An AC generator has a maximum output emf of 4.20 x 102 V. What is the rms emf? (Remember AVms = 0.707Vmax)
a. 3 V
c.
297 V
b.
594 V
d.
30 V
Please select the best answer from the choices provided
Ο Α
Mark this and return
4

Answer:

a) F₂₁ = 0.02 N, attracting.

b) F₁₂ = 0.02 N, attracting.

Explanation:

a)

      \(F_{21} = k * \frac{q_{1} *q_{2}}{r_{12}^{2} } = 9e9 N.m2/C2 * \frac{(25e-9C)*(75e-9C)}{(0.03m)^{2}} = 0.02 N (1)\)

b)

We are given:

initial velocity (u) = 0 m/s

time (t) = 4 seconds

acceleration (a) = 1.3 m/s²

displacement (s) = x m

Solving for the displacement:

from the second equation of motion

s = ut + 1/2 at²

replacing the variables with the given values

x = (0)(4) + 1/2 (1.3)(4)(4)

x = 1/2 * 20.8

x = 10.4 m

When the bus is fully loaded, the center of gravity rises to 1.73 m, and the critical angles are 35.3 degrees for the unloaded bus and 30.6 degrees for the loaded bus, respectively.

Theoretically, the body's total weight is concentrated at a location called the center of gravity.

The main distinction between the centers of mass and gravity is that the center of gravity refers to the location where the total weight of the body is balanced, whereas the center of mass refers to the location where the body's complete mass is directed.

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Answer:

The frequency of light can be calculated using the formula:

`c = λv`

Where `c` is the speed of light in a vacuum, `λ` is the wavelength of light, and `v` is the frequency of light.

The speed of light in a vacuum is `3.00 × 10^8 m/s`.

To convert the wavelength from nanometers to meters, we need to divide by `1 × 10^9`.

Thus, the frequency of light with a wavelength of 655 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(655 × 10^-9 m)`

`v = 4.58 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`.

Similarly, the frequency of light with a wavelength of 515 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(515 × 10^-9 m)`

`v = 5.83 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz`.

Finally, the frequency of light with a wavelength of 475 nm is:

`v = c/λ`

`v = (3.00 × 10^8 m/s)/(475 × 10^-9 m)`

`v = 6.32 × 10^14 Hz`

Therefore, the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

So, the frequency of light with a wavelength of 655 nm is `4.58 × 10^14 Hz`, the frequency of light with a wavelength of 515 nm is `5.83 × 10^14 Hz` and the frequency of light with a wavelength of 475 nm is `6.32 × 10^14 Hz`.

This question involves the concepts of the equations of motion, kinetic energy, and potential energy.

a. The kinetic energy of the rocket at launch is "3.6 J".

b. maximum gravitational potential energy of the rocket is "3.6 J".

The kinetic energy of the rocket at launch is given by the following formula:

\(K.E=\frac{1}{2} m v_i^2\)

where,

Therefore,

\(K.E=\frac{1}{2}(0.05\ kg)(12\ m/s)^2\)

K.E = 3.6 J

First, we will use the third equation of motion to find the maximum height reached by rocket:

\(2gh=v_f^2-v_i^2\)

where,

Therefore,

2(-9.81 m/s²)h = (0 m/s)² - (12 m/s)²

h = 7.34 m

Hence, the maximum gravitational potential energy will be:

P.E = mgh

P.E = (0.05 kg)(9.81 m/s²)(7.34 m)

P.E = 3.6 J

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\(\\ \sf\longmapsto (101010)_2\)

\(\\ \sf\longmapsto 0\times 2^0+1\times 2^1+0\times 2^2+1\times 2^3+0\times 2^4+1\times 2^5\)

\(\\ \sf\longmapsto 2+8+32\)

\(\\ \sf\longmapsto (42)_{10}\)

Answer:

In the taller, skinnier cylinder, each tick mark represents 2, while in the larger one, each tick mark represents 50.

Explanation:

If you count the tick marks between each big number, then you can divide, and figure out what amount each tick mark represents.

The separation distance in B is a number times greater than in A, then the force of attraction in B is less than A.

The force of attraction between two charged object is determined by applying Coulomb's law.

Coulomb's law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of the charges and inversely proportional to the square of the distance between the charges.

Mathematically, this law is written as;

F = Kq₁q₂/r²

where;

From the formula given above, as the distance of separation increases, the magnitude of the force of attraction between the charges decreases. Also, as the distance of separation between the charges decreases, the magnitude of the force of attraction between the charges increases.

Thus, the magnitude of the force of attraction between charged objects is a function of the magnitude of the charges and distance of separation between the charges.

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The question is incomplete. The complete question is :

First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:

The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number

What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.

Solution :

a). Given :

\($Q_1 = 42 \ J$\) , \($Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $\)

Net heat transfer

\($Q= Q_1+Q_2$\)

   = 42 + (-22)

   = 20 J

Total work

\($W= W_1+W_2$\)

   = 6 + (-6)

   = 0 J    

∵ ΔU = Q - W

       = 20 - 0

        = 20 J

This is the net change in the internal energy of the system.

b). ΔU = Q + W

           = (-140) + (-165)

           = -305 J

c). ΔU = Q + W

           = (-440) + (645)

           = 205 J

d). ΔU = Q + W

           = (-450) + (690)

           = 240 J

Answer:

Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.

Explanation:

hope it help

The net force on particle q₂, located between particles q₁ and q₃, is approximately 189000 N. The force exerted by particle q₁ on q₂ is positive and equals 252000 N, while the force exerted by particle q₃ on q₂ is negative and equals -63000 N.

To find the net force on particle q₂, we need to calculate the individual forces exerted on q₂ by particles q₁ and q₃ and then determine their sum.

The force between two charged particles can be calculated using Coulomb's law:

F = k * |q₁ * q₂| / r²

Where F is the force between the particles, k is the electrostatic constant (k ≈ 9.0 x \(10^9\) Nm²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

First, let's calculate the force exerted on q₂ by q₁:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9.0 x \(10^9\) Nm²/C²) * |(8.0 μC) * (3.5 μC)| / (0.10 m)²

F₁₂ ≈ 252000 N

The force is positive because q₁ and q₂ have opposite charges.

Next, let's calculate the force exerted on q₂ by q₃:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9.0 x \(10^9\)Nm²/C²) * |(3.5 μC) * (-2.5 μC)| / (0.15 m)²

F₂₃ ≈ -63000 N

The force is negative because q₂ and q₃ have the same charge.

Finally, we can find the net force on q₂ by summing the individual forces:

Net force = F₁₂ + F₂₃

Net force = 252000 N + (-63000 N)

Net force ≈ 189000 N

The net force on particle q₂ is approximately 189000 N.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

A car is moving north at 5.2 m/s². SI unit in this value is m/s² (meter per second square), expressing Acceleration. Thus, Option D is the correct answer.

Here Acceleration of any object is given by the Rate of change in velocity in relation to time.

\(Acceleration = \frac{velocity}{time}\) ............(i)

The standard Indian (SI) unit of Acceleration is meter per second square(m/s²).

The velocity of any object is displacement per unit time.

\(velocity = \frac{Displacement}{Time}\)............(ii)

The standard Indian (SI) unit of Velocity is meter per second(m/s)

The Displacement of any object is the shortest distance covered by any object considering the direction of motion also.

The standard Indian (SI) unit of displacement is meter(m).

The standard Indian (SI) unit of Time is Second(s).

We can find standard Indian (SI) unit of Acceleration using formula as follows:

\(Acceleration= \frac{Velocity}{time*time}\)  (we got this formula from (i), (ii) )

The standard Indian (SI)  units of acceleration are \(=\frac{m}{s*s}\)

\(=m/s^{2}\)

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Answer:

B. Atomic number minus mass number

Explanation:

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

please mark me as the brainliest :)

use your brain :)

Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.

Amphiphilic molecules typically have a hydrophilic (water-attracting) or polar end and a hydrophobic (water-repelling) or nonpolar end. This dual nature allows them to interact with and dissolve in both polar and nonpolar solvents. Let's explore the reasons behind this ability:

Polar interactions: The hydrophilic end of the amphiphilic molecule can form hydrogen bonds and electrostatic interactions with polar solvents (like water) due to the presence of charged or partially charged atoms (such as oxygen or nitrogen). This makes them soluble in polar substances.

Nonpolar interactions: The hydrophobic end of the amphiphilic molecule is typically a long hydrocarbon chain or a nonpolar group. These nonpolar regions can interact with other nonpolar substances (like oils, fats, or hydrophobic portions of biomolecules) through van der Waals forces and London dispersion forces.

Emulsification: The presence of both polar and nonpolar parts in an amphiphilic molecule allows it to act as an emulsifier. It can stabilize the interface between polar and nonpolar substances, preventing them from separating. For example, in the formation of an oil-in-water or water-in-oil emulsion, amphiphilic molecules form a barrier between the two immiscible substances, keeping them dispersed and forming a stable mixture.

Examples of amphiphilic molecules include surfactants (detergents), phospholipids (essential components of cell membranes), and lipoproteins (transporters of lipids in the bloodstream). These substances play vital roles in various biological processes and industrial applications due to their ability to interact with both polar and nonpolar substances.

Hence, Substances that can dissolve both polar and nonpolar substances are called amphiphilic or amphipathic. This unique property arises due to their molecular structure, which contains both polar and nonpolar regions.

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Answer:

\(4,760\:\mathrm{N}\)

Explanation:

From Newton's 2nd Law, we have \(\Sigma F=ma\). Substituting given values, we get:

\(F=1700\cdot 2.8,F=\boxed{4,760\:\mathrm{N}}\)

The electric field of magnitude 16 N/C will be at a distance of 6.0 meters from the charged object.

The magnitude of the electric field due to a point charge object follows the inverse square law, which states that the magnitude of the electric field is inversely proportional to the square of the distance from the charged object. Mathematically, this is expressed as:

\(E = k*q/r^2\)

where E is the electric field, k is Coulomb's constant (\(k = 9 x 10^9 N*m^2/C^2\)), q is the charge of the object, and r is the distance from the object.

We can use this formula to find the distance at which the electric field has a magnitude of 16 N/C. Let's call this distance x:

16 = \(k*q/x^2\)

We can rearrange this equation to solve for x:

x = \(\sqrt(k*q/16)\)

To find q, we need another piece of information. We know that the electric field has a magnitude of 9 N/C at a distance of 4.0 m. Using the same formula as before, we can solve for q:

9 = \(k*q/4^2\)

q = \(9*4^2/k\)

Now we can substitute this value for q into the equation for x:

x =\(\sqrt(k*(9*4^2/k)/16)\)

x =\(\sqrt(9*4^2/16)\)

x = \(\sqrt(36)\)

x = 6.0 meters

Therefore, the electric field of magnitude 16 N/C will be at a distance of 6.0 meters from the charged object.

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The time interval between the man clapping and hearing his echo is approximately 1.75 seconds.

An echo is a repetition or reflection of a sound or signal. It can be caused by sound waves bouncing off a surface, signal interference, or the repetition of a message in communication.

The speed of sound in air at room temperature is approximately 343 meters per second. When a person claps, the sound waves propagate outward in all directions and reach the school building, where they bounce off and return to the person as an echo. The time it takes for the sound to travel the distance to the building and back to the person is the time interval between the clap and the echo.

To calculate the time interval, we can use the following formula:

time = distance / speed

where distance is the total distance traveled by the sound (twice the distance from the person to the school building), and speed is the speed of sound in air.

distance = 2 x 300m = 600m

speed = 343 m/s

time = 600m / 343 m/s = 1.75 seconds (rounded to two decimal places)

Therefore, the time interval between the man clapping and hearing his echo is approximately 1.75 seconds.

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If a swimmer pushes water backward, the direction of force of water will be forward, because action-reaction forces are directed in opposite directions.

option C.

Newton's third law states that for every action there is an equal and opposite reaction. As you a rocket emits gases or hot air downwards, the downward action of these gases pushes the rocket upward. This is an example of Newton's third law.

Thus, based on the Newton's third law of motion, if a swimmer pushes water backward, the water exert a force on the swimmer in opposite direction. The magnitude of this force will be equal to the magnitude of force of swimmer.

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Answer:

10kj should be the answer so I think

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = \(\mathbf{\sqrt{R^2+L^2}}\)

\(\mathbf{z = \sqrt{R^2+(Lw)^2}}\)

Current amplitude

\(\mathbf{I_o = \dfrac{V_o}{z}} \\ \\ \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}\)

a)

Given that:

\(V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\ L = 0.52 H\)

∴

\(I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}\)

\(\mathbf{I_o= 0.0562} \\ \\ \mathbf{I_o = 56.2 \ mA}\)

b)

Phase constant :

\(tan \ \phi = \dfrac{L \omega}{R } \\ \\ tan \ \phi = \dfrac{0.52 \times 51}{21} \\ \\ tan \phi = 1.263\)

\(\text{Phase constant : }\phi = tan^{-1} (1.263) \\ \\ \phi = 51.6^0\\ \\\text{To radians} \phi = 51.6 \times \dfrac{\pi}{180} \\ \\ \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}\)

(a)The value of current amplitude will be 56.2 milliamperes.

(b)The value of the phase constant will be 0.9 rad.

RL Circuits often known as RL networks or RL filters are a form of a circuit that uses a mix of inductors and resistors and is powered by a source of electricity.

The given data in the problem is;

I₀ is the current amplitude =?

V₀= 1.9 V

ω is the angular velocity = 51 rad/s

R is the resistance in the circuit = 21 Ω

Ф is the phase constant=?

(a) The value of current amplitude will be 56.2 milliamperes.

The formula for the current amplitude is given as;

\(\rm I_0= \frac{V_0}{\sqrt{R^2+L^2 \omega^2} } \\\\\)

\(\rm I_0= \frac{1.9}{\sqrt{(21)^2+(0.52\times 51)^2} } \\\\\)

\(\rm I_0 = 0.0562 \ A \\\\ \rm I_0 =56.2 mA\)

Hence the value of the current amplitude will be 56.2 milliamperes.

(b)The value of the phase constant will be 0.9 rad.

The formula for phase constant is given by;

\(\rm tan\phi=\frac{L\omega}{R} \\\\ \rm tan\phi=\frac{0.52\times 51}{21} \\\\ \rm tan\phi=1.263 \\\\ \rm \phi = tan^{-1}(1.263) \\\\ \phi= 51.6 ^0\)

To convert degree into radian the following formula is used;

\(1 \pi =180^0 \\\\ 1^0=\frac{\pi}{180} \\\\ 51.6^0=51.6 \times \frac{\pi}{180} \\\\ \phi=0.9 \ rad\)

Hence the value of the phase constant will be 0.9 rad.

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Grouping data refers to the process of categorizing or organizing data based on specific criteria or attributes.

It involves grouping similar data points together to gain a better understanding of patterns, relationships, and trends within the dataset. By grouping data, you can simplify complex information and derive meaningful insights from large amounts of data. The purpose of grouping data is to create subsets or clusters that share common characteristics.

This enables easier analysis, summarization, and comparison of data within each group. Grouping can be performed on various types of data, such as numerical, categorical, or time-based data. Grouping data allows for the exploration of data at different levels of granularity.

For example, you can group sales data by region to analyze regional performance, or group customer data by demographics to identify specific customer segments. This process helps in identifying outliers, detecting patterns, and making data-driven decisions.

Common techniques for grouping data include using functions like GROUP BY in SQL or utilizing data visualization tools to create charts or graphs that illustrate the grouped data. Grouping can be applied in various fields, such as marketing, finance, healthcare, and research, to uncover insights and support decision-making processes.

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Answer:

true

Explanation:

noble gases are octet meaning they they have eight electrons in their outer shell so the are stable

Answer:

35 N to the left

Explanation:

Do 65 minus 30, and you get 35 N to the left

Because force is a vector quantity, you have to mention both magnitude and direction in your answer.

Answer:

ke=648J

Explanation:

ke=1/2mv²

m=9kg

v=12m/s

1/2(9)(12)²

=648J

Answer:

Explanation:

Based on the observations during the experiment, the car's velocity can be described as follows. .

The car had a   constant speed of approximately 60 km/h throughout the experiment,indicating a consistent rate of motion.

In terms of   direction, the car initially traveledin a straight line towards the east.

However, after a certain point,   it made a sharp turn towards the north, changing its direction but maintaining thesame speed.

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Answer: Z

Explanation: I took the quiz!! :3

During a solar eclipse as seen from Earth, the moon will be at the position Z.

When the Moon moves in front of the Sun, obstructing the Sun's view for a small portion of the Earth either completely or partially, this is known as a solar eclipse.

Here,

When the Moon is in its new moon phase, when its orbital plane is closest to that of the Earth's, and during eclipse season, such an alignment takes place roughly every six months. The Sun's disc is completely covered by the Moon during a total eclipse. The Sun is only partially covered by annular and partial eclipses. A solar eclipse can only be seen from a relatively restricted area of the earth, in contrast to a lunar eclipse, which can be seen from anyplace on the night side of Earth.

The Moon is parallel to the line connecting the Earth and the Sun at its first and final quarters. Only one-half of the Moon is visible to us as being illuminated by the Sun; the other one-half is hidden in shadow.

Hence,

During a solar eclipse as seen from Earth, the moon will be at the position Z.

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Answer:

If you are at sea level, each square inch of your surface is subjected to a force of 14.6 pounds. The pressure increases about one atmosphere for every 10 meters of water depth. At a depth of 5,000 meters the pressure will be approximately 500 atmospheres or 500 times greater than the pressure at sea level.

Explanation:

At sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.

Pressure is defined as the force applied perpendicularly to an object's surface divided by the surface area over which it is applied.

Pressure is the physical amount of force exerted on a particular area.

Pressure can be expressed as

Pressure = Force / area

There are three types  of pressure.

Thus, at sea level, a force of 14.6 pounds is applied to every square inch of your surface. For every 10 meters of sea depth, the pressure rises by approximately one atmosphere. The pressure will be about 500 atmospheres, or 500 times more than the pressure at sea level, at a depth of 5,000 meters.

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(i) The total force acting on the train is the driving force minus the total resistance to motion. The total resistance to motion is the sum of the resistances of the three trucks. Therefore, the total force acting on the train is:

F = 37,000 N - (1,200 kg + 900 kg + 600 kg) ₓ 9.8 m/s² = 37,000 N - 25,740 N = 11,260 N

The acceleration of the train is given by the formula:

a = F / (m1 + m2 + m3) = 11,260 N / (1,200 kg + 900 kg + 600 kg) = 0.3 m/s²

Therefore, the acceleration of the train is 0.3 m/s².

(ii) The forces acting on truck Z are the driving force, the force in the coupling between trucks Y and Z, and the resistance to motion of truck Z. The diagram showing all the forces acting on truck Z in the line of its motion is:

Driving force ≥ Truck Z ≤ Force in coupling Y and Z ≤ Resistance to motion of truck Z

(iii) With the driving force removed and brakes applied, the total resistance to motion is the sum of the resistances of the three trucks and the additional resistance due to the brakes. Therefore, the total resistance to motion is:

R = (1,200 kg + 900 kg + 600 kg) ₓ 9.8 m/s²+ 11,000 N = 25,740 N + 11,000 N = 36,740 N

The total force acting on the train is the total resistance to motion. Therefore, the acceleration of the train is:

a = F / (m1 + m2 + m3) = 0 / (1,200 kg + 900 kg + 600 kg) = 0 m/s²

Therefore, the new acceleration of the train is 0 m/s².

(iv) When the brakes are applied to the engine, the force in the coupling between trucks Y and Z is equal to the resistance to motion of truck Z. Therefore, the force in the coupling between trucks Y and Z is:

F = 600 kg ² 9.8 m/s² + 11,000 N = 5,880 N + 11,000 N = 16,880 N

The force in the coupling between trucks Y and Z is a tension.

When the brakes are applied to truck Z, the force in the coupling between trucks Y and Z is equal to the resistance to motion of truck Z plus the resistance to motion of the engine and the trucks in front of truck Y. Therefore, the force in the coupling between trucks Y and Z is:

F = (600 kg + 900 kg + 1,200 kg) ₓ9.8 m/s² + 11,000 N = 17,640 N + 11,000 N = 28,640 N

The force in the coupling between trucks Y and Z is a thrust.

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