To stay submerged in salt water, a 53.0 kg grouper with a body density of 1013 kg/m3 must exert a force of approximately 7.27 N. This is calculated using the principle of buoyancy to determine the buoyant force in relation to the grouper's weight.
The force a 53.0 kg grouper must exert to stay submerged in salt water with a body density of 1013 kg/m3 can be found by applying the principle of buoyancy (Archimedes' principle), which states that the buoyant force on a submerged object is equal to the weight of the fluid that is displaced by the object.
First, calculate the volume of the grouper. Since density ( ) equals mass (m) divided by volume (V), we have V = m / . For a 53.0 kg grouper with a density of 1013 kg/m³, the volume V would be 53.0 kg / 1013 kg/m³ = 0.05234 m³.
Next, calculate the weight of the volume of salt water displaced. The density of salt water is approximately 1027 kg/m3. The weight of the displaced water (Ww) is the product of its volume (V), its density ( water), and the acceleration due to gravity (g). So, Ww = V × water × g = 0.05234 m³ × 1027 kg/m³ × 9.81 m/s² = 527.3 N.
Finally, the force the grouper must exert to stay submerged (F) is the difference between the buoyant force and the grouper's weight. The weight of the grouper (Wg) is calculated as mass times gravitational acceleration, Wg = 53.0 kg ×9.81 m/s2 = 520.03 N. Thus, F = Ww - Wg = 527.3 N - 520.03 N = 7.27 N.
Therefore, a 53.0 kg grouper must exert a force of approximately 7.27 N to stay submerged in salt water.
The grouper must exert a force of approximately [tex]\( 516.88 \, \text{N} \)[/tex] to stay submerged in salt water.
To determine the force that the grouper must exert to stay submerged in salt water, we can use Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.
The buoyant force [tex](\( F_b \))[/tex] can be calculated using the formula:
[tex]\[ F_b = V \times \rho_{\text{fluid}} \times g \][/tex]
Where:
[tex]\( V \)[/tex] is the volume of the object submerged in the fluid
[tex]\( \rho_{\text{fluid}} \)[/tex] is the density of the fluid
[tex]\( g \)[/tex] is the acceleration due to gravity
The weight of the grouper [tex](\( F_g \))[/tex] can be calculated using the formula:
[tex]\[ F_g = m \times g \][/tex]
For the grouper to stay submerged, the buoyant force must be equal to the weight of the grouper. Therefore:
[tex]\[ F_b = F_g \][/tex]
[tex]\[ V \times \rho_{\text{fluid}} \times g = m \times g \][/tex]
[tex]\[ V \times \rho_{\text{fluid}} = m \][/tex]
[tex]\[ V = \frac{m}{\rho_{\text{fluid}}} \][/tex]
Now, we can calculate the volume of the grouper submerged in the fluid:
[tex]\[ V = \frac{53.0 \, \text{kg}}{1013 \, \text{kg/m}^3} \][/tex]
[tex]\[ V = \frac{53.0 \, \text{kg}}{1013 \, \text{kg/m}^3} \][/tex]
[tex]\[ V = 0.05236 \, \text{m}^3 \][/tex]
Now, we can use this volume to calculate the buoyant force:
[tex]\[ F_b = V \times \rho_{\text{fluid}} \times g \][/tex]
[tex]\[ F_b = 0.05236 \, \text{m}^3 \times 1013 \, \text{kg/m}^3 \times 9.8 \, \text{m/s}^2 \][/tex]
[tex]\[ F_b = 516.88 \, \text{N} \][/tex]
Therefore, the grouper must exert a force of approximately [tex]\( 516.88 \, \text{N} \)[/tex] to stay submerged in salt water.
___ cables are continually being improved to increase speed. the fastest cables under development today use multiple cores and/or multiple wavelengths to transmit data as fast as 255 tbps.
In general, the rate of evaporation increases with ___________. in general, the rate of evaporation increases with ___________. decreasing intermolecular forces decreasing volatility decreasing temperature decreasing surface area
In general, the rate of evaporation of liquid increases with increase in temperature and increases with increase in surface area of the liquid exposed to air. The rate of evaporation decreases with increasing of the intermolecular forces. The bigger intermolecular forces , the harder the molecules can escape from the liquid and become gas.
the length of a pipe will weigh the most when
If three devices are plugged into one port on a switch and two devices are plugged into a different port, how many collision domains are on the switch?
A howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 4.6 km. assume the acoustic output of a howler to be uniform in all directions and that the threshold of hearing is 1.0 × 10-12 w/m2. what is the acoustic power emitted by the howler?
You stand by the railroad tracks as a train passes by. you hear a 1 000-hz frequency when the train approaches, which changes to 800 hz as it goes away. how fast is the train moving? the speed of sound in air is 340 m/s.
Answer:
[tex]v_s= 37.8 m/s[/tex]
Explanation:
As per Doppler's effect when source and observer moves relative to each other then the frequency of the sound observed is different from the real frequency
When source is moving towards the stationary observer then we have
[tex]f_1 = f_o(\frac{v}{v - v_s})[/tex]
now when source of sound moving away from stationary observer then we have
[tex]f_2 = f_o(\frac{v}{v + v_s}[/tex]
now from above two equations
[tex]\frac{f_1}{f_2} = \frac{v + v_s}{v - v_s}[/tex]
here we know that
[tex]f_1 = 1000 hz[/tex]
[tex]f_2 = 800 hz[/tex]
v = 340 m/s
now we have
[tex]\frac{1000}{800} = \frac{340 + v_s}{340 - v_s}[/tex]
[tex]5(340 - v_s) = 4(340 + v_s)[/tex]
[tex]340 = 9 v_s[/tex]
[tex]v_s = 37.8 m/s[/tex]
A red blood cell is placed in a hypertonic solution. this means the concentration of solutes in the solution is ________ than the concentration of solutes in the intracellular fluid, and will cause the cell to ________.
A hypertonic solution has a higher concentration of solutes than the inside of a red blood cell. This causes water to move out of the cell, causing the cell to shrink or crenate.
Explanation:When a red blood cell is placed in a hypertonic solution, the concentration of solutes in the solution is higher than that in the intracellular fluid. As a result of this difference in solute concentration, water will move out of the cell and into the solution in a process called osmosis. This will cause the cell to shrink or create due to water loss.
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Water in the oceans may become fresh water available to humans through the processes of
Ocean water can be made into fresh water via desalination, a process which removes salt and minerals. This technology is essential in areas with scarce water supplies but is energy-intensive and contributes to environmental issues with its saline byproduct.
Water in the oceans can become fresh water available to humans primarily through the process of desalination. Desalination is a method that increases the amount of fresh water on Earth by removing dissolved salt and minerals from seawater or saline groundwater. Despite the virtually unlimited supply of saltwater, desalination techniques such as boiling, filtration, electrodialysis, and reverse osmosis are energy-intensive and costly. They also produce highly saline wastewater, which poses significant environmental challenges and must be managed appropriately. Desalination is a vital technology, especially in regions like the Middle East where water is scarce but energy resources are abundant.
When considering the water (hydrologic) cycle, it's important to note that while the Earth has an abundance of water, less than 1 percent of it is accessible fresh water. Through the water cycle, processes such as evaporation and condensation contribute to the natural distribution of water, but they do not desalinate it. For desalination to occur and produce fresh water for human use, technological intervention is required. The importance of these technologies cannot be overstated, as the availability of freshwater is critical to the survival of many living organisms, including humans.
If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, how far would it have fallen (in m) after 2 s? (g = 10 m/s2) 20.0
Neglecting air resistance, the rock would have fallen a distance of 20 meters after 2 seconds, calculated using the formula d = gt²/2 with g as the acceleration due to gravity (10 m/s²).
If you drop a rock from a tall building neglecting air resistance, the distance it has fallen after 2 seconds can be calculated using the formula for the distance fallen under constant acceleration, which is d = gt²/2, where g is the acceleration due to gravity (approximately 10 m/s²) and t is the time in seconds the object has been falling.
Plugging in the values, we get:
d = 10 m/s² × (2 s)² / 2 = 20 m
Therefore, the rock would have fallen 20 meters after 2 seconds.
The yellow wire connected to a power supply carries 12v. if the power supply provides 60w of power to the yellow wire, how much current is passing through the yellow wire?
If you observe an angular unconformity, you could infer that the region had experienced ________.
The rotating nozzle sprays a large circular area and turns with the constant angular rate theta overscript dot endscripts = 2.2 rad/s. particles of water move along the tube at the constant rate l overscript dot endscripts = 1.8 m/s relative to the tube. find the magnitudes of the velocity v and acceleration a of a water particle p when the distance l = 0.85 m if the angle beta = 64°.
The magnitude of the velocity of the water particle is approximately 2.6 m/s, and the magnitude of the acceleration of the water particle is approximately 7.5 m/s^2.
Explanation:The magnitudes of the velocity v and acceleration a of a water particle can be calculated using the concepts of angular velocity, and the relationships between linear and angular motion.
The velocity of the water particle v has two components: a tangential component due to the rotation (equal to the angular velocity times the distance from the center of rotation), and a radial component due to the flow of water along the tube.
The tangential velocity Vt = ωr, where r = 0.85 m is the distance of the water particle from the center of rotation and ω = 2.2 rad/s is the angular velocity. Substituting these values we find that Vt = 2.2 rad/s * 0.85 m = 1.87 m/s. The radial velocity Vr = l_dot = 1.8 m/s is given in the problem.
The total velocity is obtained by combining the radial and tangential velocities. Its magnitude is given by v = sqrt(Vr^2 + Vt^2). Substituting the known values, we find that the magnitude of the velocity v of the water particle is approximately 2.6 m/s.
The acceleration of the particle also has two components: a radial component equal to the square of the tangential velocity divided by the radius, and a tangential component equal to the product of the angular velocity and the radial velocity. Therefore, the magnitude of the acceleration a is given by a = sqrt((Vt^2 / r)^2 + (ωVr)^2), which gives us a value of approximately 7.5 m/s^2.
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The magnitude of the velocity [tex]v[/tex] of the water particle is approximately 2.6 m/s. The magnitude of the acceleration [tex]a[/tex] of the water particle is 4.114 m/s².
Velocity due to rotation (tangential velocity):
The tangential velocity [tex]v_{tangential}[/tex] is given by
[tex]v_{tangential} = l \cdot \dot{\theta}[/tex]
Substituting the values:
[tex]v_{tangential} = 0.85 \text{ m} \cdot 2.2 \text{ rad/s} = 1.87 \text{ m/s}[/tex]
Velocity along the tube:
The velocity along the tube [tex]v_{tube}[/tex] is given by [tex]\dot{l}[/tex], which is already provided as 1.8 m/s.
Resultant velocity:
The total velocity of the particle [tex]v[/tex] is the vector sum of the tangential and tube velocities. Since the angle [tex]\beta[/tex] between these two components is given, we can use the Pythagorean theorem:
[tex]v = \sqrt{( v_{tangential} )^2 + ( v_{tube} )^2}[/tex]
Substituting the values:
[tex]v = \sqrt{(1.87 \text{ m/s})^2 + (1.8 \text{ m/s})^2} \approx 2.6 \text{ m/s}[/tex]
Acceleration
Centripetal acceleration:
The centripetal (radial) acceleration [tex]a_c[/tex] is given by:
[tex]a_c = l \cdot \dot{\theta}^2[/tex]
Substituting the values:
[tex]a_c = 0.85 \text{ m} \cdot (2.2 \text{ rad/s})^2 = 4.114 \text{ m/s}^2[/tex]
Tangential acceleration:
The tangential acceleration [tex]a_t[/tex] will be zero because the angular velocity is constant (no angular acceleration).
Total acceleration:
The total acceleration [tex]a[/tex] is due to the centripetal acceleration alone, as there is no tangential component. Hence:
[tex]a = a_c = 4.114 \text{ m/s}^2[/tex]
Which expression is equivalent to 3m+1-m
Answer:
Th equivalent of [tex]3m+1-m[/tex] is [tex]2m+1[/tex]
Explanation:
We need to write the equivalent expression of [tex]3m+1-m[/tex]
first, combine the like terms,
[tex]3m-m+1[/tex]
take out the common variable,
[tex](3-1)m+1[/tex]
[tex]2m+1[/tex]
Therefore, the equivalent of [tex]3m+1-m[/tex] is [tex]2m+1[/tex]
Is it true or false that air pollution only occurs as a result of human activity
Answer:
This is false
Explanation:
The given statement "Air pollution only occurs as a result of human activity" is false because air pollution can occur as a result of both human activity and natural processes.
Human activities that contribute to air pollution include:
Burning fossil fuels: Burning fossil fuels, such as coal, oil, and natural gas, releases pollutants into the air. These pollutants include carbon dioxide, sulfur dioxide, nitrogen oxides, and particulate matter.Transportation: Transportation is another major source of air pollution. Cars, trucks, buses, and airplanes emit pollutants into the air, including carbon monoxide, hydrocarbons, and nitrogen oxides.Industry: Industrial activities, such as power plants, factories, and refineries, also emit pollutants into the air. These pollutants include sulfur dioxide, nitrogen oxides, and particulate matter.Natural processes that can contribute to air pollution include:
Volcanic eruptions: Volcanic eruptions can release large amounts of pollutants into the air, including sulfur dioxide, carbon dioxide, and particulate matter.Wildfires: Wildfires can also release large amounts of pollutants into the air, including carbon dioxide, particulate matter, and ozone.Dust storms: Dust storms can stir up dust and other particles into the air, which can contribute to air pollution.To know more about the Air pollution, here
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Calculate the work energy, w, gained or lost by the system when a gas expands from 15 l to 35 l against a constant external pressure of 1.5 atm
The work done by the gas is about 30 L.atm
[tex]\texttt{ }[/tex]
Further explanationThe Ideal Gas Law and Work formula that needs to be recalled is:
[tex]\boxed {PV = nRT}[/tex]
[tex]\boxed { W = P \Delta V }[/tex]
where:
W = Work ( J )
P = Pressure ( Pa )
V = Volume ( m³ )
n = number of moles ( moles )
R = Gas Constant ( 8.314 J/mol K )
T = Absolute Temperature ( K )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
initial volume = Vo = 15 L
final volume = V = 35 L
constant pressure = P = 1.5 atm
Asked:
word done by the gas = W = ?
Solution:
[tex]W = P \Delta V[/tex]
[tex]W = P \times ( V - Vo )[/tex]
[tex]W = 1.5 \times ( 35 - 15 )[/tex]
[tex]W = 1.5 \times 20[/tex]
[tex]W = 30 \texttt{ L.atm }[/tex]
[tex]\texttt{ }[/tex]
Conclusion:The work done by the gas is about 30 L.atm
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Pressure
The work energy, w, gained or lost by the system when the gas expands is 30 Joules.
Given the following data:
Pressure = 1.5 atmInitial volume = 15 litersFinal volume = 35 litersTo find the work energy, w, gained or lost by the system when the gas expands:
Mathematically, the work energy by a system is given by the formula:
[tex]Work = P \delta V[/tex]
Where:
P is the pressure.V is the volume.Substituting the given parameters into the formula, we have;
[tex]Work\;energy = 1.5(35-15)\\\\Work\;energy = 1.5(20)[/tex]
Work energy = 30 Joules.
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Why is it that the two possible faults wit inductors are short circuit and open circuit
Although the class has no water, the students want to know if the cube will float or sink in water. Explain, in detail, the steps that must be taken to determine if the cube will float or sink. Be sure to use the following key terms: triple beam balance, metric ruler ,and density.
Investigators are most likely to use the case history method when they study
Find the number of certifications held by people grouped by planet. this should have two columns the first, "name" will be the names of planets that have at least one certification. the second column should be "certcount" an
The tables are the following:
CREATE TABLE `bsg_cert` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB
CREATE TABLE `bsg_cert_people` (
`cid` int(11) NOT NULL DEFAULT '0',
`pid` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`cid`,`pid`),
KEY `pid` (`pid`),
CONSTRAINT `bsg_cert_people_ibfk_1` FOREIGN KEY (`cid`) REFERENCES `bsg_cert` (`id`),
CONSTRAINT `bsg_cert_people_ibfk_2` FOREIGN KEY (`pid`) REFERENCES `bsg_people` (`id`)
) ENGINE=InnoDB
CREATE TABLE `bsg_people` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`fname` varchar(255) NOT NULL,
`lname` varchar(255) DEFAULT NULL,
`homeworld` int(11) DEFAULT NULL,
`age` int(11) DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `homeworld` (`homeworld`),
CONSTRAINT `bsg_people_ibfk_1` FOREIGN KEY (`homeworld`) REFERENCES `bsg_planets` (`id`) ON DELETE SET NULL ON UPDATE CASCADE
) ENGINE=InnoDB
CREATE TABLE `bsg_planets` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`population` bigint(20) DEFAULT NULL,
`language` varchar(255) DEFAULT NULL,
`capital` varchar(255) DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `name` (`name`)
) ENGINE=InnoDB
Joining them all up doing a count with a group by should do the trick:
SELECT planet.name ,
COUNT(*) AS cert_count
FROM bsg_cert_people people_cert
JOIN bsg_people people ON people.id = people_cert.pid
JOIN bsg_planet planet ON people.homeworld = planet.id
GROUP BY planet.name
Or we can also use this syntax to get the same result:
SELECT pl.name, count(cert) AS "CertCount"
FROM bsg_planets pl
JOIN bsg_people pe ON pl.id = pe.homeworld
JOIN bsg_cert_people cp ON cp.pid = pe.id
GROUP BY pl.id
Which transmits the most light?
A. Tinted glass
B. Opaque glass
C. Clear glass
D. All types transmit light equally
Resistance is a measure of how well a material conducts an electric charge. Would you expect to find a high or low, level of resistance in a copper wire and why
The famous leaning tower of pisa doesn't topple over because its center of gravity is
A certain string on a piano is tuned to produce middle c (f = 261.63 hz) by carefully adjusting the tension in the string. for a fixed wavelength, what is the frequency when this tension is tripled?
The final frequency is 453.16 Hz when the tension is tripled.
We know that:
f = v / λ
Where,
f = frequency of sound wave.
v= speed of sound wave
λ = wavelength of sound wave
So, for a fixed wavelength ( let it be λ), frequency is directly proportional to velocity of the sound wave.
Again for a certain tension T, velocity of sound wave in a string is given by, v = √(T/μ) where, μ is mass per unit length of the string.
So, when this tension is tripled; velocity of sound wave becomes √3 times of its initial value and as frequency is directly proportional to the velocity, final frequency becomes = √3 × initial frequency
= √3 × 261.63 Hz. ( as given in the question, initial frequency = 261.63 Hz)
= 453.16 Hz.
Hence, when this tension is tripled, the frequency will be 453.16 Hz.
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The ____ button on the rehearsal toolbar clears the slide time box and resets the timer to 0:00:00.
Cube C1 has sides of length L, has an area A, and a volume V. Cube C2 has side that is 1/4 L. What is the area and the volume of cube C2?
The point of origin of an epileptic seizure is called the ____.â
Different units used to describe average velocity can be changed from one to another by the use of
If the atomic radius of lead is 0.175 nm, calculate the volume of its unit cell in cubic meters.
The volume of a unit cell of lead, which is a face-centered cubic, can be determined by calculating the edge length of the cell using its atomic radius and then cubing the edge length. The volume of a lead unit cell with an atomic radius of 0.175 nm is 1.205 * 10^-29 m^3.
Explanation:To calculate the volume of a unit cell of lead, we need to understand that lead crystalizes in a face-centered cubic lattice, where the edge length of the unit cell equals the square root of 2 times the atomic radius, doubled (since the diameter is 2 times the radius). So we first calculate the edge length, a = 2 * atomic radius * sqrt(2), and then calculate the volume of the unit cell, which is a cubic volume, V = a^3.
Given the atomic radius of lead is 0.175 nm or 0.175 * 10^-9 meters, we calculate:
Edge length, a = 2 * 0.175 * 10^-9 meters * sqrt(2) = 0.494 * 10^-9 meters.
The volume V = (0.494 * 10^-9 meters)^3 = 0.1205 * 10^-27 m^3 or 1.205 * 10^-29 m^3.
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Koi is climbing through a crevice and places her feet so that her body is perpendicular as shown below she then rest in this spot for twenty minutes
Answer:
static friction
Explanation:
she's not moving so she has static friction
A cyclist does work at the rate of 500 w while riding. how much force does her foot push with when she is traveling at 8.0 m/s?
Answer:
The foot push of the cyclist exerted a force of 63 Newton.
Explanation:
Given data
Power (P): 500 WattVelocity (v): 8.0 m/sForce (F): ?We can find the force exerted by the foot push of the cyclist using the following expression.
P = F × v
v = P/F = 500 W/ (8.0 m/s) = 63 N
The foot push of the cyclist exerted a force of 63 Newton.