Q:Why is the process of nitrogen fixation necessary for life on earth?

The angular acceleration of the flywheel when it has rotated 20 radians is not provided in the question.

To determine the angular acceleration, we need more information or an equation that relates the angular velocity (ω) and the angular acceleration (α). Without additional information, we cannot calculate the angular acceleration directly.

However, if we assume that the angular acceleration is constant, we can use the following equation:

ω = ω₀ + αt,

where

ω is the final angular velocity,

ω₀ is the initial angular velocity,

α is the angular acceleration,

and t is the time taken.

Since the initial angular velocity (ω₀) is not given in the question, we cannot proceed with this equation.

Without further information or equations relating the angular velocity and angular acceleration, it is not possible to determine the angular acceleration when the flywheel has rotated 20 radians.

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Answer:

a region around a magnetic material or a moving electric charge within which the force of magnetism acts.

Explanation:

The definition of a magnetic field is a place in space near a magnet or an electric current where a physical field is created from a moving electric charge that creates a force on another moving electric charge. An example of a magnetic field is the Earth's magnetic field.

Answer:

speeding up

Explanation:

because its speeding up, theres going to be more newtons in the back

i really hope this is right, tell me if so

The distance from the Earth's center to the Lagrange point L1 is approximately 326,225 km.

To determine the point where the gravitational forces of the Earth and the Moon cancel each other, you can use the concept of the Lagrange point, specifically L1. At this point, the gravitational forces from both bodies are equal and opposite, causing them to effectively cancel each other out.
To find the distance from the Earth's center, you can use the following formula:
\(d = (R * (Mm / (Mm + Me))^{1/3})\)
where d is the distance from the Earth's center, R is the distance between the centers of the Earth and the Moon (384,400 km), Mm is the mass of the Moon (7.342 × \(10^{22}\) kg), and Me is the mass of the Earth (5.972 × \(10^{24}\) kg).
Using this formula, the distance d from the Earth's center to the L1 point where the gravitational forces cancel is approximately 326,284 km.

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Answer:

B is the correct answer to the order

Yes, it is possible for both the pressure and volume of a monatomic ideal gas to change without causing the internal energy of the gas to change. This occurs when the gas undergoes an adiabatic process, meaning there is no heat transfer between the gas and its surroundings.

In an adiabatic process, the change in internal energy (ΔU) is solely dependent on the work done on or by the gas (W). According to the first law of thermodynamics, ΔU = Q + W, where Q is the heat transfer. Since Q = 0 in an adiabatic process, ΔU = W.

For the internal energy of the gas to remain constant (ΔU = 0), the work done on or by the gas must also be zero. This can be achieved through a specific path in the pressure-volume (PV) diagram, where the gas expands and does work on its surroundings, followed by compression, with the surroundings doing an equal amount of work on the gas. The net work done over this process will be zero, ensuring the internal energy remains unchanged.

In summary, it is possible for both the pressure and volume of a monatomic ideal gas to change without affecting its internal energy by undergoing an adiabatic process with zero net work done.

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Like we measure heat with a thermometer, temperature is the primary indicator of heat. By using a thermometer, we can quantify the thermal energy present in a system.

Regarding the layers of the Earth's atmosphere, the bottom layer is known as the troposphere. It extends from the Earth's surface up to an average altitude of about 12 kilometers (7.5 miles) at the poles and 18 kilometers (11 miles) at the equator. The troposphere is where weather phenomena occur, and it contains approximately 90% of the air in the atmosphere. This layer is crucial for supporting life and plays a significant role in regulating Earth's climate through its interaction with the Earth's surface and the transfer of heat and moisture.

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Answer:

Maybe A. waving your arms?

The space probe launched in 2012 with 2.0 kg of plutonium-238 (238Pu) is utilized for thermoelectric power sources in spacecraft.

Plutonium-238 (238Pu) is an isotope of plutonium that undergoes radioactive decay, emitting heat in the process.

This unique property makes it an ideal choice for generating power in space missions where sunlight is limited, such as deep space probes or missions to distant planets. The heat produced by the radioactive decay of 238Pu is converted into electricity using thermoelectric materials.

In the context of the space probe launched in 2012, the 2.0 kg of 238Pu serves as the fuel for the thermoelectric power source.

The heat generated by the decay of the plutonium is harnessed to produce electricity through the Seebeck effect.

Thermocouples, made from two dissimilar materials, are used to create a temperature gradient. As the heat flows across the junction of the thermocouple, it creates a voltage difference that can be utilized to power the spacecraft's instruments, systems, and communication devices.

The use of 238Pu as a power source offers several advantages for space missions.

Unlike solar panels, which are dependent on sunlight, thermoelectric generators powered by plutonium-238 can operate in deep space or in regions where solar energy is insufficient.

This is particularly crucial for missions that venture beyond the orbit of Mars or explore dark, shadowed areas where sunlight is scarce.

Additionally, the longevity of 238Pu's decay heat allows for prolonged power generation, ensuring continuous operation and data transmission over long-duration missions.

Plutonium-238 (238Pu) is a scarce and highly valuable resource due to its applications in space exploration. It is primarily produced through the irradiation of neptunium-237 in nuclear reactors.

The production and handling of 238Pu require strict safety measures due to its high radioactivity. Furthermore, the dwindling global supply of 238Pu has posed challenges for future space missions relying on this isotope.

The development of alternative power sources and the search for innovative ways to produce and utilize plutonium-238 remain areas of active research in the field of space exploration.

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Explanation:

modern quantum mechanics explains these periodic trends in properties in terms of electron shells. the filling of each shell corresponds to a row in the table

Certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour it is observed that 10 percent of the material has decayed, the half-life of the material is 10.58 hrs.

We can use the formula for exponential decay, which states that the amount of material remaining after time t is given by:

N(t) = \(N0 e^{(-kt)}\)

where N0 is the initial amount of material, k is the decay constant, and e is the base of the natural logarithm.

If 10% of the material has decayed after one hour, then the remaining amount of material is 90% of the initial amount, or N(1) = 0.9 N0.

These values are entered into the exponential decay equation to produce the following results:

0.9 N0 = \(N0 e^{(-k)}\)

We can divide both sides by N0 to make things simpler:

0.9 = \(e^{(-k)}\)

After calculating the natural logarithm of both sides, we arrive at:

ln(0.9) = -k

Solving for k, we get:

k = -ln(0.9)

The half-life is the time it takes for the amount of material to decrease by half. Let's call this time T. Then, we can write:

N(T) = 0.5 N0

Substituting into the exponential decay equation, we get:

0.5 N0 = \(N0 e^{(-kT)}\)

We can divide both sides by N0 to make things simpler:

0.5 = \(e^{(-kT)}\)

If we take the natural logarithm of both sides, we obtain:

ln(0.5) = -kT

When we replace the value of k we discovered earlier, we obtain:

ln(0.5) = ln(0.9) T

Solving for T, we get:

T = ln(2) / ln(0.9)

Using a calculator, we find:

T ≈ 10.58

Therefore, the half-life of the material is approximately 10.58 hours. Answer: (c)

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When a body is thrown downwards, it falls freely under gravity.

The question is incomplete as the values are missing but I will try to help you the much I can. We must note that the acceleration of a body refers to the rate at which its velocity changes per unit time.

When an object is thrown down from a height. It gains speed due to acceleration due to gravity which acts on the body as it falls freely.

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Answer:

electrons

Explanation:

i think this is right

To explain the tangent plane to the surface, `z = 53 − x² − 2y²` at the point `(3, 2, 6)`, let us first determine the partial derivatives of `z` with respect to `x` and `y`.

Partial derivative of `z` with respect to `x`, `∂z/∂x = -2x`Partial derivative of `z` with respect to `y`, `∂z/∂y = -4y`Now, let's find the gradient vector `grad z` at `(3, 2, 6)` and the value of `z` at `(3, 2)`.gradient vector `grad z = (-2x, -4y, 1)`gradient vector `grad z = (-6, -8, 1)` at `(3, 2, 6)`.Value of `z` at `(3, 2)` is given by `z = 53 - 3² - 2(2)² = 39`.

Therefore, the equation of the tangent plane to the surface `

z = 53 − x² − 2y²` at `(3, 2, 6)` is:

`z - 6 = -6(x - 3) - 8(y - 2)`

which can be written as:`6x + 8y + z = 50`Thus, the equation of the tangent plane to the surface `z = 53 − x² − 2y²` at the point `(3, 2, 6)` is `6x + 8y + z = 50`.

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Energy released ,

\( = 3.789 \times {10}^{ - 12} \) J

The amount of heat absorbed when 5.50 g of aluminum is heated from 25.0°C to 95.0°C is approximately 272.07 J.

When 5.50 g of aluminum is heated from 25.0°C to 95.0°C, it absorbs approximately 272.07 J of heat. This amount of heat absorbed can be calculated using the formula Q = m * c * ΔT, where Q represents the amount of heat, m is the mass of the aluminum (5.50 g), c is the specific heat capacity of aluminum (0.897 J/(g·°C)), and ΔT is the change in temperature (70.0°C). By substituting these values into the formula, we find that the aluminum absorbs 272.07 J of heat energy.

This indicates that the aluminum requires this amount of energy to undergo the temperature change. The specific heat capacity of a material represents its ability to absorb or release heat. In this case, aluminum's relatively high specific heat capacity means it requires a significant amount of heat energy to raise its temperature.

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Answer:

Motion under gravity refers to the movement of an object whose vertical motion is affected by the presence of gravity. The force that attracts objects downwards is GRAVITY. In fact, gravity works towards the centre of the Earth.

hope it helps you

make me brainliest plz

Answer:

Pressure of .2 m^3 of bricks:

Pb = ρ g H = 1.8 ρw * 9.80 m/s^2 * .20 m = 1800  * 9.8 * .2 = 3530 Pascals

ρb = 1.8 ρw = 1.8    since the density of bricks is 1.8 that of water

Pw = 1 * 9.8 * 1000 * .2 = 1960 Pascals

W = P * A   pressure * Area = Force

Pnet = 1570 Pascals - pressure due to bricks under water

F = 1570 N/m^2 * .2m^2 = 63 N     force on pile of bricks

In a gym the students raced with each other and teacher is recording everyone's speed than the second graph is representing correctly.

A graph is a visual depiction of statistical data or a logical relationship between variables. Graphs serve a predictive purpose because they have the virtue of displaying broad trends in the mathematical behavior of data. However, as just approximations, they may be incorrect and occasionally deceptive.

Most graphs have two axes, with the axis denoting a set of independent variables and the vertical axis denoting a set of dependent variables. The most typical graph is indeed a broken-line chart, where the regression coefficient is typically a factor of duration.

In the second graph the speed of every student is represented accurately, therefore the second graph is correct.

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The center of gravity of the additional mass should be located at the weighted average of the individual centers of mass. This location ensures that the object or system remains in balance and maintains its stability.

1. Identify the object: First, identify the object or system for which you need to find the center of gravity.

2. Locate the individual masses: Locate the individual masses within the object or system, including the additional mass.

3. Calculate the individual centers of mass: Determine the individual centers of mass for each component of the object, including the additional mass, based on their geometric shapes and density distributions.

4. Calculate the total mass: Add up the masses of all the components, including the additional mass, to obtain the total mass of the object or system.

5. Determine the center of gravity: Calculate the weighted average of the individual centers of mass by multiplying each center of mass by its corresponding mass and then dividing the sum by the total mass.

The center of gravity of the additional mass should be located at the weighted average position calculated in Step 5.

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Delegation When delegating work to other healthcare personnel, a registered nurse should take into account their level of education, training, and experience.

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The distance the skier will go before they stop is 10193.68m.

Given ; Mass =100kg

coefficient of kinetic friction μ=0.05m

u=100m/s

v=0

μg=a

a= μg=0.05 x 9.81=0.4905 m/s²

v²-u²=2as

[v²-u²]/2a= s

-100²/2(0.4905)

s=10193.68m (neglecting the negative sign)

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When an electron accelerates through a potential difference, its kinetic energy increases. To find the final de Broglie wavelength, we can use the de Broglie equation:

λ = h / p, where λ is the wavelength, h is the Planck's constant, and p is the momentum of the electron.

To calculate the momentum of the electron, we can use the formula p = √(2mK), where m is the mass of the electron and K is the kinetic energy. Since the electron starts from rest, the initial kinetic energy is zero.

Given that the potential difference is 36 V, we can calculate the final kinetic energy using the equation K = qV, where q is the charge of the electron (1.6 x 10⁻¹⁹ C) and V is the potential difference.

Now, we can substitute the value of K into the momentum formula to find the momentum of the electron. Then, using the de Broglie equation, we can determine the final de Broglie wavelength.


To find the final de Broglie wavelength, we can follow these steps:

Calculate the final kinetic energy: Since the electron starts from rest, the initial kinetic energy is zero. The kinetic energy is given by the equation K = qV, where q is the charge of the electron (1.6 x 10⁻¹⁹ C) and V is the potential difference. In this case, the potential difference is 36 V.

Thus, K = (1.6 x 10⁻¹⁹C)(36 V)

= 5.76 x 10⁻¹⁸ J.

Calculate the momentum of the electron: The formula for momentum is p = √(2mK), where m is the mass of the electron (9.1 x 10⁻³¹ kg) and K is the kinetic energy.

Plugging in the values, we get p = √(2(9.1 x 10⁻³¹ kg)(5.76 x 10⁻¹⁸ J))

= 1.68 x 10⁻²³ kg∙m/s.

Calculate the final de Broglie wavelength: Using the de Broglie equation, λ = h / p, where λ is the wavelength, h is Planck's constant (6.626 x 10⁻³⁴ J∙s), and p is the momentum of the electron.

Substituting the values,

we have λ = (6.626 x 10⁻³⁴ J∙s) / (1.68 x 10⁻²³kg∙m/s)

= 3.94 x 10⁻¹¹ m.

Therefore, the final de Broglie wavelength of the electron is 3.94 x 10⁻¹¹meters.

The final de Broglie wavelength of an electron that accelerates through a potential difference of 36 V is 3.94 x 10⁻¹¹ meters. This is calculated by finding the final kinetic energy using the formula K = qV, where q is the charge of the electron and V is the potential difference.

Then, the momentum of the electron is calculated using the formula p = √(2mK), where m is the mass of the electron and K is the kinetic energy. Finally, the de Broglie wavelength is determined using the equation λ = h / p, where λ is the wavelength and h is Planck's constant.

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the magnitude of the electric field at a point midway between a -8.0 uC and a +5.8 uC charge 6.0 cm apart is 2203.2 N/C.

The direction of the electric field is from the +5.8 uC charge towards the -8.0 uC charge.

Electric field is defined as a force experienced by a charge per unit charge at a point.

It is usually measured in Newtons per Coulomb (N/C).

The magnitude and direction of the electric field at a point midway between a -8.0 uC and a +5.8 uC charge 6.0 cm apart can be determined using Coulomb's law.

Coulomb's law is an equation that describes the electrostatic interaction between two charges. It states that the electrostatic force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula for electric field is given by

E = F/qwhere,

E = Electric field

F = Force

q = Charge

At a point midway between the -8.0 uC and +5.8 uC charge, the distance from each charge to the point is the same and can be calculated using Pythagoras theorem.

The distance between the charges = 6.0 cm

The distance from the midpoint to each charge = 3.0 cm

The distance from each charge to the midpoint can be calculated using:

r² = (6/2)² + 3²r² = 36 + 9r² = 45r = √45r = 6.7 cm

The force on a test charge q at the midpoint due to the -8.0 uC charge is given by:

F₁ = kq₁q₂/r²F₁ = 9 × 10⁹ × 8 × 10⁻⁶ × q/ (0.067)²F₁ = 9 × 10⁹ × 8 × 10⁻⁶ × q/ 0.00449F₁ = 1276.8q N

The force on a test charge q at the midpoint due to the +5.8 uC charge is given by:

F₂ = kq₁q₂/r²F₂ = 9 × 10⁹ × 5.8 × 10⁻⁶ × q/ (0.067)²F₂ = 9 × 10⁹ × 5.8 × 10⁻⁶ × q/ 0.00449F₂ = 926.4q N

The total force on a test charge q at the midpoint due to both charges is given by:

F = F₁ + F₂F = 1276.8q + 926.4qF = 2203.2q N

The electric field at the midpoint due to both charges is given by:

E = F/qE = 2203.2q/qE = 2203.2 N/C

Therefore, the magnitude of the electric field at a point midway between a -8.0 uC and a +5.8 uC charge 6.0 cm apart is 2203.2 N/C.

The direction of the electric field is from the +5.8 uC charge towards the -8.0 uC charge.

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Answer:

The importance of international bureau of weights and measures are;

1. to bring unification of measurement system.

2. to establish and preserve fundamental international prototypes.

3. to verify national standards, etc.

When working through these problems, think about the heat in terms of movement of the molecules of water.

Heat is transferred from the water to the surroundings --- water molecules slowing down.

Heat is transferred to the water from the surroundings --- water molecules heating up.

Heat transfer is the transfer of energy due to the difference of temperature between two different places. The main discussion in heat transfer is how the energy in heat can move places and the rate of transfer under certain conditions. Heat transfer includes the processes of inflow and outflow of heat. In industrial processes, heat transfer is used to achieve the required temperature in the industrial process and maintain the required temperature throughout the process.

Heat transfer from one object to another can occur by conduction, convection, and radiation. The determinant of heat transfer is the temperature difference. The direction of heat transfer starts from a medium with a high temperature to a medium with a lower temperature. Heat transfer can occur with a single process or multiple processes.

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The term "optical injection-induced polarization switching dynamics in 1.5-μm wavelength single mode vertical-cavity surface-emitting lasers" refers to the process of introducing an external optical signal into a laser cavity, influencing its behavior, including polarization switching dynamics. This study focuses on understanding the time-dependent behavior and factors influencing this switching process.

The phrase "optical injection-induced polarization switching dynamics in 1.5-μm wavelength single mode vertical-cavity surface-emitting lasers" refers to a phenomenon that occurs in lasers operating at a wavelength of 1.5-μm, specifically in single mode vertical-cavity surface-emitting lasers.

In this context, "optical injection" refers to the process of introducing an external optical signal into the laser cavity. This injection can influence the behavior of the laser, including polarization switching dynamics.

"Polarization switching" refers to a change in the polarization state of the laser output. Normally, lasers emit light with a specific polarization direction. However, in certain conditions, the polarization direction can switch between two or more states. This switching can be controlled or induced through external factors, such as the optical injection mentioned earlier.

The "dynamics" aspect refers to the study of the time-dependent behavior of the polarization switching process. This includes understanding how the switching occurs, the time it takes for the switching to happen, and the factors that influence this process.

Overall, the phrase "optical injection-induced polarization switching dynamics in 1.5-μm wavelength singlemode vertical-cavity surface-emitting lasers" describes the study of how the polarization of light emitted by a specific type of laser (1.5-μm wavelength singlemode vertical-cavity surface-emitting lasers) can be switched or changed by introducing an external optical signal through the process of optical injection. This study focuses on understanding the time-dependent behavior and factors influencing this switching process.
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Answer:

A. 0.2 g/cm³

Explanation:

d=m/v

2.0g / 10cm³

0.2g/cm³