The acceleration of a 300,000-kg jumbo jet with a thrust force of 120,000 N is calculated using Newton's second law of motion to be 0.4 m/s².
Explanation:The student has asked a Physics question related to calculating the acceleration of an object given its mass and the force applied to it. The subject of this question falls under Newton's second law of motion, which states that the acceleration (a) of an object is directly proportional to the net force (F) acting on it and inversely proportional to its mass (m), which can be represented by the equation a = F / m.
In the case of the jumbo jet with a mass of 300,000 kg experiencing a thrust force of 120,000 N, we can find the acceleration of the jet by using Newton's second law:
a = F / m = 120,000 N / 300,000 kg = 0.4 m/s²
The acceleration of the jumbo jet is 0.4 m/s² just before takeoff.
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]
The ____ button on the rehearsal toolbar clears the slide time box and resets the timer to 0:00:00.
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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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 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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What number does a bit that is turned on represent? what number does a bit that is turned off represent?
Imagine that you're standing in a large room when a loud noise is made. You begin hearing a series of echoes. Which characteristic of sound best explains what just happened?
Imagine that you're standing in a large room when a loud noise is made. You begin hearing a series of echoes. The characteristic of sound that best explains what just happened is: Reverberation.
Which transmits the most light?
A. Tinted glass
B. Opaque glass
C. Clear glass
D. All types transmit light equally
The famous leaning tower of pisa doesn't topple over because its center of gravity is
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?
Different units used to describe average velocity can be changed from one to another by the use of
The _________ specifies the station that sent the frame.
Ans: The source Address field specifies the station that sends the frame.
The format of an Ethernet frame includes a destination address at the beginning which contains the address of the device which is sending the frame. And, the source address tells us which station the information is received from.
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]
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
How many ohms of resistance are in a 120–volt hair dryer that draws 7.6 amps of current?
A. 16 ohms
B. 910 ohms
C. 0.063 ohms
D. 6,900 ohms
The concept Ohm's law is used here to determine the resistance. The resistance is found to be 16 ohm .The correct option is A.
What is Ohm's law?The relationship between the electric current and the potential difference is given by the Ohm's law. The current which flows through the conductors is directly proportional to the voltage applied. Mathematically the relationship is given as:
V = IR
V - Potential difference
R - Resistance
I - Current
R = V / I
R = 120 / 7.6
R = 15.7 ohm ≈ 16 ohm
The ohm's law holds true if the provided temperature and the other physical factors remain constant. In certain components, increasing the current raises the temperature. In this case Ohm's law is violated.
It is the Ohm's law which maintains the desired voltage drop across the electronic components. It helps to determine the voltage, resistance or current of an electric circuit.
Thus the correct option is A.
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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.
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]
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 point of origin of an epileptic seizure is called the ____.â
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.
how does bare soil surface type affect the amount of runoff?
the length of a pipe will weigh the most when
If you observe an angular unconformity, you could infer that the region had experienced ________.
___ 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.
When you measure the current and voltage in a ac circuit with a multimeter, is your result the peak value or the rms value?
Why is it that the two possible faults wit inductors are short circuit and open circuit
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?
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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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.
Investigators are most likely to use the case history method when they study