ben(55 kg) is standing on very slippery ice when junior(25kg) bumps into him. junior was moving at a speed of 8m/s before the collision and ben and junior embrace after the collision. find the speed of ben and junior as they move across the ice after the collision .

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.

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.

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 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.

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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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.

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]

Answer:

The foot push of the cyclist exerted a force of 63 Newton.

Explanation:

Given data

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 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 work done by the gas is about 30 L.atm

[tex]\texttt{ }[/tex]

The 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]

The work done by the gas is about 30 L.atm

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: 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:

To 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:

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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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]

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

Answer:

static friction

Explanation:

she's not moving so she has static friction

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.

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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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.

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:

Natural processes that can contribute to air pollution include:

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