The work done is given by 742.5 J while the coefficient of kinetic friction between the block and the surface is 0.46.
What is the work done?The work done is given by the use of the formula;
W = F * x
Where;
F = force applied
x = distance covered
W = 150 N * 4.95 m = 742.5 J
Now;
The coefficient of kinetic friction is given by;
μ = F/mg
μ = 150/ 33 * 9.8
μ = 0.46
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Which of the following would have the least amount of inertia? Assume all the bags are the same size.
bag of rocks
bag of feathers
bag of bricks
bag of sand
Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
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
Earth’s atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by Earth’s atmosphere?
Answer:
Earth's atmosphere traps energy from the sun. Which is a direct result of the trapping of energy by earth's atmosphere? Earth has moderate temperatures.
Explanation:
What current is needed in the solenoid's wires?
A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.8 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field.
What current is needed in the solenoid's wires? Express your answer with the appropriate units.
6th grade science I mark as brainliest
Answer:
8. organelle
Explanation:
9. Epithelial tissue
am i correct?
Determine the voltage Vab for the first circuit and also determine the voltages Vab and Vcd for the second circuit
Vab= E = 20V
because I = 0 and the voltage drop across the resistances R1 and R2 is also 0.
Second circuit:
Vab = 10V (no voltage drop across R1)
Vcd= E2-E1 = 20V
Series connection of voltage sources. But the sources are connected to the contrary and voltage drop across R1 or R2 is 0 V.
What is the mass of 9.11 moles of
ozone, 03?
The molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
What is molecular mass?Molecular mass exists as a number equivalent to the totality of the atomic masses of the atoms in a molecule. The totality of the atomic masses of all atoms in a molecule is established on a scale in which the atomic masses of hydrogen, carbon, nitrogen, and oxygen exist 1, 12, 14, and 16, respectively.
To compute the Molecular Mass of [tex]$O_{3}[/tex]
Atomic mass of oxygen(O) = 16
As [tex]$O_{3}[/tex] contains 3 atoms,
The molecular mass of [tex]$O_{3}[/tex]
= (16 x 3) = 48g/mol
Hence the mass of 9.11 moles O3
= 9.11 mol x 48g/mol
= 437.28g
= 0.43728kg
Therefore, the molecular mass of [tex]$O_{3}[/tex] is 0.43728kg
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Tell types of mirros and
each
one
Answer: We can identify the different types of mirrors without touching them by looking at the image it produces. Look into each mirror, the nature of the image produced will tell you the type of mirror it is.
- A plane mirror will produce an image of the same size as your face.
- A concave mirror will produce a magnified image of your face.
- A convex mirror will produce a diminished image of your face.
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A toy car can go 5 mph. How long would it take to go 12 miles?
Plzzz answer this question correctly
Answer:
changing the direction in which a force is exerted
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.
QUESTION 9 / 10
What is the first step you should take when you want to open a savings account?
A. Present your photo ID to the bank representative,
B. Make your initial deposit.
C. Review the different savings account options that your
bank offers.
D. Go to the bank and fill out an application.
Answer:
A
Explanation:
AWNSER:
awnser:
C
explanation:
Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.96 Hz. The amplitude of the motion is 5.95 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits
Answer:
a) A' = 0.345 m, b) f = 2,800 Hz
Explanation:
b) The angular velocity of a simple harmonic motion is
w =[tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and frequency are related
w = 2π f
we substitute
f = 1 /2π √k/m
indicates that the initial frequency value f = 3.96 Hz
in this case the mass is reduced by half
m ’= m / 2
we substitute
f = 2π [tex]\sqrt{\frac{k}{m} }[/tex]
f = √1/2 (2π √k/m)
f = 1 /√2 3.96
f = 2,800 Hz
a) The amplitude of the movement is defined by the value of the initial depalzamienot before an external force that initiates the movement.
When the block is divided into two parts of equal masses as if it were exploding, for which we can use the conservation of moment
initial instant. Right before the division
p₀ = (m₁ + m₁) v
final instant. Right after the split
p_f = m₁ v '
p₀ = p_f
(2 m₁) v = m₁ v ’
v ’= 2v
At this point we can use conservation of energy for the system with only half the block.
Starting point. Where the block divides
Em₀o = K = ½ m v'²
Final point. Point of maximum elongation
Em_f = Ke = ½ k A²
how energy is conserved
Em₀ = Em_f
½ m’ v’² = ½ k A’²
we substitute the previous expressions
½ m/2 (2v)² = ½ k A’²
A’² = 2 m v² / k (1)
Let's use the conservation of energy with the initial conditions, before dividing the block
½ m v2 = ½ k A2
A² = mv² / k = 5.95 10⁻² m²
we substitute in 1
A'² = 2 A²
A ’²= 2 5.95 10⁻²
A ’²= 11.9 10⁻² m
A' = 0.345 m
Help plz I’ll mark brainliest
Answer:
The second option- a substance that a wave can travel through.
Explanation:
Hope This Helps!!
(brainliest please)
Two train whistles have identical frequencies of 180 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the train platform hears beats with a frequency of 2.00 beats/s when the whistles sound at the same time. What are the two possible speeds and directions that the moving train can have?
Actual answers :3.85 m/s away from the station and 3.77 m/s towards the station from the book. I just need to know how to get to the answers.
Answer:
-3.77 m/s
3.85 m/s
Explanation:
given that
Frequency at stationary = 180 Hz
Beat frequency = 2 Hz
Using Doppler effect, we know that
f' = f[(v ± v0) / (v ± vs)], where
v = speed of sound, 343 m/s
v0 = speed of the observer, 0
vs = speed of light, ?
f = stationary frequency, 180 Hz
f' = stationary ± beat frequency, 180 ± 2
Applying the formula, we have
f' = f[(v ± v0) / (v ± vs)]
182 = 180 [(343 + 0) / (343 + vs)]
182/180 = 343 / 343 + vs
343 + vs = 343 * 180/182
343 + vs = 339.23
vs = 339.23 - 343
vs = -3.77 m/s
Again, using
f' = f[(v ± v0) / (v ± vs)]
178 = 180 [(343 + 0) / (343 + vs)]
178/180 = 343 / 343 + vs
343 + vs = 343 * 180/178
343+ vs = 346.85
vs = 346.85 - 343
vs = 3.85 m/s
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
To fully describe velocity you must have a _____
A. Magnitude and unit
B. Speed and unit
C. Average speed and position
D. Magnitude and direction
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.
An astronaut named Sandra Bullock has drifted too far away from her spaceshuttle while attempting to repair the Hubble Space telescope. She realizes that theshuttle is moving away from her at 3 m/s. On her back is a 10 kg jetpack which consistsof an 8 kg holding tank filled with 2 kg of pressurized gas. Without the jetpack, sheand her space suit have a mass of 80 kg.
Required:
a. She is able to use the gas to propel herself in the same direction as the shuttle. The gas exits the tank at a uniform rate with a constant velocity of 100 m/s, relative to the tank (and her). After the gas in the tank has been released, what is her velocity?
b. After this, she throws her empty tank into space and relies on the conservation of momentum to increase her speed to match that of the shuttle. With what velocity (in her frame of reference!) will she have to throw the tank?
Answer:
a) v_f = 0.898 m / s, b) v₂ = -6.286 m / s
Explanation:
a) For this exercise we use the conservation of momentum, we define a system formed by the astronaut, her equipment and the expelled gases. We must also define a stationary frame of reference, let's place the system on the platform, so the speed of the subject is v = -3 m / s
Initial instant. Before you start to pass gas
p₀ = (M + Δm) v
M is the mass of the astronaut M = 80Kg and Δm the masses of the gases
Final moment. When you expel the gases
p_f = M (v + Δv) + Δm (v-v_e)
where v_e is the gas velocity v_e = 100 m / s
momentum is conserved
p₀ = p_f
M v + Δm v = Mv + M Δv + Δm v -Δm ve
0 = M Δv - Δm v_e
if we make the very small quantities Δv → dv and Δm → dm, furthermore the quantity of output gas is equal to the decrease in the total mass dm = -dM
M dv = -v_e dM
∫ dv = - v_e ∫ dM / M
We solve, between the lower limits v₀ = v with M = M₀ and the upper limit v = v_f for M = M_f
v_f - v₀ = - v_e (ln M_f - Ln M₀)
v_f - v₀ = v_e ln ([tex]\frac{M_o}{M_f}[/tex])
v_f = v₀ + v_e ln (\frac{M_o}{M_f})
let's calculate
v_f = -1.3 + 100 ln (80 + 10 + 2/80 + 10)
v_f = -1.3 +2.20
v_f = 0.898 m / s
b) launch the jetpack to increase its speed up to the speed of the platform
initial instant. Before launching the tanks
p₀ = (M + m') v_f
final instnte. After launching the tanks
p_f = M v₁ + m' v₂
indicate that the final velocity of the astronaut is the platform velocity v₁=0 m / s, since the reference system is fixed on it
p₀ = p_f
(M+ m) v_f = M v₁ + m v₂2
v₂ = [tex]\frac{ M ( v_f - v_o) + m' v_f}{m'}[/tex]
v₂ = [tex]\frac{M}{m}[/tex] (v_f -v₁) + v_f
let's calculate
v₂ = 80/10 (0.898 - 0) + 0.898
v₂ = -7.1874 + 0.898
v₂ = -6.286 m / s
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
What is the velocity of the cart in these sections?
a-b
c-d
e-f
f-g
What force causes a resistance in motion
when two surfaces are touching?
Answer:
FRICTION
Explanation:
Friction is a force, the resistance of motion when one object rubs against another.
Frictional force
Explanation:
Its the opposing force against horizontal motion
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
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3. Consider a large windmill 30m in diameter. On a windy day, suppose that the windmill entrains a stream of air at a speed of 40 mph. Downstream of the windmill, the entrained stream exits over a large diameter at a speed of 20 mph. The pressure is 2atm at the inlet and equals atmospheric pressure at the outlet. Find the power (in megawatts) generated by the windmill. Density of air is 1.2 kg/m3
Answer:
The power generated by the windmill is approximately 1.364 MW
Explanation:
The diameter of the windmill, d = 30 m
The inlet speed of the wind, [tex]V_e[/tex] = 40 mph = 17.88 m/s
The exit stream velocity, [tex]V_i[/tex] = 20 mph = 8.94 m/s
The pressure at the inlet, P₁ = 2 atm
The pressure at the outlet, P₂ = 1 atm
The density of air, ρ = 1.2 kg/m³
The power obtained from the windmill, 'P', is given as follows;
[tex]P =\dfrac{1}{4 \cdot g_c} \cdot \rho \cdot A \cdot (V_i + V_e)\cdot (V_i^2 - V_e^2)[/tex]
Where;
[tex]g_c[/tex] = 1.0 kg/(N·s²)
A = Cross-sectional rea of the the windmill = π·D²/4 = π×(30 m)²/4 = 706.858347 m²
Plugging in the values, we get;
[tex]P =\dfrac{1}{4 \times 1.0} \times1.2 \times 706.858347 \times (17.88 + 8.94)\cdot (17.88^2 - 8.94^2) = 1363668.19438[/tex]
The power generated by the windmill, P ≈ 1363668.19438 W ≈ 1.364 MW.
An electron, tial well may be anywhere within the interval 2a. So the uncertainty in its position is Δx= 2a. There must be a corresponding uncertainty in the momentum of the electron and hence it must have a certain kinetic energy. Calculate this energy from the uncertainty relationship and compare it.
Answer:
[tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]K
Explanation:
The Heisenberg uncertainty principle is
Δx Δp ≥ h' / 2
h’ =[tex]\frac{h}{2\pi }[/tex]
The kinetic energy of a particle is
K = ½ m v²
p = mv
v = [tex]\frac{p}{m}[/tex]
substitute
K = [tex]\frac{1}{2} \frac{p^2}{m}[/tex]
from the uncertainty principle,
Δp = [tex]\frac{h'}{2 \ \Delta x}[/tex]
we substitute
K = [tex]\frac{1}{2m} ( \frac{h'}{2 \ \Delta x})^2[/tex]
[tex]K = \frac{h'}{8 m \ \Delta x^2}[/tex]
Which part of the water cycle is where vapor from plants leaves the plants as they breath?
condensation
Transpiration
evaporation
Answer:
I think it is transpiration
Answer:
transpiration is the right answer
what happens when a wave passes through a medium ?
Answer:
When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.
Explanation:
Which best explains how fiber-optic technology has improved communication?
It has eliminated the need to send audio data through telephones.
It has allowed for faster transmission of Internet signals.
It has increased the speed at which light travels through space.
It has reduced society’s reliance on devices such as computers and cell phones.
Answer:
B. It has allowed for faster transmission of Internet signals.
Explanation:
i took the test on engenuity
Fiber-optic technology has allowed for faster transmission of Internet signals.
What is meant by fiber-optics?The term fiber optics, often known as optical fiber, describes the technique used to transport data via light pulses travelling along a glass or plastic fiber.
Here,
In fiber-optic communications, optical fibres are widely used to send light over longer distances and at higher bandwidths (data transfer rates) than electrical cables. They are most frequently used to convey light between the two ends of the fiber.
The concept of complete total internal reflection governs the operation of optical fibres. When a light beam strikes the interior surface of an optical fiber cable with an incidence angle greater than the critical angle, the incident light beam reflects in the same medium, and the occurrence is repeated.
Hence,
Fiber-optic technology has allowed for faster transmission of Internet signals.
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1. Three centimeters of water evaporated from a 200-hectare vertical walled reservoir during 24 hours. Storm water was added to the reservoir at a constant rate of 3 m3/s during this period. Determine the volume in ha-cm of water released during the period (through the bottom of the reservoir) if the water level was the same at the beginning and the end of the day.
Answer:
25920 ha-cm
Explanation:
Since water evaporates from the reservoir at a rate of 3 cm in 24 hours, its height changes at a rate of 3 cm/24 × 3600 s = 3 cm/86400s = 3.472 10⁻⁵ cm/s.
Now, the volume loss is dV/dt = dV/dh × -dh/dt
= dV/dt × -3.472 × 10⁻⁵ cm/s
= -3.472 × 10⁻⁵ cm/sdV/dh
The reservoir increases in volume at a rate of 3 m³/s = 3 × 10⁶ cm³/s in 24 hours.
So, the net rate of volume change per unit time of the reservoir is
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt where A = area of vertical walled reservoir and dh/dt = change in height of the reservoir with respect to time
So, 3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt
Since dh/dt = 0 in 24 hours(since the water level remains the same after 24 hours, that is dh = 0)
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = Adh/dt
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = A × 0
3 × 10⁶ cm³/s - 3.472 × 10⁻⁵ cm/sdV/dh = 0
3.472 × 10⁻⁵ cm/sdV/dh = 3 × 10⁶ cm³/s
dV/dh = 3 × 10⁶ cm³/s ÷ 3.472 × 10⁻⁵ cm/s
dV/dh = 8.64 × 10¹¹ cm²
dV = (8.64 × 10¹¹ cm²)dh
Integrating both sides with V from 0 to V and h from h = 0 to h = 3 cm, we have
∫dV = ∫(8.64 × 10¹¹ cm²)dh
∫dV = (8.64 × 10¹¹ cm²)∫dh
V = (8.64 × 10¹¹ cm²)[h]₀³
V = (8.64 × 10¹¹ cm²)[3 cm - 0 cm]
V = (8.64 × 10¹¹ cm²)(3 cm)
V = 25.92 × 10¹¹ cm³
V = 2.592 × 10¹² cm³
V = 2.592 × 10¹² cm² × 1 cm
Since 1 ha = 10⁸ cm²,
V = 2.592 × 10¹² cm² × 1 ha/10⁸ cm² × 1 cm
V = 2.592 × 10⁴ ha-cm
V = 25920 ha-cm