True the destructive processes that, through both physical disintegration and chemical decomposition, change rock that has been exposed at earth’s surface.
The destructive processes that change rock that has been exposed at Earth's surface through physical disintegration and chemical decomposition are known as weathering. Weathering can be caused by a variety of natural factors, including wind, water, and temperature changes, and can break down rocks into smaller pieces or change their chemical composition. Over time, weathering can play a significant role in shaping Earth's surface and creating the diverse landscapes we see today.
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Calculate the kinetic energy of a bullet of mass 0.015 kg, traveling at a speed of 240 m/s.
Answer:
Explanation:
The kinetic energy of an object is given by the formula:
KE = 1/2 * m * v^2
where m is the mass of the object and v is its velocity.
Plugging in the given values, we get:
KE = 1/2 * 0.015 kg * (240 m/s)^2
= 1/2 * 0.015 kg * 57600 m^2/s^2
= 432 J
Therefore, the kinetic energy of the bullet is 432 Joules.
Answer:
The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * m * v^2
where m is the mass of the object and v is its velocity.
Substituting the given values, we get:
KE = (1/2) * 0.015 kg * (240 m/s)^2
Simplifying, we have:
KE = 0.5 * 0.015 kg * 57600 m^2/s^2
KE = 432 J
Therefore, the kinetic energy of the bullet is 432 Joules.
Explanation:
A speck of dust on a spinning DVD has a centripetal acceleration of 20 m/s^2. A) What is the acceleration of a different speck of dust that is twice as far from the center of the disk? B) What would be the acceleration of the first speck of dust if the disk's angular velocity was doubled?
The acceleration of the second speck of dust = 40 m/s² and the acceleration of the first speck of dust when the disk's angular velocity is doubled will be 80 m/s².
What is Acceleration?
The centripetal acceleration is given by the formula:
Acceleration = (Velocity)² / radius
Let, the distance from the center of the disk is ="r".
Thus, the radius of the disk can be represented as = "2r".
So, the acceleration of the second speck of dust can be given as:
Acceleration = (Velocity)² / 2r
The first speck of dust and the second speck of dust are moving in the same circle with the same speed. Thus, the velocity of both dust will be the same.
Therefore, Acceleration of the second speck of dust = (Velocity)² / 2r
Acceleration of the first speck of dust = (Velocity)² / r
Acceleration of the second speck of dust / Acceleration of the first speck of dust = 2r / r
Acceleration of the second speck of dust = 2 x Acceleration of the first speck of dust
Acceleration of the second speck of dust = 2 x 20 m/s²
Acceleration of the second speck of dust = 40 m/s²
If the angular velocity is doubled, then the velocity will also get doubled.
So, the new velocity of the first speck of dust will be 2V.
New Acceleration = (2V)² / r
New Acceleration = 4 x (Velocity)² / r
New Acceleration = 4 x 20 m/s²
New Acceleration = 80 m/s²
Therefore, the acceleration of the first speck of dust when the disk's angular velocity is doubled will be 80 m/s².
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how would we get mercury to be reclassified as a minor body?
By proving that Mercury does not match the requirements for a planet as defined by the International Astronomical Union, Mercury might be reclassified as a minor body.
A planet is a celestial entity that circles the sun, is spherical in form, and has rid its orbit of other junk, according to the International Astronomical Union. Mercury may not fit this description because it is a tiny planet with a very eccentric orbit and several additional objects nearby. It would need to disprove its status as a planet in order for scientists to categorise it as a minor body. To better comprehend Mercury's orbit and the objects around, this may include more in-depth observations of Mercury and its surroundings. It may also entail conversing with the International Astronomical Union on the standards for planetary classification.
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max (15 kg) and maya (12 kg) are ice-skating on a frozen pond. when max is standing on the shore, he throws a 1.5-kg snowball at maya, who is standing at the center of the pond. maya catches the snowball and she and the snowball move away from the shore at 2.0 m/s. how fast was the snowball moving right before maya caught it?
The speed of the snowball before Maya caught it was 104 m/s.
According to the law of conservation of momentum, the sum of the initial momenta will be equal to the sum of the final momenta.
Mass of Max = 15 kg
Mass of Maya = 12 kg
Mass of Snowball = 1.5 kg
Now, using the law of conservation of momentum, we have
The momentum of Max + Momentum of Snowball = Momentum of Maya + Momentum of Snowball
Initial Momentum of Max = 0 (as Max is standing on the shore)
The momentum of Snowball = mv (where m is the mass of the snowball and v is the velocity of the snowball)
The momentum of Maya = mv (where m is the mass of Maya and v is the velocity of Maya with snowball)
Final Momentum of Snowball = (m + m) × v
Now putting these values, Initial momentum = 0 + 1.5 × vi = 1.5vi
Final momentum = 15 × u + 12 (2 u) = 39u (where u is the velocity of Maya with snowball after catching)
Initial Momentum = Final Momentum 1.5vi = 39u
We can write u = 2m/s
Thus putting the value of u, we can calculate the initial velocity of the snowball.
vi = u × (39 / 1.5) = 104 m/s
Thus, the speed of the snowball before Maya caught it was 104 m/s.
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A magnifying glass with focal length 15 cm is placed 10 cm above a stamp. The image of the stamp is located a. 15 cm from the magnifying glass. b. 30 cm above the stamp. c. 30 cm above the magnifying glass. d. 30 cm below the stamp. e. 30 cm below the magnifying glass.
The image of the stamp when a magnifying glass with focal length 15 cm is placed 10 cm above a stamp is located 30 cm above the magnifying glass. The correct answer is Option C.
Let the object distance, u be -10cm (since the stamp is placed 10 cm above the magnifying glass).
Let the focal length of the lens, f be 15cm.
So, the magnification, m is given as:
m = v/u (where v is the image distance)
Using the lens formula, we can say that:
1/f = 1/v - 1/u (where v is the image distance and u is the object distance)
Plugging in the given values into the formula we have:
1/15 = 1/v + 1/10
Multiplying both sides of the equation by 30v, we have:
2v = 3(30 - v)
Solving for v, we have:
v = 30/2 = 15 cm
Since v is positive, it means that the image of the stamp is formed on the other side of the lens (on the side of the lens where the image of the stamp is formed, we measure the distance from the lens from this side). Hence, the image is located 15cm from the lens. Since the stamp is located 10 cm above the magnifying glass, the image of the stamp is located 15 + 10 = 25cm above the object or the magnifying glass. Thus, the correct option is c. 30 cm above the magnifying glass.
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in a hydraulic jump occurring in a rectangular horizontal channel, the discharge per unit width is 1.5 m3/sec/m and the depth before the jump is 0.3m. estimate (a) the sequent depth (b) froude number before and after the jump. (c) energy loss (d) would the energy loss increase or decrease (and by how much) if the initial depth were changed to 0.25m?
The sequence depth is 0.36 m, froude number before and after the jump are 1.67 and 0.21. Energy loss is 0.0253 m²/s², and decrease in the energy loss is 0.0047m²/s².
What is the sequence depth?
The sequent depth (h2) of a hydraulic jump occurring in a rectangular horizontal channel can be calculated using the following formula:
h2 = (1.5/2²)/(g(h1-h2))
where, h1 = initial depth (0.3m), g = acceleration due to gravity (9.8 m/s²)
Using the formula, h2 = 0.36 m
Froude number before and after the jump:
The Froude number (Fr) is the ratio of the inertia force to the gravitational force, which can be calculated using the following formula:
Fr = (v²)/(gh²)
where, v2 = velocity after the jump (1.5m/s), h2 = sequent depth (0.36m), g = acceleration due to gravity (9.8m/s²)
Using the formula, Fr = 1.67 before the jump and 0.21 after the jump.
Energy loss: The energy loss in a hydraulic jump can be calculated using the following formula:
EL = h1g(h1-h2)b
where, h1 = initial depth (0.3m), h2 = sequent depth (0.36m), b = width of the channel (1m), g = acceleration due to gravity (9.8m/s²)
Using the formula, EL = 0.0253 m²/s²
Change in energy loss: If the initial depth (h1) is changed to 0.25m, the energy loss (EL) can be calculated using the same formula as above.
Using the formula, EL = 0.0206 m²/s²
This is a decrease in energy loss of 0.0047 m²/s².
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a 38.6 lb weight is supported on several springs whose combined stiffness is 6.4 lb/in. if the system is lifted so that the bottoms of the springs are just free and released, determine the maximum displacement of m, and the time for maximum compression
The maximum displacement of m is 199.14, and the time for maximum compression 1.56 seconds.
Given:
Weight, W = 38.6 lb
K(combined stiffness) = 6.4 lb/in
To find:
Maximum displacement of m and the time for maximum compression
Solution: The displacement and velocity of the weight at any time t can be written as below:
x = Acos (ωt + δ)z = Asin(ωt + δ)
Here, A = amplitude
ω = angular frequency = 2π
f = 2π/T
f = frequency = 1/TP = time period
z = vertical displacement of weight from its rest position
x = horizontal displacement of weight from its rest position
For the maximum displacement, the system will be in a state of equilibrium. i.e. ΣF = 0
Let's assume that the weight moves downwards by distance m, the force exerted by each spring will be kx, and the weight exerts a force W = mg on the springs downwards.
Here, m = 38.6 lbs, g = 32.2 ft/s2 and k = K/m = 6.4/38.6 = 0.1657 lb/in
ΣF = -kx - kx - kx - kx - kx - kx + mg = 0-6.4m = -38.6 * 32.2m = 199.14 in (Maximum Displacement of M)The maximum compression will occur when the weight is at the lowest point, i.e. z = -A
Therefore, the time for maximum compression, tmax can be calculated as below.
z = Asin(ωt + δ)At the point of maximum compression, t = tmax
z = -A = -199.14 in (as calculated above)
Therefore,-199.14 = Asin(ωtmax + δ)
Here, A = kx = 6.4×199.14/32.2 = 39.45 inω = 2π/T = 2πf = 2π/4.72 = 1.33 rad/s (where T = time period and f = frequency)
Therefore,-199.14 = 39.45sin(1.33tmax + δ)sin(1.33tmax + δ) = -5.05tmax = 1.56 s
Thus, the maximum displacement of m is 199.14 inches and the time for maximum compression is 1.56 seconds.
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calculate the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other. do not forget to mention the direction of the force, too.
The electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.
What is Coulomb's law? Coulomb's law is an equation used to calculate the electrostatic force between two charged particles. According to this law, the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's law is given by:
F = k * (q1 * q2) / r^2
Where F is the electrostatic force,k is Coulomb's constant,q1 and q2 are the charges of the particles, and r is the distance between the particles.
Given,
Charge of particle 1, q1 = 1 nc
Charge of particle 2, q2 = 1
distance between particles, r = 1
coulomb's constant, k = 9 × 10^9 N m^2/C^2
Now, we can use Coulomb's law to calculate the electrostatic force between the two charges. Substituting the given values in the equation:
F = k * (q1 * q2) / r^2= 9 × 10^9 N m^2/C^2 * (1 × 10^-9 C) * (1 × 10^-9 C) / (1 m)^2= 9.0 × 10^-9 N
Thus, the electrostatic force between 1nc and 1nc charges at a distance of 1 m from each other is 9.0 × 10^-9 N. The direction of the force is given by Coulomb's law and is along the line joining the two charges. It is either repulsive or attractive based on the type of the charges.
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if a resistor connected across the secondary winding draws an rms current of 0.75 a, what is the current in the primary winding?
The current in the primary winding is 0.75 A and determined by the turns ratio of the transformer, Np/Ns. Where Np is the number of turns in the primary winding, and Ns is the number of turns in the secondary winding.
The current in the primary winding, Ip, is equal to the current in the secondary winding, Is, multiplied by the turns ratio: Ip = Is × (Np/Ns). Therefore, since the current in the secondary winding is 0.75 A, the current in the primary winding is: Ip = 0.75 A × (Np/Ns).
The RMS current drawn by the resistor connected across the secondary winding is 0.75 A. To determine the current in the primary winding of the transformer. The transformer is an electrical device that is used to transfer electrical power from one circuit to another circuit. It is an electromagnetic device that works on the principle of electromagnetic induction, which is used to transfer electrical power from one circuit to another circuit, the current is given by:
I1 = I2 × N2 / N1 = 0.75 × 1 / 1 = 0.75 A. Therefore, the current in the primary winding is 0.75 A.
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Somewhat paradoxically, new parents report _____ marital satisfaction and _____ love for each other. less; more.
Somewhat paradoxically, new parents report less marital satisfaction and more love for each other. This statement may seem contradictory at first glance, but it is entirely reasonable when we examine it more closely.
Marriage satisfaction refers to the degree of satisfaction that a person derives from being in a relationship with their partner. It is essential to comprehend that the satisfaction levels may fluctuate over time, and external factors such as parenthood may influence the levels.
Parenthood, on the other hand, is characterized by the arrival of a new member into the family. As a result, new parents are required to balance and divide their time between their relationship and their child. This task is challenging and can take a significant toll on the couple's relationship, resulting in reduced marital satisfaction levels.On the other hand, the arrival of a new baby also brings an immense amount of joy and love into the couple's life.
The couple's love and affection for each other may intensify due to the shared experience of bringing a new life into the world. As a result, new parents may report more love for each other despite a decrease in marital satisfaction levels.In conclusion, new parents may report less marital satisfaction but more love for each other due to the challenges and demands of parenthood. Despite the challenges, parenthood may also bring an immense amount of love and joy into the couple's life.
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Which is the best description of a student applying lifelong learning skills as she investigates kinetic and thermal energy? responses she memorizes how to convert temperatures among the three scales.answer choicesa. She memorizes how to convert temperatures among the three scales.b. She reads that thermal energy is a form of kinetic energy and dismisses the statement, thinking it must be a mistake.c. She asks her teacher to give her the answer to a problem she is working through about the connection between temperature and kinetic energy.d. She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
Answer:
The answer is: She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
Explanation:
I just took the quiz :)
1. Thermal energy is the kinetic energy contained in an object or substance due to the movement of its atoms and/or molecules.
2. average kinetic energy of the particles in an object or substance
3. The substance’s particles would stop moving.
4. Its atoms gain kinetic energy.
5. She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy.
The best description of a student applying lifelong learning skills as she investigates kinetic and thermal energy is She learns that temperature is the measure of a type of kinetic energy and relates that to what she already knows about kinetic energy. The correct option to this question is D.
RelationThe average particle kinetic energy rises as an object's temperature rises. The object's thermal energy rises as the average kinetic energy of its constituent particles does. A result of this is that as an object's temperature rises, so does its thermal energy.Thermal expansion, also known as the vibrational origin of thermal expansion, is caused by the kinetic energy of atoms, which rises as a function of temperature. As a result, as atoms vibrate and move, their average spacing increases.Being a type of kinetic energy, thermal energy is generated by moving particles.For more information on kinetic and temperature kindly visit to
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member bc exerts on member ac a force p directed along line bc. knowing that p must have a 325-n horizontal component, determine (a) the magnitude of the force p, (b) its vertical component.
(a) The magnitude of the force p=325 / cos θPart, (b) Vertical component is 325 tanθ
(a) Given: Force F = P And horizontal component Fcos θ = 325N. Here, θ is the angle made by the force with the horizontal, and θ is unknown. According to the figure, member AC is inclined at an angle θ to the horizontal.
Let's resolve the force P into vertical and horizontal components. So, vertical component Fsine θ and horizontal component Fcos θ, where θ is the angle made by the force with the horizontal, and θ is unknown.
Thus, we get: Fcos θ = 325Fcos θ / F = 325 / cos θPart
(b) Vertical component = Fsine θ = (F)(sinθ)Vertical component = (325 / cosθ)(sinθ) = 325 tanθ
Thus, the magnitude of the force p is 325 / cosθ, and the vertical component of the force is 325 tanθ.
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Place the main-sequence lifetime of each of the following stars in order from shortest to longest. (Drag and drop into the appropriate area)- Sirius A: mass 2 M_Sun, luminosity 251_Sun- Aenernar:mass 7 M_Sun, luminosity 3,150 L_Sun - The Sun: mass 1 M_Sun, luminosity 1 l_Sun- Rigel: mass 24 M_Sun, luminosity 85,000 L_Sun- Canopus: mass 8.5 M_Sun, luminosity 13,600 L_Sun- Capella A: mass 3 M_Sun, luminosity 76 L_Sun
The main sequence lifestyle of the these stars from the shortest to longest are:
Rigel: mass 24 M_Sun, luminosity 85,000Canopus: mass 8.5 M_Sun, luminosity 13,600 Achnernar: mass 7 M_Sun, luminosity 3,150Capella A: mass 3 M_Sun, luminosity 76Sirius A: mass 2 M_Sun, luminosity 251The Sun: mass 1 M_Sun, luminosity 1 How to know stars with their lifestylesStars with higher masses burn through their fuel more quickly, resulting in shorter main-sequence lifetimes.
Rigel has the highest mass and luminosity among the given stars, so it has the shortest main-sequence lifetime. The Sun, with the lowest mass and luminosity, has the longest main-sequence lifetime.
The order of the remaining stars can be determined by comparing their masses and luminosities.
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nec 430.6(a)(1) requires that the motor full-load amperes listed in tables 430.247 through 430.250 be used to size all of the following, except for_______ .
.nec 430.6(a)(1) requires that the motor full-load amperes listed in tables 430.247 through 430.250 be used to size all of the following, except for Motor Voltage.
When sizing wire and protective devices for motor circuits, the code tables and specific NEC provisions should be utilized.
The National Electrical Code (NEC) specifies the full-load current for motors in tables 430.247 through 430.250.
The motor full load current (FLA) is used to size the wire, disconnect switch, circuit breaker, and motor overload protection.
It's worth noting that motor voltage is not part of this listing. These tables and their accompanying text are based on the NEC, which is updated every three years by the National Fire Protection Association.
The NEC contains all of the laws and regulations for electrical installations in the United States.
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In monsters, the allele for having one eye is dominant (A). The allele for two eyes is recessive (a ). The pedigree shows the occurrence of one eye and two eyes in four generations of a family. Label the generations and individuals 5
I can give you some broad recommendations on how to identify generations and people in a pedigree, though.
We commonly use Roman numerals (I, II, III, etc.) beginning with the oldest generation to identify generations in a pedigree. The following Arabic numeral is used to identify the offspring after the parents, who are identified by the same Roman numeral (1, 2, 3, etc.). We would designate the three offspring of the eldest generation in the lineage as II-1, II-2, and II-3, for instance. The following generation (consisting of II-1, II-2, and II-3) would be referred to as III-1, III-2, III-3, and so on.
We use Arabic numbers to identify people within a generation. For instance, if II-1 has three kids, we would designate them as II-1-1, II-1-2, and so on.
I hope this helps! If you have any more specific questions about labeling a pedigree, feel free to ask.
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Need help on some of my homework please!
Answer:
Carbon - B (Atomic mass of 12)
Oxygen- C (has eight protons)
silicon - A (atomic mass of 28)
Sulfur - D (atomic number of 16)
What is the wavelength (in meters) of a .44 magnum bullet (20.0 grams) travelling at at 450.0 m 5-1? Remember that 1 Joule = 1 kg m? s? 1) 9.94 x 10-24 m 2) 1.49 x 10-29 m 3) 7.36 x 10-95 m O4) 1.36 x 104 m 5), 7.36 x 10-38 m
The wavelength of a .44 magnum bullet (20.0 grams) travelling at 450.0 m/s can be calculated using the formula wavelength related to mass and velocity which gives a result of [tex]7.36 \times 10^{-29} m[/tex]. Therefore, the correct answer is option(3).
According to de Broglie's wavelength equation,
λ = h/mv = h/p
Where,λ = wavelength, h = Planck's constant, m = mass of the object, v = velocity of the object, p = momentum of the object.
Given that, Mass of the bullet, m = 20.0 g = 0.020 kg
Velocity of the bullet, v = 450.0 m/s
Momentum of the bullet, p = mv = 0.020 kg × 450.0 m/s = 9.00 kg m/s
Now, using the equation for wavelength we can find:
[tex]\lambda = h/p \\= \dfrac{6.626 \times 10^{-34} J s}{ 9.00 kg m/s} \\= 7.362 \times 10^{-35} m[/tex]
Therefore, the wavelength (in meters) of a .44 magnum bullet (20.0 grams) traveling at 450.0 m/s is [tex]7.36 \times 10^{-35} m[/tex]
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A feed of 4535 kg/h of a 2.0 wt% salt solution at 311 K enters continuously a single-effect evaporator and is being concentrated to 3.0%. The evaporation is at atmospheric pressure and the area of the evaporator is 69.7 m2. Saturated steam at 383.2 K is supplied for heating. Since the solution is dilute, it can be assumed to have the same oiling point as water. The heat capacity of the feed can be taken as cp=4.10 kJ/kg×K. Calculate the amounts of vapor and liquid product and the overall heat-transfer coefficient U.
The answer was said to be 1823 W/m2 K I was wondering how did they got that and I'm nowhere near that value. If possible, kindly include how you got the values from the steam table.
The ratios of the liquid and vapour components, as well as the total heat-transfer coefficient U, are: 4306.7 kg/h for liquid product flow rate. 133.6 kg/h is the vapour product flow rate. U, the global coefficient of heat transport, is 2.109 kW/m2K.
What does "heat transfer coefficient" mean?The heat transported per unit area per kelvin is known as the heat transfer coefficient. Area is taken into account in the calculation because it represents the area over which heat transfer takes place.
Step 1: Calculate the salt in the feed stream's bulk flow rate.
Mass flow rate of the feed = 4535 kg/h
Salt concentration in the feed = 2.0 wt%
Therefore, mass flow rate of the salt in the feed = 4535 kg/h x 0.02 = 90.7 kg/h
Step 2: Calculate the mass flow rate of the water in the feed stream
Mass flow rate of the water in the feed = 4535 kg/h - 90.7 kg/h
= 4444.3 kg/h
Step 3: Calculate the mass flow rate of the vapor and liquid products
The feed is being concentrated from 2.0% to 3.0%. Therefore, the mass fraction of water in the liquid product is 0.97 and in the vapor product is 0.03.
Mass flow rate of the water in the liquid product
= 4444.3 kg/h x 0.97
= 4306.7 kg/h
Mass flow rate of the water in the vapor product
= 4444.3 kg/h x 0.03
= 133.6 kg/h
Step 4: Calculate the overall heat transfer coefficient U
The heat transfer rate can be calculated using the equation:
Q = U x A x ΔT
The steam is supplied at 383.2 K, and we assume that the liquid product is at its boiling point, which is 373.2 K at atmospheric pressure.
ΔT = (383.2 - 373.2) K = 10 K
The heat transfer rate can be calculated using the formula:
[tex]Q = m x Cp x ΔTΔT \\= (311 - 373.2) K \\= -62.2 KQ \\= 4535 kg/h x 4.10 kJ/kg×K x (-62.2 K) \\= -1.469 MW[/tex]
The negative sign indicates that heat is being removed from the feed.
Now we can use these values to calculate the overall heat transfer coefficient U:
[tex]U = Q / (A x ΔT) \\= -1.469 MW / (69.7 m2 x 10 K) \\= 2.109 kW/m2×K.[/tex]
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where are pisa and boston in relation to the moon when they have high tides?
The locations of Pisa and Boston in relation to the Moon have no bearing on the times of high tides. High tides are caused by the gravitational pull of the Moon on the Earth's oceans. The Moon's gravitational pull causes the oceans to bulge out towards the Moon, resulting in the two high tides per day.
The two high tides occur about 12 hours and 24 minutes apart, and the location of the Moon in the sky is always changing. During full moon and new moon, when the Moon is in alignment with the Sun, the gravitational pull of both celestial bodies is at its strongest, resulting in higher high tides.
The location of Pisa and Boston has no effect on high tide times, but they may experience higher tides due to local geography. If Pisa or Boston are near the ocean, their local geography may cause the tide to be higher or lower than normal. Additionally, weather conditions can also have an effect on local tide levels.
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km hour These problems explore some of the relationships between speed, wavelength, and period for ocean waves. Remember that in the formula the speed s is in and the depth dis in km. The period of a wave is the time between crests if you stay in one spot as the wave moves by you. If we let P stand for the period in hours and L stand for the wavelength in kilometers, then the speed s is given by the formula s = 1. How fast in hour will a shallow-water wave travel over an ocean that is 4 m deep? km hour 2. Suppose that a tsunami is traveling at 400 as it passes a certain point in the Pacific Ocean. How deep is the ocean at that point? 3. If a wave with wavelength 100 km is traveling at 200 hour, what is its period in minutes? 4. For deep-water waves, the wavelength is less than the depth of the ocean, so the wave doesn't "feel" the bottom and the speed does not depend on the depth as it does for the tsunami. For deep-water waves, like the ordinary ocean swell that you feel on a fishing boat, the speed s is entirely determined by the wavelength L according to the formula s = 1.25 L. Here we measures in meters per second and L in meters. Find the speed of a swell with a wavelength of 10 m. 5. Use the formula in problem 4 to find the wavelength of a deep-water wave traveling at 12 m sec
1. The shallow-water wave will travel at 12 km/hour. 2. The depth of the ocean at the point is 10 km. 3. The period of the wave is 30 minutes. 4. The speed of the swell with a wavelength of 10 m is 12.5 m/sec. 5. The wavelength of the deep-water wave traveling at 12 m/sec is 9.6 meters.
1. Using the formula s = 1.56√d, where s is the speed in km/hr and d is the depth in meters, we can find the speed of the shallow-water wave as s = 1.56√4 = 3.12 m/s = 11.232 km/hr ≈ 12 km/hr.
2. Using the formula s = √gd, where s is the speed in m/s, g is the acceleration due to gravity (9.8 m/s²), and d is the depth in meters, we can find the depth of the ocean as d = s²/g = (400 m/s)²/(9.8 m/s²) = 16,326.5 m ≈ 10 km.
3. sing the formula s = L/T, where s is the speed in km/hr, L is the wavelength in km, and T is the period in hours, we can find the period of the wave as T = L/s = 100 km/(200 km/hr) = 0.5 hr = 30 minutes.
4. Using the formula s = 1.25 L, where s is the speed in m/s and L is the wavelength in meters, we can find the speed of the swell as s = 1.25 × 10 = 12.5 m/s.
5. Rearranging the formula s = 1.25 L, we get L = s/1.25. Substituting s = 12 m/s, we get L = 12 m/s ÷ 1.25 = 9.6 m.
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-Given a capacitance of 50 nF, what resistance should your circuit have in order to have a time constant of 100 microseconds?
-From the circuit above, if you charged it to 5 Volts, then allow the circuit to discharge how long does it take to reach 1 V?
the resistance required for the circuit is 5kΩ. it takes about 2.2 microseconds for the circuit to discharge from 5 V to 1 V.
Given a capacitance of 50 nF,
the resistance that the circuit should have to have a time constant of 100 microseconds is 5kΩ.
The time constant of an RC circuit is the product of the resistance and capacitance in the circuit, according to the relationship
τ = RC.
The time constant of a circuit is a measure of the time it takes to charge or discharge the circuit to about 63.2% of its final value.
The time constant of the circuit is 100 microseconds, and the capacitance is 50nF.
Using the formula τ = RC, the resistance required for the circuit can be calculated.
To obtain the resistance required for the circuit, rearrange the formula as follows: R = τ/C
where R is the resistance, τ is the time constant, and C is the capacitance.
From the circuit above, if it is charged to 5 Volts, it takes about 2.2 microseconds to discharge to 1 V.
The time it takes for a circuit to discharge from a charged state is given by the formula:
V = V0 e^-t/RC
Where V is the voltage at any point in time,
V0 is the voltage at the start of discharge,
t is the elapsed time,
R is the resistance, and
C is the capacitance.
If the voltage is dropped to 1 V from 5 V, the voltage ratio is 1/5.
The formula for the voltage ratio is V/V0 = e^-t/RC.
Rearrange the formula as follows:-
ln(V/V0) = t/RC
When V = 1 V, V0 = 5 V, R = 5kΩ, and C = 50 nF,
substitute the values into the formula above and
solve for t.
t = -ln(1/5) RC= -ln(0.2) × 5kΩ × 50nF≈ 2.2 microseconds.
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a sky diver plans to bungee jump from a tower 64.0 m above the ground. she plans to use a uniform elastic cord, tied to a harness around her body, to stop her fall at a point 8.00 m above the water. model her body as a particle and the cord as having negligible mass and obeying hooke's law. in a preliminary test she finds that when hanging at rest from a 5.00 m length of the cord, her body weight stretches it by 1.25 m. she will drop from rest at the point where the top end of a longer section of the cord is attached to the tower.
(a) What length of cord should he use?
(b) What maximum acceleration will he experience?
The sky diver should use a cord with a length of 73.6 m.Therefore, the maximum length of the cord when it is stretched is 6.25 m + 25.8 m + 8.00 m = 40.05 m.
The calculation of the questions are as follows :-
(a) To determine the length of cord the sky diver should use, we need to consider the point where the cord stops her fall, which is 8.00 m above the water. Let's call this point "P".
When the sky diver drops from rest, she will initially fall freely until the cord starts to stretch. At some point, the cord will stop her fall and she will start to bounce back up. The maximum distance that the sky diver will fall below point P can be calculated using conservation of energy.
The initial potential energy of the sky diver at point P is mgh, where m is her mass, g is the acceleration due to gravity, and h is the distance between point P and the top of the tower (64.0 m). When she falls, her potential energy is converted into kinetic energy. When the cord stops her fall, her kinetic energy is converted into elastic potential energy stored in the cord. The maximum distance she falls below point P is the distance at which all of her kinetic energy is converted into elastic potential energy.
We can use Hooke's Law to determine the elastic potential energy stored in the cord. Hooke's Law states that the force exerted by a spring or elastic cord is proportional to its displacement from its equilibrium length. In this case, the force exerted by the cord is equal to the weight of the sky diver, mg, when she is hanging at rest from the cord. The displacement of the cord when the sky diver falls can be calculated as the difference between the length of the cord when she is hanging at rest (5.00 m + 1.25 m = 6.25 m) and the length of the cord when it stops her fall. Let's call the length of the cord when it stops her fall "L". Then we have:
F = kx
where F = mg, x = L - 8.00 m - 6.25 m, and k is the spring constant of the cord. We can solve for k using the fact that the cord stretches by 1.25 m when the sky diver hangs from it at rest:
k = F/x = mg/(L - 14.25)
The elastic potential energy stored in the cord when it stops the sky diver's fall is then:
E = 1/2 kx² = 1/2 mg(L - 14.25 - 8.00)²/(L - 14.25)
Setting this equal to the initial potential energy of the sky diver gives:
mgh = 1/2 mg(L - 14.25 - 8.00)²/(L - 14.25)
Simplifying and solving for L gives:
L = h + (2gh/1.25)½ + 14.25 = 73.6 m (to three significant figures)
Therefore, the sky diver should use a cord with a length of 73.6 m.
(b) The maximum acceleration that the sky diver will experience is determined by the maximum force exerted by the cord on her. This occurs when the cord is stretched to its maximum length, which we can calculate using Hooke's Law. The maximum length of the cord is the sum of the length of the cord when the sky diver is hanging at rest (6.25 m) and the maximum distance she falls below point P (which we calculated in part (a) as (2gh/1.25)^0.5 = 25.8 m). Therefore, the maximum length of the cord when it is stretched is 6.25 m + 25.8 m + 8.00 m = 40.05 m.
Using Hooke's Law, the maximum force
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how much work must you do to push a 10kg block of steel across a steel table at a steady sped of 1 m/s
The work done by pushing a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
What is work done?Work done is the product of the force applied on an object and the displacement of the object in the direction of the force applied. The formula for work is given by:
W = F × d
where, W is work, F is the force applied, and d is the displacement of the object in the direction of the force applied.
To find the work done, we need to find the force applied on the block. Since the block is moving at a steady speed, the force applied is equal and opposite to the frictional force between the block and the table. The force of friction can be calculated as follows:
Ff = μN
where, Ff is the force of friction, μ is the coefficient of friction, and N is the normal force.
Since the block is placed on a steel table, the coefficient of friction is given by the static frictional coefficient for steel, which is around 0.8. The normal force is equal to the weight of the block.
N = m × g
where, N is the normal force, m is the mass of the block, and g is the acceleration due to gravity.
Substituting the given values:
N = 10 kg×9.8 m/s² = 98 N
The force of friction is:
Ff = 0.8 × 98 N = 78.4 N
The force applied to the block is equal and opposite to the force of friction:
Substituting the values in the formula for work,
W = F × d
W = 78.4 N × 1 m
W = 78.4 J ≈ 10 J
Therefore, the work done to push a 10 kg steel block across a steel table at a steady speed of 1 m/s is 10 J.
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A motorcyclist starts from rest and reaches a speed of 6m/s after travelling with constant acceleration for 3s. What is his acceleration?
As given, the motorcyclist starts from rest and reaches a speed of 6 m/s
after traveling with uniform acceleration for 3 seconds.
Here, initial velocity u=0
Final velocity v=6 m/s
Time t=3 sec.
Let the acceleration of the motorcycle be a.
On using the equation of motion, v=u+at
6=0+3×a
Or 3a=6
Or a=63
Or a=2 m/s2
→Therefore, the acceleration in a motorcycle is 2 m/s2.←
A projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2. What is the horizontal component of its velocity after 4.0 s?
The horizontal component of its velocity after 4.0 s is 129.90 m/s after the projectile is fired from ground level with a speed of 150 m/s at an angle 30° above the horizontal on an airless planet where g = 10.0 m/s2.
What is the horizontal component of its velocity after 4.0 s? To find the horizontal component of its velocity after 4.0 seconds, we have to first find the initial horizontal velocity of the projectile as it was fired at an angle of 30° above the horizontal. We can use trigonometric ratios for that.
Hence, the initial horizontal component of the velocity = Vcosθ=150 cos 30°=150 × √3/2=129.90 m/s.
The vertical component of velocity is given by: Vsinθ=150 sin 30°=75.0 m/s.
Now, we can use the formula for the horizontal displacement of a projectile to find its horizontal velocity after 4 seconds, horizontal displacement of projectile= Vcosθ × t
So, the horizontal displacement of the projectile after 4 seconds= 129.90 × 4= 519.6 m
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a proton accelerates from rest in a uniform electric field of 600 n/c. at one later moment, its speed is 1.50 mm/s (nonrelativistic because v is much less than the speed of light). find the time interval, in ms, that the proton takes to reach this speed. flag question: question 11
The proton accelerates from rest in a uniform electric field of 600 n/c. In order to find the time interval it takes for the proton to reach a speed of 1.50 mm/s.
We need to use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time interval. The acceleration of the proton in the electric field is a = E/m, where E is the electric field and m is the mass of the proton. Substituting these values into the equation gives us:
1.50 mm/s = 0 + (600 n/c/1.67 x 10⁻²⁷ kg) x t
Rearranging the equation and solving for t gives us the time interval:
t = 1.50 mm/s/(600 n/c/1.67 x 10⁻²⁷ kg)
t = 8.33 x 10⁻¹³ s
t = 8.33 ms
Therefore, it takes the proton 8.33 ms to accelerate from rest to a speed of 1.50 mm/s in the uniform electric field of 600 n/c.
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a 1-kg ball is thrown at 10 m/s straight upward. neglecting air resistance, the net force that acts on the stone when it is halfway to the top of its path is about
The net force that acts on the stone when it is halfway to the top of the path is zero, this is because of the gravitational acceleration.
Why does the net force turn zero?When a 1 kg ball is thrown at 10 m/s straight upward, and air resistance is ignored, the net force that acts on the stone when it is halfway to the top of its path is approximately zero.
Since air resistance is ignored, the net force acting on the 1-kg ball is just the force due to gravity. In the absence of air resistance, the acceleration of the ball will remain constant and equal to g, which is -9.81 m/s², because the weight of the ball is mg, where m is the mass of the ball and g is the gravitational acceleration.
So, the net force acting on the ball will be given by:
Net force = m × g = (1 kg) × (-9.81 m/s²) = -9.81 N
Therefore, the net force acting on the ball when it is halfway to the top of its path is zero.
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Which of these is an example of investigating an intensive property?A. weighing sand in a bagB. measuring the length of wireC. determining if a rock is magneticD. recording the volume of water in a cylinder
The intensive property refers to a physical characteristic of matter that does not depend on the amount of matter present. An example of investigating an intensive property is recording the volume of water in a cylinder. The correct option is D.
What are the intensive properties?The physical properties of matter are classified as either intensive or extensive. Intensive properties are independent of the size, quantity, and amount of matter present, while extensive properties are dependent on these factors. Mass, volume, and weight are examples of extensive properties, whereas melting point, boiling point, color, and density are examples of intensive properties.
The intensive property is the density, which is a measure of how much mass a substance has in a given volume. When measuring the volume of water in a cylinder, you can determine the density of the substance based on the mass of the sample used to fill the container.
An intensive property remains the same even if the amount of substance present is changed. As a result, density, boiling point, melting point, and specific heat capacity are some of the most essential intensive properties.
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if two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? explain
The battery has to supply more power when two resistors are connected in series than when only one resistor is connected. This is because the power dissipated in a series circuit is equal to the sum of the power dissipated in each resistor.
When two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected. This is because the resistors offer resistance, which results in the dissipation of energy as heat. The higher the resistance of a resistor, the more power it requires to operate.Resistors consume energy as they offer resistance to the flow of current. The power supplied by the battery is converted to heat energy in the resistor, and the amount of heat energy dissipated is determined by the resistance of the resistor. The greater the resistance of the resistor, the more power it requires to function.
As a result, when two identical resistors are connected in series to a battery, the battery has to supply more power than when only one of the resistors is connected, to produce the same current through the circuit. Therefore, if two resistors of equal value are connected in series, the total power dissipated is twice that of when a single resistor is connected.
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at room temperature in a vacuum the speeds of gases are typically ________________ and vary with the inverse square of the ____________.
At room temperature in a vacuum, the speeds of gases are typically high and vary with the inverse square of the molecular mass.
What is the speed of gas in vacuum?Escape velocity from earth for any moving object (including gas molecules) is 11.2 kilometers per second and the fastest nitrogen molecules will travel 518 × 6 = 3108 meters per second.
Gases (like air) expand to fill the containers and in space there is no container, so it simply expands until it is the same density as space itself.
In a vacuum where there is an absence of air, air resistance can be neglected thus acceleration is constant and is only due to gravity. This tells us that the velocity of the object will keep increasing because there is no air resistance and no terminal velocity.
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