 N.) answer in N
2.On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total weight of the balloon, including its load and the hot air in it, is 17000 N.
(a) What is the weight of the displaced air?
answer in N
(b) What is the volume of the displaced air?
answer in m^3
Solution :
1. We know that : Buoyant force = weight of the liquid displace
= volume displaced x density of the fluid
Now volume of the man = [tex]$\frac{\text{mass}}{\text{density}}$[/tex]
Mass = weight / g
[tex]$=\frac{940}{9.8}$[/tex]
= 95.92 kg
And density = 1000 [tex]kg/m^3[/tex]
Therefore,
[tex]$\text{volume} = \frac{\text{mass}}{\text{density}}$[/tex]
[tex]$=\frac{95.92}{1000}$[/tex]
= 0.0959 [tex]m^3[/tex]
We know density of air = 1.225 [tex]kg/m^3[/tex]
∴ Mass of air displaced = 0.0959 x 1.225
= 0.1175 kg
Weight of the air displaced = 1.1515 N
Therefore, the buoyant force = 1.1515 N
2). As the balloon is not accelerated, the net force acting on it is zero.
Thus the weight that acts downwards = buoyant force upwards
So, the weight of the air displaced = weight of the balloon
= 17000 N
Therefore, the mass of the air displaced = volume of the air displaced (volume of the balloon) x density of air
[tex]$\frac{17000}{9.8} = \text{volume of air} \times 1.225$[/tex]
[tex]$\text{Volume of air displaced} = \frac{1700}{9.8 \times 1.225}$[/tex]
= 1416.0766 [tex]m^3[/tex]
In first case a mass M is split into two parts with one part being 1/6.334 th of the original mass. In second case M is split into two equal parts. In both the cases the two parts are separated by same distance. What ratio of the magnitude of the gravitational force in first case to the magnitude of the gravitational force in the second case
Answer:
[tex]F_r=0.132:0.25[/tex]
Explanation:
From the question we are told that:
[tex]M_1=M*\frac{1}{6.334}[/tex]
Therefore
[tex]M_2=M-M*\frac_{1}{6.334}[/tex]
[tex]M_2=M*\frac{5.334}{6.334}[/tex]
Generally the equation for Gravitational force of attraction is mathematically given by
For Unequal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(M*\frac_{1}{6.334})(M*\frac{5.334}{6.334})}{d^2}[/tex]
[tex]F=\frac{GM^2}{d^2}*(0.132)[/tex]
For equal split
[tex]F=\frac{GM_1M_2}{d^2}[/tex]
[tex]F=\frac{G(\frac{M}{2})((\frac{M}{2}}{d^2}[/tex]
[tex]F=0.25 \frac{GM^2}{d^2}[/tex]
Therefore the ratio of the gravitational force is
[tex]F_r=0.132:0.25[/tex]
Calculate the magnitude of a gravitational force between two object 400kg and 800kg separated by a distance of of 45m (take G =6.67 * 10^-11 Nm^2 kg^-2)
Answer:
Explanation:
The formula is
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:
[tex]F_g=\frac{(6.67*10^{11})(400)(800)}{(45^2)}[/tex] and multiply and divide all that out to get
[tex]F_g=1.1*10^{-8}[/tex] It should really only be 1 significant digit since 400 and 800 both have only 1 significant digit, but I used 2. It should be
[tex]F_g=[/tex] 1 × 10⁻⁸ N
A student wants to start a small business in school. Write down six items that
he/she can sell in school at a profit.
Answer:
packets of pen
packets of pencil
copies
books
bottles
mask
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
What is a profit?Profit is the difference between the revenue earned by a business or individual and the costs incurred to produce the goods or services sold.
It is an important measure of financial success for companies and is often used to determine the value of a business.
We have,
Here are six items that a student can sell in school at a profit:
Homemade baked goods - cupcakes, cookies, brownies, and other treats can be sold individually or as a pack.
School supplies - items such as pens, pencils, erasers, rulers, notebooks, and binders are always in demand.
Drinks - bottled water, juices, and sodas are popular beverages that students may purchase during the school day.
Healthy snacks - fresh fruit, granola bars, and trail mix are nutritious snacks that many students are interested in buying.
Personalized accessories - items like keychains, bracelets, and bookmarks with unique designs or student names can be popular among peers.
Stickers - fun and colorful stickers can be sold individually or in packs and are often a favorite of younger students.
Thus,
Six items that a student can sell in school at a profit:
- Homemade baked goods
- School supplies
-Drinks
- Healthy snacks
- Personalized accessories
- Stickers
Learn more about profit here:
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why meter cube is called derived unit
Answer:
Because it is the result of two more fundamental units, a derived unit is termed that. For volume, the cubic meter (m³) is the fundamental unit of area. Any number that cannot be measured directly with any equipment is referred to as a derived unit. For example, we can't quantify a substance's density using a rule, scale, or bucket.
OAmalOHopeO
An observer on Earth sees Planet X to be stationary and also sees a rocket traveling toward Planet X at 0.5c. The rocket emits a pulse of light that travels outward in all directions. According to an observer on Planet X, how fast is the light pulse traveling toward them?
a) 2c/3
b) c/2
c) 2c/3
d) 5c/6
e) c
(E) c
Explanation:
The speed of light is always equal to c regardless of the relative motion of the light source.
The graph below shows a cycle of a heat engine. Add the following labels to the graph. Some labels are used more than once.
Labels: Isobaric process; W= 0J; Work done on the system; Work done by the system.
I will give brainliest!
P.S. AL2006 if you see this please help!
I'm not very good at this material. I'll try it, but if I were you, I wouldn't bet money on these answers.
"Isobaric" means constant pressure. So those are the horizontal lines, where every point on the line is at the same pressure. Those are the processes 1>2 and 3>4 .
I'm going around and around in my mind with the other labels, and I can't decide. So I'm afraid I can't answer any more of them ... they might be wrong.
Answer:
1 -> 2 & 3 -> 4: Isobaric process
4 -> 1: Work done BY the system
2 -> 3: Work done ON the system
W(total): W = 0J
Explanation:
The two horizontal lines (1 -> 2 & 3 -> 4) are "Isobaric" since isobaric processes take place at constant pressure. I believe 4 -> 1 is "Work done BY the system" since pressure increases when there is an increase of thermal energy, in other words, the system is absorbing heat. This is why the volume increases from 1 -> 2 after the system has absorbed heat in 4 -> 1. Following the directions of the arrows, 2 -> 3 would be "Work done ON the system" since pressure is DECREASING, meaning temperature is also exiting the system. That's why the next step (3 -> 4) shows a decrease in volume. This model depicts a process that has a W(total) of 0 J because this is a cycle.
I hope this helps :))
