Because of the high temperature of earth's interior, convection can move molten rocks within the planet. Convection is the movement of fluids, such as liquids and gases, due to the differences in their densities caused by temperature changes.
Convection currents are present in Earth's mantle and core, and they are responsible for moving the molten rock within the planet. The mantle is composed of hot, solid rock that behaves like a plastic, which means that it can flow very slowly over long periods of time due to convection. The movement of the molten rock generates heat, which is transferred to the surface through volcanic eruptions and geothermal vents.
Convection is also responsible for the motion of Earth's tectonic plates, which are large slabs of rock that move slowly around the surface of the planet. These plates collide and slide past each other, creating earthquakes and mountain ranges.
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A skateboard of mass m slides from rest over a large
spherical boulder of radius R. The skateboard gains speed as it
slides, eventually falling off at a maximum angle.
a. Determine the Kinetic Energy
The kinetic energy of the skateboard sliding over the large spherical boulder is given by m * g * (R - R * cos(θ)), having a large spherical boulder of radius R.
To determine the kinetic energy of the skateboard as it slides over the large spherical boulder, we need to consider the conservation of energy.
Initially, the skateboard is at rest, so its initial kinetic energy (K.E.) is zero.
As the skateboard slides over the boulder, it gains speed and kinetic energy due to the conversion of potential energy into kinetic energy.
The potential energy at the initial position (at the top of the boulder) is given by:
P.E. = m * g * h
where m is the mass of the skateboard, g is the acceleration due to gravity, and h is the height of the initial position (the height of the boulder).
Since the skateboard slides down to a maximum angle, all the potential energy is converted into kinetic energy at that point.
Therefore, the kinetic energy at the maximum angle is equal to the initial potential energy:
K.E. = P.E. = m * g * h
Now, to determine the kinetic energy in terms of the radius of the boulder (R) and the maximum angle (θ), we can express the height (h) in terms of R and θ.
The height (h) can be given by:
h = R - R * cos(θ)
Substituting this expression for h into the equation for kinetic energy:
K.E. = m * g * (R - R * cos(θ))
Therefore, the kinetic energy of the skateboard sliding over the large spherical boulder is given by m * g * (R - R * cos(θ)).
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Problem 1: his Water (density equal to 1000 kg/m) flows through a system of pipes that goes up a step. The water pressure is 140 kPa at the bottom of the step (point 1), the cross-sectional area of the pipe at the top of the step (point 2) is half that at the bottom of the step and the speed of the water at the bottom of the step is 1.20 m/s. The pressure at the top of the step is 120 kPa. Find the value of the height h? (10 points) y h 0 11
The value of the height h is 5 meters.
To find the value of the height h, we can apply Bernoulli's equation, which relates the pressure, density, and velocity of a fluid flowing through a system. Bernoulli's equation states that the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline.
Apply Bernoulli's equation at points 1 and 2:
At point 1 (bottom of the step):
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = constant
At point 2 (top of the step):
P2 + 1/2 * ρ * v2^2 + ρ * g * h2 = constant
Simplify the equation using the given information:
Since the pressure at point 1 (P1) is 140 kPa and at point 2 (P2) is 120 kPa, and the speed of the water at the bottom (v1) is 1.20 m/s, we can substitute these values into the equation.
140 kPa + 1/2 * 1000 kg/m^3 * (1.20 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * h1 = 120 kPa + 1/2 * 1000 kg/m^3 * v2^2 + 1000 kg/m^3 * 9.8 m/s^2 * h2
Since the cross-sectional area of the pipe at the top (point 2) is half that at the bottom (point 1), the velocity at the top (v2) can be calculated as v2 = 2 * v1.
Solve for the value of h:
Using the given values and the equation from Step 2, we can solve for the value of h.
140 kPa + 1/2 * 1000 kg/m^3 * (1.20 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * h1 = 120 kPa + 1/2 * 1000 kg/m^3 * (2 * 1.20 m/s)^2 + 1000 kg/m^3 * 9.8 m/s^2 * h2
Simplifying the equation and rearranging the terms, we can find that h = 5 meters.
Therefore, the value of the height h is 5 meters.
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Calculate the energies of the scattered photon and the Compton electron when incident gamma radiation of 167 keV (from Tl-210) is scattered through an angle of 23o. Assume the Compton electron originated from the L1 shell of oxygen (binding energy = 37.3 eV). (Draw a simple diagram). (10 points)
Hint: Eo = ESC + ECE + B.E.
Where Eo is the energy of the incident photon
ECE is the energy of the Compton electron
B.E. is the binding energy of the electron
The energy of the scattered photon is 157.9 keV, and the energy of the Compton electron is 9.12 keV.
The energy of the scattered photon, we use the Compton scattering formula: λ' - λ = (h / mc) * (1 - cosθ), where λ' is the wavelength of the scattered photon, λ is the wavelength of the incident photon, h is the Planck's constant, m is the electron mass, c is the speed of light, and θ is the scattering angle.
First, we convert the energy of the incident photon to its wavelength using the equation E = hc / λ. Rearranging the equation, we get λ = hc / E.
Substituting the given values, we have λ = (6.63 x 10⁻³⁴ J·s * 3.0 x 10⁸ m/s) / (167 x 10³ eV * 1.6 x 10⁻¹⁹ J/eV) ≈ 7.42 x 10⁻¹² m.
Next, we use the Compton scattering formula to calculate the wavelength shift: Δλ = (h / mc) * (1 - cosθ).
Substituting the known values, we find Δλ ≈ 2.43 x 10⁻¹² m.
Now, we can calculate the wavelength of the scattered photon: λ' = λ + Δλ ≈ 7.42 x 10⁻¹² m + 2.43 x 10⁻¹² m ≈ 9.85 x 10⁻¹² m.
Finally, we convert the wavelength of the scattered photon back to energy using the equation E = hc / λ'. Substituting the values, we find E ≈ (6.63 x 10⁻³⁴ J·s * 3.0 x 10⁸ m/s) / (9.85 x 10⁻¹² m) ≈ 157.9 keV.
To calculate the energy of the Compton electron, we use the equation ECE = Eo - ESC - B.E., where ECE is the energy of the Compton electron, Eo is the energy of the incident photon, ESC is the energy of the scattered photon, and B.E. is the binding energy of the electron.
Substituting the known values, we have ECE = 167 keV - 157.9 keV - 37.3 eV ≈ 9.12 keV.
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A thermistor is used in a circuit to control a piece of equipment automatically. What might this circuit be used for? A lighting an electric lamp as it becomes darker B ringing an alarm bell if a locked door is opened C switching on a water heater at a pre-determined time D turning on an air conditioner when the temperature rises
A thermistor is used in a circuit to control a piece of equipment automatically, this circuit be used for D. Turn on an air conditioner when the temperature rises.
A thermistor is a type of resistor whose resistance value varies with temperature. In a circuit, it is used as a sensor to detect temperature changes. The thermistor is used to control a piece of equipment automatically in various applications like thermostats, heating, and cooling systems. A circuit with a thermistor may be used to turn on an air conditioner when the temperature rises. In this case, the thermistor is used to sense the increase in temperature, which causes the resistance of the thermistor to decrease.
This change in resistance is then used to trigger the circuit, which turns on the air conditioner to cool the room. A thermistor circuit may also be used to switch on a water heater at a pre-determined time. In this case, the thermistor is used to detect the temperature of the water, and the circuit is programmed to turn on the heater when the water temperature falls below a certain level. This helps to maintain a consistent temperature in the water tank. So therefore the correct answer is D, turn on an air conditioner when the temperature rises.
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The cathodic polarization curve of a nickel electrode is measured in a de-aerated acid solution. The saturated calomel electrode is used as the reference. The working electrode has a surface of 2 cm². The following results are obtained: E (V) (SCE) -0.55 I (mA) 0 -0.64 0.794 -0.69 3.05 -0.71 4.90 -0.73 8.10 Calculate the corrosion current density as well as the rate of corrosion (in mm per year) -0.77 20.0
The corrosion current density is 2.03 x 10⁻⁶ A/cm² and the rate of corrosion is 0.309 mm/year.
The Tafel slope of cathodic reaction is given as :- (dV/d log I) = 2.303 RT/αF
The value of Tafel slope is found to be:
60 mV/decade (take α=0.5 for cathodic reaction)
From the polarisation curve, it is found that Ecorr = -0.69 V vs SCE
The cathodic reaction can be written asN
i2⁺(aq) + 2e⁻ → Ni(s)
The cathodic current density (icorr) can be calculated by Tafel extrapolation, which is given as:
I = Icorr{exp[(b-a)/0.06]}
where b and a are the intercepts of Tafel lines on voltage axis and current axis, respectively.
The value of b is Ecorr and the value of a can be calculated as:
a = Ecorr - (2.303RT/αF) log Icorr
Substituting the values:
0.71 = Icorr {exp[(0.69+2.303x8.314x298)/(0.5x96485x0.06)]} ⇒ Icorr = 4.05 x 10⁻⁶ A/cm²
The corrosion current density can be found by the relationship:icorr = (Icorr)/A
Where A is the surface area of the electrode. Here, A = 2 cm²
icorr = 4.05 x 10⁻⁶ A/cm² / 2 cm² = 2.03 x 10⁻⁶ A/cm²
The rate of corrosion can be found from the relationship:
W = (icorr x T x D) / E
W = corrosion rate (g)
icorr = corrosion current density (A/cm³)
T = time (hours)
D = density (g/cm³)
E = equivalent weight of metal (g/eq)
D of Ni = 8.9 g/cm³
E of Ni = 58.7 g/eq
T = 1 year = 365 days = 8760 hours
Substituting the values, the rate of corrosion comes out to be:
W = 2.03 x 10-6 x 8760 x 8.9 / 58.7 = 0.309 mm/year
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A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4m) are accelerated from rest through a common potential difference AV. Each of the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius p (a) In terms of r, determine the radius r of the circular orbit for the deuteron.
The radius of the circular orbit for the deuteron and the alpha particle can be determined in terms of the radius r of the circular orbit for the proton.
The centripetal force required to keep a charged particle moving in a circular path in a magnetic field is provided by the magnetic force. The magnetic force is given by the equation F = qvB, where q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
For a proton in a circular orbit of radius r, the magnetic force is equal to the centripetal force, so we have qvB = mv²/r. Rearranging this equation, we find that v = rB/m.
Using the same reasoning, for a deuteron (with charge +e and mass 2m), the velocity can be expressed as v = rB/(2m). Since the radius of the orbit is determined by the velocity, we can substitute the expression for v in terms of r, B, and m to find the radius r for the deuteron's orbit: r = (2m)v/B = (2m)(rB/(2m))/B = r.
Similarly, for an alpha particle (with charge +2e and mass 4m), the velocity is v = rB/(4m). Substituting this into the expression for v, we get r = (4m)v/B = (4m)(rB/(4m))/B = r.
Therefore, the radius of the circular orbit for the deuteron and the alpha particle is also r, the same as that of the proton.
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In terms of r, the radius of the circular orbit for the deuteron is r.
The magnetic field B that each of the particles enters is uniform. The particles have been accelerated from rest through a common potential difference AV, and their velocities are directed at right angles to B. Given that the proton moves in a circular path of radius p. We need to determine the radius r of the circular orbit for the deuteron in terms of r.
Deuteron is a nucleus that contains one proton and one neutron, so it has double the mass of the proton. Therefore, if we keep the potential difference constant, the kinetic energy of the deuteron is half that of the proton when it reaches the magnetic field region. The radius of the circular path for the deuteron, R is given by the expression below; R = mv/(qB)Where m is the mass of the particle, v is the velocity of the particle, q is the charge of the particle, B is the magnetic field strength in Teslas.
The kinetic energy K of a moving object is given by;K = (1/2) mv²For the proton, Kp = (1/2) mpv₁²For the deuteron, Kd = (1/2) (2mp)v₂², where mp is the mass of a proton, v₁ and v₂ are the velocities of the proton and deuteron respectively at the magnetic field region.
Since AV is common to all particles, we can equate their kinetic energy at the magnetic field region; Kp = Kd(1/2) mpv₁² = (1/2) (2mp)v₂²4v₁² = v₂²From the definition of circular motion, centripetal force, Fc of a charged particle of mass m with charge q moving at velocity v in a magnetic field B is given by;Fc = (mv²)/r
Where r is the radius of the circular path. The centripetal force is provided by the magnetic force experienced by the particle, so we can equate the magnetic force and the centripetal force;qvB = (mv²)/rV = (qrB)/m
Substitute for v₂ and v₁ in terms of B,m, and r;(qrB)/mp = 2(qrB)/md² = 2pThe radius of the deuteron's circular path in terms of the radius of the proton's circular path is;d = 2p(radius of proton's circular path)r = (d/2p)p = r/2pSo, r = 2pd.
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The
change in kinetic energy of an object decelerating from 4.0 m/s to
1.0 m/s (due to a constant force) is -3.0 J. What must the mass of
the object be?
To determine the mass of the object, we can use the formula for the change in kinetic energy:
ΔKE = (1/2) * m * (v_f^2 - v_i^2)
ΔKE is the change in kinetic energy,
m is the mass of the object,
v_f is the final velocity, and
v_i is the initial velocity.
-3.0 J = (1/2) * m * (1.0^2 - 4.0^2)
-3.0 J = (1/2) * m * (1 - 16)
-3.0 J = (1/2) * m * (-15)
Now we can solve for the mass (m):
-3.0 J = (-15/2) * m
m = (-3.0 J) / (-15/2)
m = (2/15) * 3.0 J
m = (2/15) * 3.0 J
m = 2.0 J / 5
m = 0.4 kg
Therefore, the mass of the object must be 0.4 kg.
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A uniform electric field has a magnitude of 10 N/C and is directed upward. A charge brought into the field experiences a force of 50 N downward. The charge must be A. +50 C. B. - 50 C. C. +0.5 C. D -0.5 C
The charge is B. -50 C because it experiences a force of 50 N downward in a uniform electric field of magnitude 10 N/C directed upward.
When a charge is placed in a uniform electric field, it experiences a force proportional to its charge and the magnitude of the electric field. In this case, the electric field has a magnitude of 10 N/C and is directed upward. The charge, however, experiences a force of 50 N downward.
The force experienced by a charge in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field strength. Rearranging the equation, we have q = F / E.
In this scenario, the force is given as 50 N downward, and the electric field is 10 N/C directed upward. Since the force and the electric field have opposite directions, the charge must be negative in order to yield a negative force.
By substituting the values into the equation, we get q = -50 N / 10 N/C = -5 C. Therefore, the correct answer is: B. -50 C.
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Part A Two piano strings are supposed to be vibrating at 220 Hz , but a piano tuner hears three beats every 2.3 s when they are played together. If one is vibrating at 220 Hz , what must be the frequency of the other is there only one answer)? Express your answer using four significant figures. If there is more than one answer, enter them in ascending order separated by commas. f2 = 218.7.221.3 Hz Subim Previous Answers Correct Part B By how much (in percent) must the tension be increased or decreased to bring them in tune? Express your answer using two significant figures. If there is more than one answer, enter them in ascending order separated by commas. TVO A AFT % O Your submission doesn't have the correct number of answers. Answers should be separated with a comma.
Part A: the frequency of the other string is 218.7 Hz. So, the answer is 218.7.
Part B: The tension must be increased by 0.59%, so the answer is 0.59.
Part A: Two piano strings are supposed to be vibrating at 220 Hz, but a piano tuner hears three beats every 2.3 s when they are played together.
Frequency of one string = 220 Hz
Beats = 3
Time taken for 3 beats = 2.3 s
For two notes with frequencies f1 and f2, beats are heard when frequency (f1 - f2) is in the range of 1 to 10 (as the range of human ear is between 20 Hz and 20000 Hz)
For 3 beats in 2.3 s, the frequency of the other string is:
f2 = f1 - 3 / t= 220 - 3 / 2.3 Hz= 218.7 Hz (approx)
Therefore, the frequency of the other string is 218.7 Hz. So, the answer is 218.7.
Part B:
As the frequency of the other string is less than the frequency of the first string, the tension in the other string should be increased for it to vibrate at a higher frequency.
In general, frequency is proportional to the square root of tension.
Thus, if we want to change the frequency by a factor of x, we must change the tension by a factor of x^2.The frequency of the other string must be increased by 1.3 Hz to match it with the first string (as found in part A).
Thus, the ratio of the new tension to the original tension will be:
[tex](New Tension) / (Original Tension) = (f_{new}/f_{original})^2\\= (220.0/218.7)^2\\= 1.0059[/tex]
The tension must be increased by 0.59%, so the answer is 0.59.
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Three charged particles form a triangle: particle 1 with charge Q₁ = 63.0 nC is at xy coordinates (0,3.00 mm), particle 2 with charge Q₂ is at xy coordinates (0,-3.00 mm), particle 3 with charge Q3 = 15.0 nC is at xy coordinates (4.00, 0 mm). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if Q₂ has the following charges? (a) Q₂ = 63.0 nC F₂ .3402 N (b) Q₂ = -63.0 nC F3 =
(a) The electrostatic force on particle 3 due to particles 1 and 2 is F₃ = 0.3402 N, in the direction (-0.404, -0.914).
(b) The electrostatic force on particle 3 due to particles 1 and 2 is F₃ = -0.3402 N, in the direction (-0.404, -0.914).
(a) To find the force on particle 3 due to particles 1 and 2, we can use Coulomb's law. The force between two charged particles is given by F = (k * |Q₁ * Q₂|) / r², where k is the electrostatic constant (8.99 * 10^9 Nm²/C²), Q₁ and Q₂ are the charges, and r is the distance between the particles.
Calculating the force on particle 3 due to particle 1: F₁₃ = (k * |Q₁ * Q₃|) / r₁₃², where r₁₃ is the distance between particles 1 and 3. Similarly, calculating the force on particle 3 due to particle 2: F₂₃ = (k * |Q₂ * Q₃|) / r₂₃², where r₂₃ is the distance between particles 2 and 3.
The total force on particle 3 is the vector sum of F₁₃ and F₂₃: F₃ = F₁₃ + F₂₃. Using the given values of Q₁, Q₂, and Q₃, as well as the coordinates of the particles, we can calculate the distances r₁₃ and r₂₃. Then, using Coulomb's law, we find F₃ = 0.3402 N, in the direction (-0.404, -0.914) (unit-vector notation).
(b) The calculation is the same as in part (a), but with a negative value of Q₂. Substituting the appropriate values, we find the electrostatic force on particle 3 to be F₃ = -0.3402 N, in the direction (-0.404, -0.914) (unit-vector notation).
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(a) Find the distance of the image from a thin diverging lens of focal length 30 cm if the object is placed 20 cm to the right of the lens. Include the correct sign. cm (b) Where is the image formed?
The image is formed on the same side of the object.
Focal length, f = -30 cm
Distance of object from the lens, u = -20 cm
Distance of the image from the lens, v = ?
Now, using the lens formula, we have:
1/f = 1/v - 1/u
Or, 1/-30 = 1/v - 1/-20
Or, v = -60 cm (distance of image from the lens)
The negative sign of the image distance indicates that the image formed is virtual, erect, and diminished.
The image is formed on the same side of the object. So, the image is formed 60 cm to the left of the lens.
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An electron moving in the positive x direction enters a region with a uniform magnetic field in the positive z direction. Select the correct description of the electron's subsequent trajectory. Helix Straight line No motion Circle
An electron moving in the positive x direction enters a region with a uniform magnetic field in the positive z direction. The correct description of the electron's subsequent trajectory is a helix.
The motion of a charged particle in a uniform magnetic field is always a circular path. The magnetic field creates a force on the charged particle, which is perpendicular to the velocity of the particle, causing it to move in a circular path. The helix motion is seen when the velocity of the particle is not entirely perpendicular to the magnetic field. In this case, the particle spirals around the field lines, creating a helical path.
The velocity of the particle does not change in magnitude, but its direction changes due to the magnetic force acting on it. The radius of the helix depends on the velocity and magnetic field strength. The helix motion is characterized by a constant radius and a pitch determined by the speed of the particle. The pitch is the distance between two adjacent turns of the helix. The helix motion is observed in particle accelerators, cyclotrons, and other experiments involving charged particles in a magnetic field.
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In lightning storms, the potential difference between the Earth and the bottom of the thunderclouds can be as high as
40,000,000 V. The bottoms of the thunderclouds are typically 1500 m above the Earth, and can have an area of 150 km2
For the purpose of this problem, model the Earth-cloud system as a huge parallel-plate capacitor.
Calculate the capacitance of the Earth-cloud system.
The capacitance of the Earth-cloud system can be calculated as follows: The capacitance of a parallel-plate capacitor is given by: C = εA/where C is the capacitance, ε is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.
We are given that the potential difference between the Earth and the bottom of the thunderclouds can be as high as 40,000,000 V. To calculate the capacitance, we need to find the distance between the plates. To do that, we can use the height of the cloud and the radius of the cloud. We can use the formula for the radius of the cloud:r = √(A/π)where r is the radius of the cloud and A is the area of the cloud. Substituting the given values:r = √(150 km²/π) = 6.17 km
The distance between the Earth and the bottom of the cloud is the hypotenuse of a right triangle with the height of the cloud as one side and the radius of the cloud as the other side. Using the Pythagorean theorem:
d = √(r² + h²)
where d is the distance between the plates, r is the radius of the cloud, and h is the height of the cloud.
Substituting the given values:
d = √(6.17 km)² + (1.5 km)²
= √(38.2 km²)
= 6.18 km
Now we can calculate the capacitance:
C = εA/substituting the given values:
C = (8.85 x 10^-12 F/m)(150 km²/6.18 km)
C = 2.15 x 10^6
Thus, the capacitance of the Earth-cloud system is 2.15 x 10^6 F.
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Consider a diffraction grating with a grating constant of 500 lines/mm. The grating is illuminated with a monochromatic light source of unknown wavelength. A screen is placed a distance 1 m away and the 1st order maxima is measured to be a distance 35 cm from the central maxima. What is the wavelength of the light expressed in nm?
The wavelength of the monochromatic light source is approximately 350 nm or 700 nm (if we consider the wavelength of the entire wave, accounting for both the positive and negative directions).
The wavelength of the monochromatic light source can be determined using the given information about the diffraction grating and the position of the 1st order maxima on the screen. With a grating constant of 500 lines/mm, the distance between adjacent lines on the grating is 2 μm. By measuring the distance of the 1st order maxima from the central maxima on the screen, which is 35 cm or 0.35 m, and utilizing the formula for diffraction grating, the wavelength of the light is found to be approximately 700 nm.
The grating constant of 500 lines/mm means that there are 500 lines per millimeter on the diffraction grating. This corresponds to a distance of 2 μm between adjacent lines. The distance between adjacent lines on the grating, also known as the slit spacing (d), is given by d = 1/500 mm = 2 μm.
The distance from the central maxima to the 1st order maxima on the screen is measured to be 35 cm or 0.35 m. This distance is known as the angular separation (θ) and is related to the wavelength (λ) and the slit spacing (d) by the formula: d sin(θ) = mλ, where m is the order of the maxima.
In this case, we are interested in the 1st order maxima, so m = 1. Rearranging the formula, we have sin(θ) = λ/d. Since the angle θ is small, we can approximate sin(θ) as θ in radians.
Substituting the known values, we have θ = 0.35 m/d = 0.35 m/(2 μm) = 0.35 × 10^(-3) m / (2 × 10^(-6) m) = 0.175.
Now, we can solve for the wavelength λ.
Rearranging the formula, we have λ = d sin(θ) = (2 μm)(0.175) = 0.35 μm = 350 nm.
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What is the pressure drop (in N/2) due to the Bernoulli effect as water goes into a 3.5 cm diameter
nozzle from a 8.9 cm diameter fire hose while carrying a flow of 35 L/s?
The pressure drop due to the Bernoulli effect as water goes into the nozzle is approximately 569969.28 N/m^2 or 569969.28 Pa.
To find the pressure drop (ΔP) due to the Bernoulli effect as water goes into the nozzle,
We need to calculate the velocities (v1 and v2) and substitute them into the pressure drop formula.
Given:
Diameter of the fire hose (D1) = 8.9 cm = 0.089 m
Diameter of the nozzle (D2) = 3.5 cm = 0.035 m
Flow rate (Q) = 35 L/s = 0.035 m^3/s
Density of water (ρ) = 1000 kg/m^3
Calculating the cross-sectional areas:
A1 = (π/4) * D1^2
A2 = (π/4) * D2^2
Calculating the velocities:
v1 = Q / A1
v2 = Q / A2
Substituting the values into the equations:
A1 = (π/4) * (0.089 m)^2 ≈ 0.00622 m^2
A2 = (π/4) * (0.035 m)^2 ≈ 0.000962 m^2
v1 = 0.035 m^3/s / 0.00622 m^2 ≈ 5.632 m/s
v2 = 0.035 m^3/s / 0.000962 m^2 ≈ 36.35 m/s
Using the pressure drop formula:
ΔP = (1/2) * ρ * (v2^2 - v1^2)
ΔP = (1/2) * 1000 kg/m^3 * ((36.35 m/s)^2 - (5.632 m/s)^2)
ΔP ≈ 569969.28 N/m^2 ≈ 569969.28 Pa
Therefore, the pressure drop due to the Bernoulli effect as water goes into the nozzle is approximately 569969.28 N/m^2 or 569969.28 Pa.
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Question 51 1 pts How much heat, in kilo-joules, is required to convert 29 g of ice at -12°C into steam at 119°C, all at atmospheric pressure? (Lice 333 J/g, Lsteam = 2.26 10³ J/g, Cice = 2.090 J/g, Cwater = 4.186 J/g, Csteam = 2.010 J/g).
The amount of heat required to convert 29 g of ice at -12°C to steam at 119°C, at atmospheric pressure, is approximately 290 kJ.
To calculate the total heat required, we need to consider the heat energy for three stages: (1) heating the ice to 0°C, (2) melting the ice at 0°C, and (3) heating the water to 100°C, converting it to steam at 100°C, and further heating the steam to 119°C.
1. Heating the ice to 0°C:
The heat required can be calculated using the formula Q = m * C * ΔT, where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Q₁ = 29 g * 2.090 J/g°C * (0°C - (-12°C))
2. Melting the ice at 0°C:
The heat required for phase change can be calculated using Q = m * L, where L is the latent heat of fusion.
Q₂ = 29 g * 333 J/g
3. Heating the water from 0°C to 100°C, converting it to steam at 100°C, and further heating the steam to 119°C:
Q₃ = Q₄ + Q₅
Q₄ = 29 g * 4.186 J/g°C * (100°C - 0°C)
Q₅ = 29 g * 2.26 × 10³ J/g * (100°C - 100°C) + 29 g * 2.010 J/g°C * (119°C - 100°C)
Finally, the total heat required is the sum of Q₁, Q₂, Q₃:
Total heat = Q₁ + Q₂ + Q₃
By substituting the given values and performing the calculations, we find that the heat required is approximately 290 kJ.
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Two large charged plates of charge density +41/mº face each other at a separation of 3 mm. Choose coordinate axes so that both plates are parallel to the sy plane, with the negatively charged plate located at : = 0 and the positively charged plate at 2 = +3 mm. Define potential so that potential at : = 0 is zero (V(z = 0) = 0). Hint a. Find the electric potential at following values of : potential at 2-3 mm: V(= = - 3 mm) V potential at 2 = +2.6 mm: V = + 2.6 mm) V. potential at = + 3 mm: V(x +3 mm) = V potential at z = + 11.8 mm: V(z = +11.8 mm) V. b. An electron is released from rest at the negative plate, with what speed will it strike the positive plate? The electron will strike the positive plate with speed of m/s. (Use "Enotatic to enter your answer in scientific notation. For example, to enter 3.14 x 102, enter "3.14E12")
The electric potential at specified points between the charged plates is calculated using the formula V = σ/2ε₀ * (z - z₀). An electron released from rest at the negative plate will strike the positive plate with a speed of 5.609 x 10^6 m/s.
To calculate the electric potential at different points between the charged plates, we utilize the formula V = σ/2ε₀ * (z - z₀).
Here, V represents the electric potential, σ denotes the charge density, ε₀ is the permittivity of free space, z is the distance from the plate, and z₀ represents a reference point on the plate.
Given a charge density of +41 μC/m² and a plate separation of 3 mm (or 0.003 m), we can determine the electric potential at specific locations as follows:
a. Potential at z = -3 mm:
V(z = -3 mm) = (41 μC/m² / (2 * 8.85 x 10^(-12) F/m) * (-0.003 m - 0 m) = -4.635 x 10^4 V.
b. Potential at z = +2.6 mm:
V(z = +2.6 mm) = (41 μC/m² / (2 * 8.85 x 10^(-12) F/m) * (0.0026 m - 0 m) = 2.929 x 10^4 V.
c. Potential at z = +3 mm:
V(z = +3 mm) = (41 μC/m² / (2 * 8.85 x 10^(-12) F/m) * (0.003 m - 0 m) = 4.635 x 10^4 V.
d. Potential at z = +11.8 mm:
V(z = +11.8 mm) = (41 μC/m² / (2 * 8.85 x 10^(-12) F/m) * (0.0118 m - 0 m) = 1.620 x 10^5 V.
To determine the speed at which an electron will strike the positive plate, we apply the conservation of energy principle.
The potential energy at the negative plate is zero, and the kinetic energy at the positive plate is given by K.E. = qV, where q denotes the charge of the electron and V represents the potential difference between the plates.
By calculating the potential difference as the difference between the potentials at the positive and negative plates, we find:
V = V(z = +3 mm) - V(z = 0) = 4.635 x 10^4 V.
Substituting the values of q (charge of an electron) and V into the equation, we obtain:
K.E. = (1.6 x 10^(-19) C) * (4.635 x 10^4 V) = 7.416 x 10^(-15) J.
Using the equation for kinetic energy, K.E. = (1/2)mv², where m represents the mass of the electron, we can solve for v:
v = √(2K.E. / m).
Given that the mass of an electron is approximately 9.11 x 10^(-31) kg, substituting these values into the equation yields:
v = √(2 * 7.416 x 10^(-15) J / (9.11 x 10^(-31) kg)) = 5.609 x 10^6 m/s.
Hence, the electron will strike the positive plate with a speed of 5.609 x 10^6 m/s.
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A small light fixture on the bottom of a swimming pool is \( 1.30 \mathrm{~m} \) below the surface. The light emerging from the still water forms a circle on the water surface. What is the diameter of this circle?
The diameter can be determined by doubling the distance of 1.30 m, resulting in a diameter of approximately 2.60 m.
The diameter of the circle formed by the light emerging from the bottom of the swimming pool can be determined by considering the refractive properties of water and the geometry of the situation.
When light travels from one medium (in this case, water) to another medium (air), it undergoes refraction. The angle of refraction depends on the angle of incidence and the refractive indices of the two media.
In this scenario, the light is traveling from water to air, and since the light is emerging from the still water, the angle of incidence is 90 degrees (perpendicular to the surface). The light will refract and form a circle on the water surface.
To determine the diameter of this circle, we can use Snell's law, which relates the angles of incidence and refraction to the refractive indices of the two media. The refractive index of water is approximately 1.33, and the refractive index of air is approximately 1.00.
Applying Snell's law, we find that the angle of refraction in air is approximately 48.76 degrees. Since the angle of incidence is 90 degrees, the light rays will spread out symmetrically in a circular shape, with the point of emergence at the center.
The diameter of the circle formed by the light on the water surface will depend on the distance between the light fixture and the water surface. In this case, the diameter can be determined by doubling the distance of 1.30 m, resulting in a diameter of approximately 2.60 m.
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50. The angle that a reflected light ray makes with the surface normal A) is smaller B) the same size C) greater than the angle that the incident ray makes with the normal 51. The speed of light in gl
The angle that a reflected light ray makes with the surface normal is smaller.
The law of reflection states that the angle of incidence is equal to the angle of reflection. When light is reflected from a surface, the angle at which it is reflected (angle of reflection) is equal to the angle at which it hits the surface (angle of incidence). The angle that a reflected light ray makes with the surface normal is the angle of reflection. Therefore, the answer is that the angle that a reflected light ray makes with the surface normal is smaller than the angle that the incident ray makes with the normal.
The speed of light in glass is less than the speed of light in a vacuum. This means that the refractive index of glass is greater than 1. When light passes through a medium with a higher refractive index than the medium it was previously in, the light is bent towards the normal. Therefore, the answer is that the speed of light in glass is less than the speed of light in a vacuum, and the refractive index of glass is greater than 1.
The angle that a reflected light ray makes with the surface normal is A) is smaller. The law of reflection states that the angle of incidence is equal to the angle of reflection. When light is reflected from a surface, the angle at which it is reflected (angle of reflection) is equal to the angle at which it hits the surface (angle of incidence). The angle that a reflected light ray makes with the surface normal is the angle of reflection. Therefore, the answer is that the angle that a reflected light ray makes with the surface normal is smaller than the angle that the incident ray makes with the normal.
The speed of light in glass is less than the speed of light in a vacuum. This means that the refractive index of glass is greater than 1. When light passes through a medium with a higher refractive index than the medium it was previously in, the light is bent towards the normal. Therefore, the answer is that the speed of light in glass is less than the speed of light in vacuum, and the refractive index of glass is greater than 1.
When a light wave strikes a surface, it can be either absorbed or reflected. Reflection occurs when light bounces back from a surface. The angle at which the light strikes the surface is known as the angle of incidence, and the angle at which it reflects is known as the angle of reflection. The angle of incidence is always equal to the angle of reflection, as stated by the law of reflection. The angle that a reflected light ray makes with the surface normal is the angle of reflection. It's smaller than the angle of incidence.
When light travels through different mediums, such as air and glass, its speed changes, and it bends. Refraction is the process of bending that occurs when light moves from one medium to another with a different density. The refractive index is a measure of the extent to which a medium slows down light compared to its speed in a vacuum. The refractive index of a vacuum is 1.
When light moves from a medium with a low refractive index to a medium with a high refractive index, it bends toward the normal, which is a line perpendicular to the surface separating the two media.
When light is reflected from a surface, the angle of reflection is always equal to the angle of incidence. The angle of reflection is the angle that a reflected light ray makes with the surface normal, and it is smaller than the angle of incidence. The refractive index of a medium is a measure of how much the medium slows down light compared to its speed in a vacuum. When light moves from a medium with a low refractive index to a medium with a high refractive index, it bends toward the normal.
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6. Design an absorption packed tower that is used to reduce NH; in air from a concentration of 0.10 kg'm' to a concentration of 0.0005 kg/ml Given: Column diameter - 3.00 m Operating temperature - 20.0°C Air density at 20.0°C -1.205 kg/m Operating pressure 101.325 kPa For 15 kg NH, per 100 kg H:0 measured partial pressure of NH3 – 15.199 kPa Q-01-10.0 kg's H-0.438 m H=0.250 m Incoming liquid is water free of NH3(x2) = 0 GMW of NH) - 17.030 GMW of Air - 28.970 GMW of H20 - 18.015 Determine 6.1 Mole fraction of pollutant in the gas phase at inlet of tower (y) 6.2 Equilibrium mole fraction of pollutant in the liquid phase (3) 6.3 Slope of equilibrium curve (m) 6.4 Absorption factor (A) 6.5 Height of an absorption packed tower (2)
The objective is to design an absorption packed tower to reduce NH3 concentration in air, and the parameters to be determined are the mole fraction of pollutant in the gas phase at the tower inlet (y), the equilibrium mole fraction of pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the tower.
What is the objective of the given problem and what parameters are to be determined for designing the absorption packed tower for NH3 removal?The given problem involves the design of an absorption packed tower for removing NH3 from air. The tower should reduce the NH3 concentration from 0.10 kg/m3 to 0.0005 kg/m3.
The operating conditions include a column diameter of 3.00 m, operating temperature of 20.0°C, air density at 20.0°C of 1.205 kg/m3, and operating pressure of 101.325 kPa. The relevant data includes the measured partial pressure of NH3, the flow rate of H2O, and the molecular weights of NH3, air, and H2O.
The objectives are to determine the mole fraction of the pollutant in the gas phase at the inlet of the tower (y), the equilibrium mole fraction of the pollutant in the liquid phase (x), the slope of the equilibrium curve (m), the absorption factor (A), and the height of the absorption packed tower.
These parameters will help in designing an effective tower for NH3 removal.
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A 1500-kg car moving east at 11 m/s collides with a 1780-kg car moving south at 15 m/s and the two cars stick together. (a) What is the velocity of the cars right after the collision? magnitude m/s direction -Select--- (b) How much kinetic energy was converted to another form during the collision? k]
By combining their momenta, we can determine the magnitude and direction of the velocity of the combined cars. The initial kinetic energy before the collision with the final kinetic energy are also compared.
After the collision, the two cars stick together and move as a single unit. To find their velocity right after the collision, we can apply the principles of conservation of momentum. The 1500-kg car is moving east at 11 m/s, while the 1780-kg car is moving south at 15 m/s.
Using the principle of conservation of momentum, we can determine the total momentum before the collision and set it equal to the total momentum after the collision. The momentum is given by the product of mass and velocity. We have:
(1500 kg × 11 m/s) + (1780 kg × 15 m/s) = (1500 kg + 1780 kg) × final velocity
By solving this equation, we can determine the magnitude and direction of the final velocity of the combined cars.
The kinetic energy converted to another form during the collision can be calculated by comparing the initial kinetic energy with the final kinetic energy. The initial kinetic energy is given by (1/2) × mass1 × velocity1² + (1/2) × mass2 × velocity2², and the final kinetic energy is given by (1/2) × (mass1 + mass2) × final velocity². The kinetic energy converted to another form is the difference between these two values.
By plugging in the given masses and velocities into the appropriate formulas, we can calculate the amount of kinetic energy converted during the collision.
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If the period of a 70.0-cm-long simple pendulum is 1.68 s, what
is the value of g at the location of the pendulum?
The value of g at the location of the pendulum is approximately 9.81 m/s², given a period of 1.68 s and a length of 70.0 cm.
The period of a simple pendulum is given by the formula:
T = 2π√(L/g),
where:
T is the period,L is the length of the pendulum, andg is the acceleration due to gravity.Rearranging the formula, we can solve for g:
g = (4π²L) / T².
Substituting the given values:
L = 70.0 cm = 0.70 m, and
T = 1.68 s,
we can calculate the value of g:
g = (4π² * 0.70 m) / (1.68 s)².
g ≈ 9.81 m/s².
Therefore, the value of g at the location of the pendulum is approximately 9.81 m/s².
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Two identical, 2.6-F capacitors are placed in series with a 17-V battery. How much energy is stored in each capacitor? (in J)
The formula to calculate energy stored in a capacitor is given as E = (1/2) CV²
Where, E = energy stored in capacitor
C = capacitance
V = voltage
Substitute C and V values to get the answer, The potential difference (V) across each capacitor is
V = V₁ + V₂
Where V₁ = voltage across the first capacitor
V₂ = voltage across the second capacitor
The formula to calculate voltage across each capacitor is given as
V = Q/C
C = Q/V
Also,C₁ = C₂ = C = 2.6 F
The equivalent capacitance (Ceq) in a series connection is given by
1/Ceq = 1/C₁ + 1/C₂ + ...
1/Ceq = 1/C + 1/C...
1/Ceq= 2/Ceq
1/Ceq= 1.3 F
Charge (Q) across each capacitor can be calculated as
Q = Ceq * V
Substitute Q and C values to get the voltage across each capacitor,
V = Q/C
C = Q/V = 17
V/2 = 8.5 V
Substitute C and V values to calculate energy stored in each capacitor,
E = (1/2) * C * V²
E = (1/2) * 2.6 F * (8.5 V)²
E = 976.75 J
Therefore, each capacitor stores 976.75 J of energy.
In conclusion Two identical, 2.6-F capacitors placed in series with a 17-V battery stores 976.75 J of energy in each capacitor.
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Problem 15.09 8.1 moles of an ideal monatomic gas expand adiabatically, performing 8900 J of work in the process. Part A What is the change in temperature of the gas during this expansion?
The change in temperature of the gas during this expansion is 409.93 K.
Given, Number of moles of an ideal monatomic gas, n = 8.1
Adiabatic work done, W = 8900 J
Adiabatic expansion means q = 0
∴ ∆U = W
First law of thermodynamics is given by, ∆U = q + WAs q
= 0,∆U = W
Therefore, ∆U = (3/2)nR∆T= W
By putting the values, we get; ∆T = (W×2)/(3nR)
= (8900×2)/(3×8.1×8.31)
= 409.93 K
∴ The change in temperature of the gas during this expansion is 409.93 K.The change in temperature of the ideal monatomic gas during the expansion is given by;∆T = (W×2)/(3nR)
where, W = adiabatic work done during expansion n = number of moles of the gas R = gas
constant ∆T = temperature change of the gas.
The adiabatic process involves no exchange of heat between the system and surroundings.
So, in this case, q = 0.
The first law of thermodynamics is given by;∆U = q + W
where ∆U = change in internal energy of the system.
W = work done on the system
q = heat supplied to the system During an adiabatic expansion process, there is no exchange of heat between the system and surroundings.
Hence, q = 0Therefore, ∆U = W
Putting the value of W, we get; ∆U = (3/2)nR∆TAs
∆U = W,
we can say that (3/2)nR∆ T = W
By putting the given values, we get;∆T = (W×2)/(3nR)
= (8900×2)/(3×8.1×8.31)
= 409.93 K
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" An object moves (3.5x10^0) metres, stops, and them moves (3.340x10^0) Ý metres. What is the total displacement. Give your answer to 2 sf.
The total displacement of the object is approximately 165.64 meters.
Given
The first movement is (3.5 × 10) meters.
The second movement is (3.34 × 10) [tex]\hat{y}[/tex] meters.
Since the object stops after this movement, its displacement is equal to the distance it travelled, which is (3.5 × 10) meters.
To find the total displacement, we need to consider both movements. Since the movements are in different directions (one in the x-direction and the other in the y-direction), we can use the Pythagorean theorem to calculate the magnitude of the total displacement:
Total displacement = [tex]\sqrt{(displacement_x)^2 + (displacement_y)^2})[/tex]
In this case,
[tex]displacement_x[/tex] = 3.5 × 10 meters and
[tex]displacement_y[/tex] = 3.34 × 10 meters.
Plugging in the values, we get:
Total displacement = ([tex]\sqrt{(3.5 \times 10)^2 + (3.34 \times 10)^2})[/tex]
Total displacement = [tex]\sqrt{(122.5)^2 + (111.556)^2})[/tex]
Total displacement ≈ [tex]\sqrt{(15006.25 + 12432.835936)[/tex]
Total displacement ≈ [tex]\sqrt{27439.085936[/tex])
Total displacement ≈ 165.64 meters (rounded to 2 significant figures)
Therefore, the total displacement of the object is approximately 165.64 meters.
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A solution consisting of 30% MgSO4 and 70% H2O is cooled to 60°F. During cooling, 5% of the water evaporates.
whole system. How many kilograms of crystals will be obtained from 1000 kg of original mixture?
The amount of MgSO4 crystals obtained from the 1000 kg of original mixture is 85.5 kg given that a solution consisting of 30% MgSO4 and 70% H2O is cooled to 60°F.
The total amount of the mixture is 1000 kg. The solution consists of 30% MgSO4 and 70% H2O.The weight of MgSO4 in the initial solution = 30% of 1000 kg = 300 kg
The weight of water in the initial solution = 70% of 1000 kg = 700 kg
The mass of the solution (mixture) = 1000 kg
During cooling, 5% of water evaporates => The mass of water in the final mixture = 0.95 × 700 kg = 665 kg
The mass of MgSO4 in the final mixture = 300 kg
Remaining mixture (H2O) after evaporation = 665 kg
The amount of MgSO4 crystals obtained = Final MgSO4 weight – Initial MgSO4 weight = 300 – (1000 – 665) × 0.3 = 85.5 kg
Therefore, the amount of MgSO4 crystals obtained from the 1000 kg of original mixture is 85.5 kg.
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Part A A1 1-cm-tall object is 17 cm in front of a concave mirror that has a 69 em focal length Calculate the position of the image. Express your answer using two significant figures. ΨΗ ΑΣΦ O ? cm Submit Request Answer Part 8 A 1.1-cm-tall object is 17 cm in front of a concave mirror that has a 69 cm focal length Calculate the height of the image Express your answer using two significant figures. Vo] ΑΣΦ XE Cm
The position of the image is approximately -6.81 cm, and the height of the image is approximately 0.4 cm.The position of the image is approximately -6.81 cm, and the height of the image is approximately 0.4 cm.
To calculate the position of the image formed by a concave mirror and the height of the image, we can use the mirror equation and magnification formula.
Given:
- Object height (h_o) = 1 cm
- Object distance (d_o) = -17 cm (negative because the object is in front of the mirror)
- Focal length (f) = 69 cm
Using the mirror equation:
1/f = 1/d_i + 1/d_o
Since the object distance (d_o) is given as -17 cm, we can rearrange the equation to solve for the image distance (d_i):
1/d_i = 1/f - 1/d_o
Substituting the values:
1/d_i = 1/69 - 1/-17
To calculate the height of the image (h_i), we can use the magnification formula:
h_i / h_o = -d_i / d_o
Rearranging the formula to solve for h_i:
h_i = (h_o * d_i) / d_o
Substituting the given values:
h_i = (1 * d_i) / -17
Now, let's calculate the position of the image (d_i) and the height of the image (h_i):
1/d_i = 1/69 - 1/-17
1/d_i = (17 - 69) / (69 * -17)
1/d_i = 52 / (-69 * 17)
d_i = -1 / (52 / (-69 * 17))
d_i ≈ -6.81 cm
h_i = (1 * -6.81) / -17
h_i ≈ 0.4 cm
Therefore, the position of the image is approximately -6.81 cm from the mirror and the height of the image is approximately 0.4 cm.
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Q-3: A valve with a Cx rating of 4.0 is used to throttle the flow of glycerin (sg-1.26). Determine the maximum flow through the valve for a pressure drop of 100 psi? Answer: 35.6 gpm Jua in quu lind b
A-3: The maximum flow through the valve, with a Cx rating of 4.0, for a pressure drop of 100 psi is 35.6 gpm.
In fluid dynamics, the Cv rating is commonly used to determine the flow capacity of a valve. However, in this question, we are given a Cx rating instead. The Cx rating is a modified version of the Cv rating and takes into account the specific gravity (sg) of the fluid being controlled by the valve.
To calculate the maximum flow through the valve, we need to use the equation:
Flow (gpm) = Cx * sqrt((Pressure drop in psi) / (Specific gravity))
In this case, the Cx rating is given as 4.0, the pressure drop is 100 psi, and the specific gravity of glycerin is 1.26. Plugging these values into the equation, we get:
Flow (gpm) = 4.0 * sqrt(100 / 1.26) = 4.0 * sqrt(79.365) ≈ 35.6 gpm
Therefore, the maximum flow through the valve for a pressure drop of 100 psi is approximately 35.6 gallons per minute.
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A CONCAVE lens has the same properties as a CONCAVE mirror.
A. true
B. False
The Given statement "A CONCAVE lens has the same properties as a CONCAVE mirror" is FALSE because A concave lens and a concave mirror have different properties and behaviors.
A concave lens is thinner at the center and thicker at the edges, causing light rays passing through it to diverge. It has a negative focal length and is used to correct nearsightedness or to create virtual images.
On the other hand, a concave mirror is a reflective surface that curves inward, causing light rays to converge towards a focal point. It has a positive focal length and can produce both real and virtual images depending on the location of the object.
So, a concave lens and a concave mirror have opposite effects on light rays and serve different purposes, making the statement "A concave lens has the same properties as a concave mirror" false.
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Write a x; in a form that includes the Kronecker delta. Now show that V.r=3.
x; = Σn=1 to ∞ δn,x vn,
where δn,x is the Kronecker delta and vn is a vector in the basis of x.
Kronecker delta is a mathematical symbol that is named after Leopold Kronecker. It is also known as the Kronecker's delta or Kronecker's symbol. It is represented by the symbol δ and is defined as δij = 1 when i = j, and 0 otherwise. Here, i and j can be any two indices in the vector x. The vector x can be expressed as a sum of vectors in the basis of x as follows: x = Σn=1 to ∞ vn, where vn is a vector in the basis of x.
Using the Kronecker delta, we can express this sum in the following form:
x; = Σn=1 to ∞ δn,x vn, where δn,x is the Kronecker delta. Now, if we take the dot product of the vector V and x, we get the following:
V·x = V·(Σn=1 to ∞ vn) = Σn=1 to ∞ (V·vn)
Since V is a 3-dimensional vector, the dot product V·vn will be zero for all but the third term, where it will be equal to 3. So, V·x = Σn=1 to ∞ (V·vn) = 3, which proves that V·x = 3.
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