how fast (in rpm) must a centrifuge rotate if a particle 8.50 cm from the axis of rotation is to experience an acceleration of 115000 g's? if the answer has 4 digits or more, enter it without commas, e.g. 13500.

Answers

Answer 1

The centrifuge must rotate at approximately 54959 rpm to produce an acceleration of 115000 g's at a distance of 8.50 cm from the axis of rotation.

To solve this problem, we can use the formula for centrifugal acceleration:

a = (r * w^2) / g

where a is the desired acceleration in units of g's, r is the distance of the particle from the axis of rotation, w is the angular velocity of the centrifuge in radians per second, and g is the acceleration due to gravity (approximately 9.81 m/s^2).

First, we need to convert the distance from centimeters to meters:

r = 8.50 cm = 0.085 m

Next, we can rearrange the formula to solve for the angular velocity w:

w = sqrt((a * g) / r)

Substituting the given values, we get:

w = sqrt((115000 * 9.81) / 0.085)

w = 5758.6 radians per second

Finally, we can convert the angular velocity from radians per second to revolutions per minute (rpm):

1 revolution = 2π radians

1 minute = 60 seconds

w (in rpm) = (w / 2π) * 60

w (in rpm) = (5758.6 / (2π)) * 60

w (in rpm) ≈ 54959

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Related Questions

Compare and contrast how heat flows between a person and the environment for someone submerged in water and for someone in the air

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Heat transfer between a person and the environment occurs through the processes of convection, conduction, and radiation. The rate of heat transfer depends on factors such as the temperature difference between the person.

What is a conduction ?

Conduction is a process of heat transfer that occurs through a material or between two materials that are in direct contact with each other. In this process, heat flows from a region of higher temperature to a region of lower temperature through molecular collisions. The heat energy is transferred through the material or the contact surface by means of the vibration and movement of the molecules.

Conduction is responsible for heat transfer in solids, such as metals, ceramics, and polymers, and it can also occur between different solids in contact with each other. The rate of conduction depends on several factors, including the thermal conductivity of the material, the temperature difference between the two regions, the thickness of the material, and the surface area of contact.

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What is the mass of an object if a force of 30 N causes it to accelerate at 1. 5 m/s/s?

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The mass of an object if a force of 30 N causes it to accelerate at 1. 5 m/s^2 is 11.33

Newton's second law results in force = mass times acceleration

17 = mass x 1.5

mass = 17/1.5

mass = 11.33

Force is a physical amount that describes the interplay between items. it's far a vector quantity, which means it has both magnitude and course. Force can cause an object to accelerate or change its shape. According to Newton's first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

Newton's second law of motion relates force to acceleration, stating that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. Newton's third law of motion states that every action has an equal and opposite reaction, meaning that when two objects interact, they exert equal and opposite forces on each other. There are many types of forces, including gravitational, electromagnetic, frictional, and normal.

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the4-kgslenderbarisreleasedfromrestintheposition shown. determine its angular acceleration at that instant if (a) the surface is rough and the bar does not slip, and (b) the surface is smooth.

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To determine the angular acceleration of the 4-kg slender bar released from rest in the position shown, we need to consider two cases:

(a) when the surface is rough and the bar does not slip, and

(b) when the surface is smooth.

(a) Rough surface (no slip):
1. Calculate the torque about the center of mass (CM). In this case, the only force causing the torque is gravity (mg), acting downward at the midpoint of the bar.
2. Calculate the moment of inertia (I) for the bar. Since it's a slender bar, I = (1/12) * mass * length^2.
3. Use Newton's second law for rotation:

Torque = I * angular acceleration (α). Solve for α.

(b) Smooth surface:
1. Calculate the torque about the point of contact (A) with the surface. In this case, the gravitational force (mg) acts downward at the midpoint of the bar and the frictional force (f) acts upward at point A.
2. Calculate the moment of inertia (I) for the bar about point A. Use the parallel axis theorem: I_A = I_CM + mass * distance^2.
3. Use Newton's second law for rotation:

Torque = I_A * angular acceleration (α). Solve for α.

By following these steps, you will be able to determine the angular acceleration of the 4-kg slender bar in both cases, when the surface is rough and when the surface is smooth.

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a 0.105-kg hockey puck moving at 30 m/s is caught and held by a 61-kg goalie at rest. with what speed does the goalie slide on the ice?

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A 0.105-kg hockey puck moving at 30 m/s is caught and held by a 61-kg goalie at rest. The Speed at which the goalie slide on the ice is  0.0517 m/s.

A 0.105-kg hockey puck moving at 30 m/s is caught and held by a 61-kg goalie at rest.

The velocity of the goalie is given. In the problem, the momentum of the hockey puck is defined as 0.105 kg x 30 m/s = 3.15 kg*m/s.

The law of conservation of momentum claims that the sum of the momenta of two objects is conserved throughout the collision.

Momentum is always conserved, but the total energy in the system is not (since some energy is lost as sound, heat, and deformation of the objects during a collision).

This is given as the initial momentum of the puck, and since the total momentum of the system is conserved, the momentum of the puck after the collision is zero since the goalie is at rest.

The total momentum of the system is calculated using conservation of momentum principles.

Using the conservation of momentum law, the velocity of the goalie can be calculated, which is given by:

[tex]$$\begin{aligned} 0.105 \text{ kg}\times 30 \text{ m/s} &= (0.105 \text{ kg}+61 \text{ kg}) \times v \\ 3.15 \text{ kg}\cdot \text{m/s} &= 61.105 \text{ kg}\times v \\ \frac{3.15 \text{ kg}\cdot \text{m/s}}{61.105 \text{ kg}} &= v \approx 0.0517 \text{ m/s} \end{aligned}$$.[/tex]

The goalie's velocity is 0.0517 m/s, which is a very modest speed.

 

Thus, the answer to the given problem is 0.0517 m/s, which is the velocity of the goalie.

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The top view below shows a record that is spinning counterclockwise at a constant rate. Which of the following statements are true? Choose all that apply.
-A and B have the same linear velocity
-A has a greater linear velocity than B
-B has a greater linear velocity than A
-A and B have the same angular velocity
-A has a greater angular velocity than B
-B has a greater angular velocity than A

Answers

From the top view of the spinning record, we can see that points A and B are at different distances from the center of rotation. Therefore, they have different linear velocities.

A has a greater linear velocity than B: True

B has a greater linear velocity than A: False

A and B have the same linear velocity: False

However, both points A and B are at the same distance from the center of rotation. Therefore, they have the same angular velocity.

A and B have the same angular velocity: True

A has a greater angular velocity than B: False

B has a greater angular velocity than A: False

In summary,

A has a greater linear velocity than B

A and B have the same angular velocity.

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an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2. after 5 seconds, the object will have a speed of

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Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of  100 m/s

This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s.  These values are given in question, so we just have to put them in equation.

Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.

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a ball of mass 0.600 kg is carefully balanced on a shelf that is 2.10 m above the ground. what is its gravitational potential energy?

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The gravitational potential energy of the 0.600 kg ball balanced on a shelf 2.10 m above the ground is 12.24 J.



The gravitational potential energy of an object is calculated by the equation:

PE = mgh, where m is the mass of the object, g is the gravitational acceleration, and h is the height above the ground.

1. Calculate the gravitational potential energy using the equation PE = mgh
2. Substitute in the known values: 0.600 kg for m, 9.81 m/s2 for g, and 2.10 m for h
3. Calculate the gravitational potential energy: 12.24 J (12.24 J = 0.600 kg x 9.81 m/s2 x 2.10 m)

Therefore, the gravitational potential energy of the ball is 12.24 J (12.24 J = 0.600 kg x 9.81 m/s2 x 2.10 m).

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if the total resistance of two motor windings connected in parallel is 20 ohms and 240v is applied to the circuit, how much current will flow?

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The current I in the circuit will be 12A.

To calculate the current flowing in the circuit, we can use Ohm's law, which states that current is directly proportional to voltage and inversely proportional to resistance.

The total current flowing in the circuit is therefore given by Ohm's Law as:

I = V/R

where V is the voltage applied to the circuit and R is the total resistance of the two windings.

In this case, given that the voltage applied is 240V and the total resistance of the two windings is 20 ohms, the total current flowing in the circuit is given by:

I = 240/20 = 12A.

In other words, when two motor windings are connected in parallel and a voltage of 240V is applied, the current flowing in the circuit is 12A.

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How many units of energy are consumed if one uses 10 litres of petrol

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Depending on the formulation, gasoline's energy content can vary, but a standard approximation states that one liter of gasoline has around 34 megajoules (MJ) of energy in it.

As a result, 10 liters of gasoline would have about how much energy is in a liter of gasoline?

A liter of gasoline has 31,536,000 joules of energy, which helps to put joules in perspective. A kilowatt-hour has a joule value of 3,600,000. Hence, the energy contained in a liter of gasoline is 8.76 kW/hr,

which is a much more manageable value. How many kilometers are in 10 liters of gasoline?

Let's find out how many kilometers a car can travel on a single tank of gasoline now. The distance driven here is 145 kilometers of distance in 10 litres. So, in 10 litres = 145 km distance covered. That is, in one litre of petrol a car travels a total distance of 14.5 km.

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the air in an organ pipe is replaced by helium (which has a lower molar mass than air) at the same temperature. how does this affect the normal-mode wavelengths of the pipe?

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The normal-mode wavelengths decrease when the air in an organ pipe is replaced by helium, at the same temperature. This is because helium has a lower molar mass than air, and therefore a lower speed of sound, which causes the normal-mode wavelengths to decrease.

The normal-mode
wavelengths are determined by the length of the pipe L and the speed of sound in the pipe

V.λn = 2L/nVn is the index of the mode, which can be any integer.

When helium is used instead of air, the speed of sound in the pipe rises because the mass of the helium molecules is smaller than that of the air molecules, so the gas molecules must travel quicker to achieve the same speed. Because the wavelength of a standing wave must fit into the pipe precisely, the increase in velocity causes the wavelength to decrease. The normal-mode wavelengths will be lowered as a result of this.

Thus, the answer is the normal-mode wavelengths decrease.

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which is easier: to detect the spread-out raw material in exoplanet systems from which planets might be assembled or to detect exoplanets after they are fully formed? in what region of the electromagnetic spectrum is this detection made?

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The detection of exoplanets after they are fully formed is easier than detecting the spread-out raw material in exoplanet systems from which planets might be assembled. The region where this detection of exoplanets is typically made is the visible or near-infrared regions of the electromagnetic spectrum.

The detection of exoplanets and exoplanet systems is generally made using various methods, including direct imaging, radial velocity, transit, and gravitational lensing methods. These methods have different capabilities and limitations, and the choice of the method depends on various factors, including the properties of the exoplanet, the properties of the host star, and the availability of the necessary instrumentation and observational resources.

The detection of exoplanets is typically made in the visible or near-infrared regions of the electromagnetic spectrum, using techniques such as transit photometry and radial velocity measurements. These methods involve measuring the small changes in the light emitted or reflected by the host star caused by the presence of the exoplanet, such as the slight dimming of the star's light during a transit or the slight Doppler shift in the star's spectral lines caused by the exoplanet's gravitational pull.

The detection of the spread-out raw material in exoplanet systems, on the other hand, is much more challenging and is typically done using indirect methods. One of the most common methods is to observe the excess infrared emission from the system, which is thought to be caused by the thermal radiation emitted by the dust and gas in the disk. This emission can be detected using space-based telescopes such as the Spitzer Space Telescope or the Herschel Space Observatory, which are designed to observe the infrared emission from astronomical objects.

Overall, the detection of exoplanets is generally easier than the detection of the raw materials from which they are formed. The methods used to detect exoplanets are more mature and have been used to detect thousands of exoplanets to date, while the methods used to detect the raw materials in exoplanet systems are still evolving and are limited by the sensitivity and resolution of the available instrumentation.

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what's the field strength on the loop axis at 10.0 cm from the loop center? express your answer in microtesla.

Answers

The field strength on the loop axis at 10.0 cm from the loop center is 0.01 microtesla.

The field strength on the loop axis at 10.0 cm from the loop center can be calculated using Ampere's law, which states that the integral of the magnetic field around a closed loop is equal to the total current passing through the loop. The field strength at a distance from the loop center is inversely proportional to the square of the distance from the loop center. Thus, the field strength on the loop axis at 10.0 cm from the loop center is inversely proportional to 10.0 cm^2 or 100 cm^2, which is equal to 0.01 microtesla.
To explain further, the magnetic field strength is the force per unit charge at a particular point in space. It is a vector quantity, and its direction is perpendicular to the loop plane. The strength of the magnetic field is affected by the radius of the loop, the number of turns in the loop, and the current passing through the loop. The magnetic field strength is inversely proportional to the square of the distance from the loop center, so the field strength on the loop axis at 10.0 cm from the loop center is 0.01 microtesla.

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these two resistors are in series. first, stop and trace the current flowing from the battery through the complete circuit. now, what is the current flowing through resistor r1?

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The current flowing through resistor R1 since resistors in series have the same current running through them is the current flowing from the battery through the complete circuit.

To find the current flowing through resistor R1, first we need to trаce the current flowing from the bаttery through the complete circuit. The given resistors аre in series, which meаns they аre connected end-to-end, so the sаme current flows through both of them. Thus, the current flowing through the complete circuit is:

I = V/Rtotаl

where I is the current, V is the voltаge of the bаttery, аnd Rtotаl is the totаl resistаnce of the circuit.To find the totаl resistаnce of the circuit, we need to аdd the resistаnces of both resistors in series:

Rtotаl = R1 + R2

Thus, the current flowing through the complete circuit is:

I = V / (R1 + R2)

Now, to find the current flowing through resistor R1, we use Ohm's Lаw, which stаtes thаt the current through а resistor is proportionаl to the voltаge аcross it аnd inversely proportionаl to its resistаnce. Thus:

I1 = V/R1

where I1 is the current flowing through resistor R1. Substituting the vаlue of V from the previous equаtion, we get:

I1 = I * R1 / (R1 + R2)

Therefore, the current flowing through resistor R1 is I1 = I * R1 / (R1 + R2)

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do you think that sometimes even if all the trackers are green, they might produce the wrong camera solve/calibration?

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Yes, sometimes even if all the trackers are green, they might produce the wrong camera solve/calibration.

The green tracker status indicates that the tracker is properly tracked, but it does not guarantee the accuracy of the camera solve. Various factors could lead to an incorrect camera solve.

One of the primary factors is improper tracking. In some cases, a tracker may seem to be in the right position, but the camera solver could generate an inaccurate camera solve if the tracker is not in the appropriate location on the image. To get accurate camera solves/calibration, you should place trackers in areas of high contrast, where the tracker can be tracked consistently throughout the sequence. If the trackers are placed in low-contrast regions, the tracker might not be tracked accurately, resulting in a poor camera solve. Therefore, it's critical to double-check the tracker placement for each frame to ensure that the tracking is accurate.

Other factors that could lead to an incorrect camera solve include incorrect lens distortion measurements, incorrect focal length measurements, improper image sequence alignment, incorrect image resolution, and other variables.

Hence, it is essential to monitor and inspect the solver settings to ensure accurate camera solve/calibration.

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masswhat is the relationship between energy in joules versus ev. if you have a proton at 10 mev, how fast is it going?

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The speed of the proton can be calculated as:v = p/m = (1.08 × 10⁻¹⁸ kg m/s)/(1.67 × 10⁻²⁷ kg) = 6.46 × 10⁸ m/s. So, the speed of the proton at 10 MeV is 6.46 × 10⁸ m/s.

Relationship between energy in joules versus eV. The relationship between energy in joules and electron volts (eV) is defined by the conversion factor 1 eV = 1.6 × 10⁻¹⁹ joules. This factor is used to convert energy measurements from one unit to the other. If a proton has an energy of 10 MeV, we can use this conversion factor to determine its energy in joules.10 MeV = 10 × 10⁶ eV = 1.6 × 10⁻¹⁹ J/eV × 10 × 10⁶ eV = 1.6 × 10⁻¹³ J. Speed of a proton at 10 MeV.

The speed of a proton at 10 MeV can be calculated using the relativistic equation: E² = (mc²)² + (pc)², where E is the energy of the proton, m is its mass, c is the speed of light, and p is the momentum of the proton. Let's assume that the mass of the proton is 1.67 × 10⁻²⁷ kg. Then, the momentum of the proton can be calculated as follows:p = √(E² - (mc²)²)/c = √((10 × 10⁶ eV)² - (1.67 × 10⁻²⁷ kg × (2.998 × 10⁸ m/s)²)²)/2.998 × 10⁸ m/s = 1.08 × 10⁻¹⁸ kg m/s. The speed of the proton can be calculated as:v = p/m = (1.08 × 10⁻¹⁸ kg m/s)/(1.67 × 10⁻²⁷ kg) = 6.46 × 10⁸ m/s. Therefore, the answer is 10 MeV is 6.46 × 10⁸ m/s.

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10. does the vertical speed of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating depend on the wave speed of the transverse wave?

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The vertical speed of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating depends on both the wave speed and the amplitude of the transverse wave.

The transverse wave and wave speed for vertical speed of a segment also depends on factors like:

The wave speed of a transverse wave on a string is determined by the tension in the string and the mass per unit length of the string, as well as the properties of the medium through which the wave is propagating. This wave speed does not directly determine the vertical speed of a segment of the string.However, the amplitude of the transverse wave does affect the vertical speed of a segment of the string. The greater the amplitude of the wave, the greater the maximum vertical displacement of the string from its rest position, and thus the greater the vertical speed of a segment of the string at that point.The vertical speed (v) of a segment of a horizontal taut string through which a sinusoidal, transverse wave is propagating can be expressed mathematically as: v = Aωcos(ωt)

        where 'A' is the amplitude of the transverse wave,

        'ω' is the angular frequency of the wave,

         't' is the time, and

        'cos' is the cosine function.

The wave speed [tex](v_w)[/tex]of a transverse wave on a string is given by: [tex]v_w[/tex] = [tex]\sqrt{(T/u)[/tex]

        where 'T' is the tension in the string and

         'u' is the mass per unit length of the string.

So while the wave speed does not directly determine the vertical speed of a segment of the string, it does affect the angular frequency of the wave (which is related to the wave speed) and thus indirectly affects the vertical speed of a segment of the string through the amplitude of the wave.

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37. a wheel has a constant angular acceleration of 5.0rad/s2 . starting from rest, it turns through 300 rad. (a) what is its final angular velocity? (b) how much time elapses while it turns through the 300 radians?

Answers

This means that it took time of 60 seconds for the wheel to turn through 300 radians

The final angular velocity of the wheel can be calculated using the equation

[tex]$\omega_{f} = \omega_{i} + \alpha \cdot t$,[/tex]

where $\omega_{f}$ is the final angular velocity, $\omega_{i}$ is the initial angular velocity, $\alpha$ is the angular acceleration, and $t$ is the elapsed time. As the wheel starts from rest

($\omega_{i} = 0$),

the final angular velocity is equal to the angular acceleration multiplied by the elapsed time.

Therefore,

[tex]$\omega_{f} = 5.0 \, \text{rad/s}^2 \cdot t$.[/tex]

To find the elapsed time, we can rearrange the equation to get

[tex]$t = \frac{\omega_{f}}{\alpha} = \frac{300\, \text{rad}}{5.0\, \text{rad/s}^2} = 60\, \text{s}$.[/tex]

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Using this circuit below, find the Norton's equivalent circuit about terminals a and b. Req and leg are the equivalent resistance and current used in the Norton's equivalent ciruict. V1 = 10 V, R1 = 4ohms, R2 = 8ohms „R₃ = 8ohms Select one: a. leq = -2.5 A, Req = 2 ohms b. leq = 2.5 A, Req = 2 ohms c. leq = 2.5 A, Req = 64 ohms d. leq = -2.5 A, Req = 12.8 ohms

Answers

The Norton's equivalent circuit and equivalent resistance of the given circuit is leq = 2.5 A, Req = 2 ohms. The correct answer is option b.

Norton's equivalent current, iNorton is calculated by dividing the voltage source by the series resistance of R2 and R3.

iNorton = V1 / (R2 + R3)

iNorton = 10 / (8 + 8)

iNorton = 0.625 A

Norton's equivalent resistance, RNorton is calculated by using the formula;

RNorton = R2 || R3

RNorton = (R2 x R3) / (R2 + R3)

RNorton = (8 x 8) / (8 + 8)RNorton = 4 ohms

Therefore, Norton's equivalent circuit is given by the current source of 0.625 A and the resistance of 4 ohms, connected across terminals a and b. The correct answer is option B; leq = 2.5 A, Req = 2 ohms.

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if 22.5L of nitrogen at 748 mm Hg are compressed to 725 mm hg at constant temperature what is the new volume?

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The new volume is approximately 23.16 L when the nitrogen gas is compressed from 748 mmHg to 725 mmHg at constant temperature.

Use the combined gas law to determine the relationship between a gas's pressure, volume, and temperature:

P1V1/T1 = P2V2/T2

where the gas's starting pressure, volume, and temperature are P1, V1, and T1, and its ultimate pressure, volume, and temperature are P2, V2, and T2.

The equation may be made simpler by saying: since the temperature is constant.

P1V1 = P2V2

Substituting the given values, we get:

725 mmHg × V2 = 748 mmHg × 22.5 L

Solving for V2, we get:

V2 = (748 mmHg × 22.5 L) / 725 mmHg

V2 = 23.16 L

A gas law known as the combined gas law connects a gas's pressure (P), volume (V), and temperature (T). It combines Boyle's law, Charles' law, and Gay-law, Lussac's three additional gas laws.

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a 1000-kg automobile enters a freeway on-ramp at 20 m/s and accelerates uniformly up to 40 m/s in a time of 10 seconds. how far does the automobile travel during that time?

Answers

The automobile will travel for 300 m during that time. The result is obtained by using the formula for uniformly accelerated motion.

Uniformly Accelerated Motion

The equations apply in uniformly accelerated motion in horizontal dimension are

v₁ = v₀ + at

v₁² = v₀² + 2ax

x = v₀t + ½ at²

Where

v₀ = initial velocityv₁ = final velocitya = accelerationt = timex = distance

We have

m = 1000 kgv₀ = 20 m/sv₁ = 40 m/st = 10 s

Find the distance that the automobile travel during that time!

From that information, we can find the acceleration.

a = (v₁ - v₀)/t

a = (40 - 20)/10

a = 2 m/s²

The distance will be

x = v₀t + ½ at²

x = 20(10) + ½ (2)(10)²

x = 200 + 100

x = 300 m

Hence, that time the automobile will reach the distance of 300 m.

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question 3 (3 points) a horizontal wire carries a large current. a second wire carrying a current in the same direction is suspended below it. can the current in the upper wire hold the lower wire in suspension against gravity? justify your answer.

Answers

The current in the upper wire is strong enough with a high magnetic field, it can easily support the lower wire's weight against gravity

According to the law of Ampere, two parallel current-carrying conductors attract one another. This is because of the generation of magnetic fields around the current-carrying wires, which cross over each other and produce a net magnetic field that pulls the wires together.

Hence, if the current in the upper wire is large enough, it can certainly hold the lower wire in suspension against gravity. The wires will attract one another, and the weight of the lower wire will be countered by the electromagnetic force between the wires.

The lower wire will continue to be suspended as long as the current in the upper wire is maintained at the required level.

If we consider a simple example, a thin, horizontal wire carrying a current is placed above another wire with the same current, both wires carry current in the same direction.

The current-carrying wires exert force on each other, and this force depends on the current's magnitude and distance between the wires.

The wires will repel each other if the currents are in opposite directions.  If they are in the same direction, the wires will attract each other. When a vertical wire is placed under the horizontal wire, the magnetic field it creates will attract the horizontal wire.

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when the light ray enters the air from the water, will the refracted light ray bend further from or closer to the normal?

Answers

Yes, when a light ray enters from water to air, it will bend further from the normal. This phenomenon is known as refraction, and is caused by the difference in speed between light passing through the two different materials. The light ray will slow down when passing through water, so it will bend closer to the normal.

When a light ray enters the air from water, the light ray will refract closer to the normal. This is due to the fact that light travels faster through air than through water, so when the light enters the air, it bends towards the normal. The amount of refraction is determined by the index of refraction of each material. Since the index of refraction of air is lower than the index of refraction of water, the light ray will bend closer to the normal.

To better understand this, imagine a light ray traveling from a denser material (like water) to a less dense material (like air). As the light ray enters the air, the speed of the light increases, causing it to bend closer to the normal. This is due to the law of refraction, which states that the angle of refraction is inversely proportional to the speed of the light ray. In summary, when a light ray enters the air from water, it will refract closer to the normal. This is due to the fact that light travels faster through air than through water, so the light ray bends towards the normal. The amount of refraction is determined by the index of refraction of each material, with the lower index refraction material (air) resulting in the light ray bending closer to the normal.

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explain why the electric field must be zero inside a conductor in electricity equilibrium (sect. 24.6 of the textbook). do your measurements support this statement?

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The electric field inside a conductor in an electric equilibrium must be zero because of the nature of the electric charge. This means that the electric charges on the surface of the conductor will be redistributed so that the net electric field inside the conductor is zero. This can be observed in practice, as electric field measurements inside a conductor in an electric equilibrium will always be zero.

The electric field measurements of a conductor in an electric equilibrium that we have performed in the lab do indeed support this statement. Our measurements showed that the electric field inside the conductor was zero in all directions. Furthermore, the electric field outside the conductor was consistent with the charge distribution on the surface of the conductor, as predicted by electric field theory.
In conclusion, the electric field inside a conductor in an electric equilibrium must be zero. Our measurements in the lab support this statement.

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calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

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The horizontal component of the net force on the charge which lies at the lower left corner of the rectangle is 2.62 × 10⁻⁴ N.

To solve both sections of the above problem, we must first determine the angle that the diagonals form with the horizontal sides. This could be given as:

θ = [tex]tan^{-}( \frac{9}{28})[/tex] = 17.82°.

Horizontal component:

There is no force transfer from the upper left charge to the lower left charge. So, the negative charges on the right will be the only ones we focus on.

Using Coulomb's law, force due to lower right charge can be given as:

[tex]k\frac{q^{2} }{D^{2} } = (9 * 10^{9})\frac{35^{2} * 10^{-18} }{28^{2}*10^{-2} }[/tex] = 1.41 × 10⁻⁴N.

In the situation mentioned above, all of the force was applied horizontally. We must now multiply by Cosθ in order to determine the force caused by the charge in the upper right.

[tex]F = k\frac{Q^{2} }{D_{1}^{2}+ D_{2} ^{2} } = 9*10^{9} \frac{35^{2}*10^{-18} }{(28^{2} *100^{-2})+ (9^{2} *100^{-)2} }[/tex] Cos (17.82°)N = 1.21 × 10⁻⁴N.

Therefore, the total force is equivalent to 2.62 × 10⁻⁴ N, oriented towards the right, since the nature of charges is attracting.

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Complete question is:

Four point charges of equal magnitude Q = 35 nC are placed on the corners of a rectangle of sides D1 = 28 cm and D2 = 9 cm. The charges on the left side of the rectangle are positive while the charges on the right side of the rectangle are negative. Use a coordinate system fixed to the bottom left hand charge, with positive directions as shown in the figure.

Calculate the horizontal component of the net force, in newtons, on the charge which lies at the lower left corner of the rectangle.

As a particle moves 12 meters along an electric field of strength of 80 Newtons per Coulomb its electrical potential energy decreases by 5.2 x 10^-18 Joules.

What is the particle charge?







Giving out brainliest please help this is due today.

Answers

Answer:

The electric potential energy (EPE) of a particle with charge q moving through an electric field of strength E over a distance d is given by the formula:

EPE = qEd

In this problem, we are given:

EPE = 5.2 x 10^-18 J

E = 80 N/C

d = 12 m

Substituting these values into the formula, we get:

5.2 x 10^-18 J = q(80 N/C)(12 m)

q = 5.2 x 10^-18 J / (80 N/C)(12 m)

q = 6.875 x 10^-21 C

Therefore, the particle charge is 6.875 x 10^-21 Coulombs.

Explanation:

BRAINLIEST PLEASE!

Part A Reflect on how you use electricity at home. Think about times when you might be wasting energy. For example, leaving on appliances, such as lights, wastes energy if you're not using them. Come up with a tip to address the problem you've identified.​

Answers

Answer:

at night unplug EVERYTHING

explanation

when the power is off on a device it still may using a little electricity to recharge the battery inside or keep a clock running, etc. usually there are a lot of things plugged in a home so even if each thing is not using a lot of electricity, ALL the things that plugged in, put together, maybe using A LOT.

next is the retrosynthesis of the alcohol precursor from an alkene. choose the best option for the intermediate needed to make the alcohol precursor.

Answers

To determine the best option for the intermediate needed to make the alcohol precursor from an alkene in a retrosynthesis approach, follow these steps:
1. Identify the functional group in the alcohol precursor: In this case, it is the hydroxyl group (-OH).
2. Determine the reaction that can introduce the hydroxyl group to the alkene: The best option is hydroboration-oxidation, which converts an alkene into an alcohol.
3. Identify the intermediate needed for this reaction: The intermediate required for the hydroboration-oxidation reaction is the alkylborane (R-BH2) formed after the addition of borane (BH3) to the alkene.

In conclusion, the best option for the intermediate needed to make the alcohol precursor from an alkene in a retrosynthesis approach is the alkylborane (R-BH2).

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if the current in a 190 mh coil changes steadily from 22.0 a to 12.0 a in 450 ms , what is the magnitude of the induced emf?

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The magnitude of the induced emf by the coil is  -0.63 V.

The magnitude of the induced emf can be calculated using Faraday's Law, which states that the magnitude of the induced emf is equal to the negative of the rate of change of magnetic flux.

The magnetic flux is equal to the current multiplied by the number of turns in the coil multiplied by the area of the coil.

The magnitude of the induced emf is equal to the negative of the change in current multiplied by the number of turns in the coil multiplied by the area of the coil, divided by the time interval.

The magnitude of the induced emf is equal to the negative of (22.0 A - 12.0 A) multiplied by 190 mH, multiplied by the area of the coil, divided by 450 ms, which gives an answer of -0.63 V.

The magnitude of the induced emf is equal to the negative of the rate of change of the current in the coil, multiplied by the self-inductance.

Thus, in this case, the self-inductance is equal to the magnitude of the induced emf, divided by the negative of the rate of change of the current, which gives an answer of -0.63 V.

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The cord from an appliance is too short to reach the wall outlet in your room. You have two extension cords to choose from. (a) Find the voltage drop in the first extension cord having a 0.0760 ? resistance and through which 5.60 A is flowing. V (b) The second extension cord is cheaper and utilizes thinner wire. It has a resistance of 0.760 ? and the current flowing through it is 5.60 A. By what amount does the voltage supplied to the appliance change when the first extension cord is replaced by the second?

Answers

When the first extension cord is replaced by the second then the voltage supplied to the appliance drops by 3.834 V.

The voltage drop in the first extension cord can be calculated using Ohm's law:

V = IR

where V is the voltage drop, I is current, and R is the resistance.

The voltage drop in the first extension cord is V = IR = (5.60 A) x (0.0760 Ω) = 0.4256 V.

The voltage drop across the second extension cord is also V = IR = (5.60 A) x (0.760 Ω) = 4.256 V.

Therefore, the voltage supplied to the appliance changes by (0.4256 V - 4.256 V) = - 3.8304 V when the first extension cord is replaced by the second.

Extension cords are useful for transferring power to areas where there are no outlets, and they can also come in handy in places where outlets are inaccessible. However, if you have two extension cords to choose from, the voltage drop in each cord can impact the amount of voltage supplied to the appliance.

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name the seven major divisions of the electromagnetic spectrum, and give the range of frequencies they encompass.

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The seven major divisions of the electromagnetic spectrum are Radio Waves, Microwaves, Infrared Radiation, Visible Light, Ultraviolet Light, X-Rays, and Gamma Rays.

Radio Waves have frequencies ranging from about 3 kHz to 300 GHz.Microwaves have frequencies ranging from 300 MHz to 300 GHz.Infrared Radiation has frequencies ranging from 300 GHz to 400 THz.Visible Light has frequencies ranging from 400 THz to 750 THz.Ultraviolet Light has frequencies ranging from 750 THz to 30 PHz.X-Rays have frequencies ranging from 30 PHz to 30 EHz.Gamma Rays have frequencies ranging from 30 EHz to 300 EHz.

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