Problem No. 1 Find the feedback coefficient 3 for sustained oscillation, give that the open-loop gain equals to :- 1. Ao 3+j4 2. Ao 10 exp(-i 2) Problem No.2 . . Prove that FSR (Free Spectral Range) = c/2n. d Find FSR for Gaz Laser of length "30 cm" Find the corresponding Αλ.

Electrical forces initiate and control nuclear fission by overcoming the repulsion between positively charged protons in the nucleus.

Nuclear fission is a process in which the nucleus of an atom splits into two or more smaller nuclei, accompanied by the release of a significant amount of energy. Electric forces, specifically the electrostatic repulsion between positively charged protons, are responsible for initiating and controlling nuclear fission. In a nucleus, protons are packed closely together, and the repulsive electric forces between them must be overcome for fission to occur. This is achieved by bombarding the nucleus with neutrons, which do not carry a charge but can interact through the strong nuclear force. When a neutron collides with a nucleus, it can be absorbed, causing the nucleus to become highly unstable and elongated. The repulsive electric forces then dominate, leading to the splitting of the nucleus into two smaller fragments.

The interplay between the strong nuclear force and the electric forces is crucial in nuclear fission. While the strong nuclear force holds the nucleus together, the electrostatic repulsion between protons needs to be overcome to induce fission. Understanding and controlling the electrical forces involved in nuclear fission is essential for harnessing this process for various applications, including energy production and nuclear reactors.

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The percentage by mass of Iodine (I) in CaI₂ is 31.3% and the percentage by mass of oxygen (O) in Al₂O₃ if it is 47.1%.

To determine the percentage by mass of Iodine in CaI₂, we first need to know the atomic mass of the constituent elements which is given as;

Atomic mass of Calcium (Ca) = 40

Atomic mass of Iodine (I) = 127

Using these atomic masses, we can find the percentage by mass of Iodine in CaI₂ as;

% Iodine by mass = (127 / (40 + (2 x 127))) x 100%= 31.3%

Therefore, the percentage by mass of Iodine in CaI₂ is 31.3% if it is 13.6% Ca by mass. The formula for the mass percentage of an element in a compound is:

% of element = (mass of an element in compound ÷ total mass of compound) × 100%

To calculate the percentage by mass of oxygen (O) in Al₂O₃ if it is 52.9% aluminum (Al), we first need to know the atomic mass of the constituent elements which is given as;

Atomic mass of Aluminium (Al) = 27

Atomic mass of Oxygen (O) = 16

Using these atomic masses, we can find the percentage by mass of oxygen (O) in Al₂O₃ as;

% of O = (2 × 16 ÷ 102) × 100% = 47.1%

Therefore, the percentage by mass of oxygen (O) in Al₂O₃, if it is 52.9% aluminum (Al), is 47.1%.

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These three equations are the differential equations for each direction.x: f(x) = A sin(kx);  kx = nπ/A, n = 1,2,3,....y, g(y) = B sin(ky); ky = mπ/A, m = 1,2,3,....z, h(z) = C sin(kz); kz = lπ/A, l = 1,2,3, the three differential equations and determine the values of kn allowed for each direction is kx = nπ/A, n = 1,2,3.

Given that the potential is

V(†) = V(x)V(y)V(z)

where each of the three directions is bound by of box of size A.

We need to solve the 3D Schrodinger equation for a 3D box using the separation of variables.

We propose a solution of the form

Y = f(x)g(y)h(z).

a. Follow the procedure to separate the differential equation into three interdependent equations. The 3D time-independent Schrödinger equation is given by:

[-(h^2/8π^2m)] [ ∂^2Ψ/∂x^2 + ∂^2Ψ/∂y^2 + ∂^2Ψ/∂z^2 ] + V(x,y,z) Ψ

= E ΨOn substituting the wave function

Y=f(x)g(y)h(z), the above equation is transformed to:

[-(h^2/8π^2m)] [f''gh + g''fh + h''fg] + V(x,y,z) fgh = Efgh

Now we divide the above equation with fgh.

Hence, it becomes: [1/f f'' + 1/g g'' + 1/h h''] + 2m(E-V(x,y,z))/h² = 0

So, we have obtained three separate ordinary differential equations as follows:

1/f f'' = kx² ;   1/g g'' = ky² ;    1/h h'' = kz² ;

where k = 2m(E-V)/h².

These three equations are the differential equations for each direction.x: f(x) = A sin(kx);  kx = nπ/A, n = 1,2,3,....y:

g(y) = B sin(ky); ky = mπ/A, m = 1,2,3,....z:

h(z) = C sin(kz); kz = lπ/A, l = 1,2,3,....

b. Solve each of the three differential equations and determine the values of kn allowed for each direction. You should have three quantum numbers at this point.

Solution to the differential equation 1/f f'' = kx² can be obtained as follows :

f(x) = A sin(nπx/A); n = 1,2,3,....

kx = nπ/A, n = 1,2,3,....

The solution of the differential equation 1/g g'' = ky² is given by :g(y) = B sin(mπy/A); m = 1,2,3,....ky = mπ/A, m = 1,2,3,....

The solution of the differential equation

1/h h'' = kz² is given by :

h(z) = C sin(lπz/A); l = 1,2,3,....

kz = lπ/A,

l = 1,2,3,....

The allowed values of k for each direction are given by:

kx = nπ/A, n = 1,2,3,....

ky = mπ/A, m = 1,2,3,....

kz = lπ/A, l = 1,2,3,...

c. Determine the total energy, by adding the three contributions.

Total energy E is given by:

E = kx² + ky² + kz² = (n² + m² + l²) π² h²/2mA

= [(n² + m² + l²) π² h²/2mA] + V(†).

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The change in entropy when the thermodynamic state of a gas changes configuration from one with 3.8x10¹⁸ microstates (W) to one with 7.9x10¹⁹ microstates (W) is 3.23x10⁻²² J/K.

The formula for entropy is:

S = KlnW

where S is the entropy of the system,

K is the Boltzmann constant,

and W is the number of microstates available.

Here, the initial number of microstates is 3.8 x 10¹⁸ and the final number of microstates is 7.9 x 10¹⁹. So, the change in entropy is:

ΔS = K ln(W₂/W₁) = (1.38 × 10⁻²³ J/K) ln(7.9 × 10¹⁹/3.8 × 10¹⁸) = (1.38 × 10⁻²³ J/K) ln(20.789) = 3.23 × 10⁻²² J/K

Given data: Number of microstates at the initial state,

W1 = 3.8x10¹⁸.

Number of microstates at the final state, W2 = 7.9x10¹⁹.

Boltzmann's constant, K = 1.38x10⁻²³ J/K.

Formula used: ΔS = Kln(W₂/W₁)

The entropy change of the system is given by the equation.

ΔS = Kln(W₂/W₁),

where W1 is the initial number of microstates,

W2 is the final number of microstates,

and K is Boltzmann's constant.

Substituting the given values in the equation, we get:

ΔS = (1.38x10⁻²³ J/K)ln(7.9x10¹⁹/3.8x10¹⁸)

ΔS = (1.38x10⁻²³ J/K)ln20.789= 3.23x10⁻²² J/K

Therefore, the change in entropy when the thermodynamic state of a gas changes configuration from one with 3.8x10¹⁸ microstates (W) to one with 7.9x10¹⁹ microstates (W) is 3.23x10⁻²² J/K.

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The magnitude of the capacitive reactance (XC) at a frequency of 10 kHz, with a capacitance (C) of 10 nF, is approximately 159.155 ohms.

The magnitude of the capacitive reactance (XC) can be calculated using the formula:

XC = 1 / (2 × π × f × C)

where:

f is the frequency in hertz,

C is the capacitance in farads, and

π is a mathematical constant (approximately 3.14159).

Given that the frequency is 10 kHz (10,000 Hz) and the capacitance is 10 nF (10 × 10⁻⁹ F), we can substitute these values into the formula:

XC = 1 / (2 × π × 10,000 Hz × 10 × 10⁻⁹ F)

XC = 1 / (2 × 3.14159 × 10,000 Hz × 10 × 10⁻⁹ F)

XC = 1 / (62,831.853 Hz × 10 × 10⁻⁹ F)

XC = 1 / (6.28318 × 10⁻³ Ω)

XC = 159.155 Ω

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The age of the universe is estimated to be approximately 13.7 billion years, making option (c) the correct answer. This age is derived from various cosmological observations and measurements.

To estimate the age of the universe:

1. Scientists use various methods, including observations and measurements, to gather data about the universe.

2. One important piece of evidence is the cosmic microwave background radiation, which is a faint glow left over from the early stages of the universe.

3. By studying this radiation, scientists can determine the expansion rate of the universe.

4. Another method involves measuring the ages of the oldest known celestial objects, such as globular clusters or white dwarf stars.

5. By analyzing the chemical composition, temperature, and other characteristics of these objects, scientists can estimate their age.

6. Combining these measurements and observations, scientists have determined that the age of the universe is approximately 13.7 billion years.

7. This value is widely accepted in the scientific community and is considered the best estimate based on current knowledge and understanding of the universe.

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True. The birthrate for teenage mothers in the United States has indeed dropped by 18% since the early 1990s when it reached its peak.

The statement is true. According to data from the Centers for Disease Control and Prevention (CDC) in the United States, the birth rate for teenage mothers has indeed dropped by 18% since the early 1990s when it reached its peak. This decline in teenage birth rates is considered a positive trend and is attributed to various factors such as increased access to contraception, improved sex education, and changes in societal norms and attitudes towards teenage pregnancy. The reduction in teenage birth rates reflects progress in addressing this issue and promoting reproductive health among teena.

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The circuit diagram of a delta source and star load is shown below:Calculation of Line Voltage and Phase Voltage of the LoadThe voltage between any line and the neutral is known as the phase voltage (Vph), and the voltage between any two line wires is known as the line voltage (Vline).If the load is connected in a star configuration, the phase voltage is the voltage across any phase winding,

while the line voltage is the voltage across any two-phase windings.Let us presume that the phase voltage at the load is 440V.Ry line voltage = phase voltage = 440VRB line voltage = phase voltage = 440VYB line voltage = phase voltage = 440VThus, the phase voltage across the load is 440V, and the line voltage is also 440V.Calculation of Line and Phase CurrentLet's presume that the current passing through one phase winding is 20 A. The total current will be the square root of 3 times the current passing through one phase winding.

IT = √3 × IphIT = √3 × 20AIT = 34.64 ALine current is the current flowing through any two line wires in a star configuration. For star loads, line current is the same as phase current.Iline = IphIline = 20 ACalculation of Total Power Active, Total Power Reactive, and Total Power ApparentWe can find the total power active, total power reactive, and total power apparent using the following formulas:P = 3 × Vline × Iline × cosφQ = 3 × Vline × Iline × sinφS = 3 × Vline × IlineP = 3 × 440 × 20 × cos(25°)P = 18,912 WattsQ = 3 × 440 × 20 × sin(25°)Q = 7,573 VARS (Volt Ampere Reactive)S = 3 × 440 × 20S = 20,491 VA (Volt-Ampere)Thus, the total power active is 18,912 Watts, the total power reactive is 7,573 VARS, and the total power apparent is 20,491 VA.

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Answer:  A) estimated electric field halfway between the plates is approximately 342.86 V/m. (in two significant figures)

              B) estimated work done by the battery to charge the plates is approximately 2.21 * 10^-8 J.  (in two significant figures)

              C) If a dielectric is inserted between the plates, the distance between the plates (d) will change, and therefore, the electric field and capacitance will also change.

A) To estimate the electric field halfway between the plates, we can use the formula E = V/d, where E is the electric field, V is the voltage, and d is the distance between the plates.

Given that the voltage is 12 V and the distance between the plates is 3.5 cm (or 0.035 m), we can substitute these values into the formula to find the electric field.

E = 12 V / 0.035 m = 342.86 V/m (rounded to two significant figures)

Therefore, the estimated electric field halfway between the plates is approximately 342.86 V/m.

B) To estimate the work done by the battery to charge the plates, we can use the formula W = 0.5 * C * V^2, where W is the work done, C is the capacitance, and V is the voltage.

Since we don't have the capacitance value, we need to estimate it. The capacitance of a parallel plate capacitor can be approximated as C = ε₀ * A / d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Given that the diameter of each pie pan is 9 inches (or 0.2286 m), the radius is half of the diameter, which is 0.1143 m. Therefore, the area of each plate is A = π * (0.1143 m)^2.

Now we can estimate the capacitance using the formula C = ε₀ * A / d.

C = (8.85 * 10^-12 F/m) * [π * (0.1143 m)^2] / 0.035 m = 3.67 * 10^-10 F (rounded to two significant figures)

Substituting the capacitance and the voltage into the formula for work done, we get:

W = 0.5 * (3.67 * 10^-10 F) * (12 V)^2 = 2.21 * 10^-8 J (rounded to two significant figures)

Therefore, the estimated work done by the battery to charge the plates is approximately 2.21 * 10^-8 J.

C) If a dielectric is inserted between the plates, the distance between the plates (d) will change, and therefore, the electric field and capacitance will also change. The electric field will decrease, and the capacitance will increase.

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The stopping distance from 100 km/h is 2 times as long as the stopping distance from 50 km/h. This is the best option. The correct option is B.

The stopping distance from 100 km/h is 2 times as long as the stopping distance from 50 km/h. It is the rate at which an object decreases speed. When you apply the brakes to your car, you are effectively causing it to decelerate. The acceleration and deceleration rates are the same, with one important difference: acceleration increases the speed of an object, while deceleration reduces it.

Factors that influence stopping distance include the reaction time of the driver and the state of the road surface. At 50 km/h and 100 km/h, drivers in two identical cars perform maximum deceleration stops. According to the formula, the stopping distance is proportional to the square of the velocity. That is, if the speed of the car is doubled, the stopping distance is quadrupled, and if the speed is halved, the stopping distance is decreased by a factor of four.

As a result, the stopping distance is proportional to the square of the velocity. The stopping distance is proportional to the square of the velocity: {stopping distance}∝{velocity²}. As a result, the stopping distance of a car traveling at 100 km/h is 4 times that of a car traveling at 50 km/h.

2² = 4.

Therefore, the stopping distance from 100 km/h is 2 times as long as the stopping distance from 50 km/h.

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1.mg = Tsinθ,2. Tcosθ = mv^2/R , 3. v 4. T = Mg + mgcosθ These equations can be used to determine the speed v of the box as it moves in the circle on the cone surface.

To find the speed v of the box as it moves in a circle on the cone surface, we can consider the forces acting on the box and apply Newton's laws of motion.

Let's analyze the forces acting on the box:

Tension force (T): The tension force in the string pulling the box towards the center of the circle.

Gravitational force (mg): The weight of the box acting vertically downwards.

Normal force (N): The normal force exerted by the cone surface on the box perpendicular to the surface.

We can decompose the gravitational force into two components:

mgcosθ: The component of gravitational force parallel to the cone surface.

mgsinθ: The component of gravitational force perpendicular to the cone surface.

Considering the circular motion, there are two acceleration components:

Centripetal acceleration (ac): The acceleration directed towards the center of the circle.

Tangential acceleration (at): The acceleration along the tangent to the circle.

Now, we can write the equations of motion for the box:

In the vertical direction:

Sum of forces in the vertical direction = ma (acceleration in the vertical direction is zero)

N - mgcosθ = 0 (Equation 1)

In the horizontal direction:

Sum of forces in the horizontal direction = ma (acceleration in the horizontal direction is the centripetal acceleration ac)

T - mgsinθ = mac (Equation 2)

Tangential acceleration:

The tangential acceleration is related to the angular speed (ω) and the radius (R) of the circle:

at = R * dω/dt = R * α (where α is the angular acceleration)

Angular acceleration:

The angular acceleration can be related to the tangential acceleration:

α = at / R

Relationship between tangential acceleration and centripetal acceleration:

Since ac = R * α, we have:

ac = at / R

Velocity:

The velocity v can be related to the angular speed ω and the radius R:

v = R * ω

These equations represent the forces and motion of the box as it moves in a circle on the cone surface. They can be used to analyze and solve for the speed v by combining and simplifying them.

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The duty ratio of a chopper cannot be negative, so we will have to flip the switch and convert it to a first-quadrant chopper with a duty ratio of 0.067.

a) Operation of first quadrant chopper:

The first-quadrant chopper operates in the first quadrant of the i-v plane. When an SCR is used as the switching component, it is generally referred to as a first-quadrant SCR chopper.

b) Principle of operation of second quadrant chopper:

When a step-down converter is used to regulate the average output voltage to less than the input voltage, it is known as a second-quadrant chopper.

Because the circuit operates in the second quadrant of the i-v plane, it is referred to as a second-quadrant chopper. It's usually used for speed control in DC motors.

A four-quadrant chopper is a combination of a first-quadrant and a second-quadrant chopper, which can operate in all four quadrants of the i-v plane.

c) Calculation of the chopper's duty ratio:

A 220 V, 1500 rev/min, 25 A permanent magnet DC motor has an armature resistance of 0.3 Q.

We know that N = (120f)/p,

where f is the frequency and p is the number of poles. If we consider the frequency to be 50Hz and the number of poles to be 4, we obtain the following:

N = (120 × 50)/4

   = 1500 rpm

We can also calculate the motor's back emf, which is given by the equation

Eb = (V - IaRa),

where V is the applied voltage, Ia is the armature current, and Ra is the armature resistance. Here, we can calculate the back emf as follows:

Eb = (220 - 25 × 0.3)

     = 212.5 V

At rated torque, the motor's speed is 1500 rpm. We can also calculate the duty ratio of the chopper, which is given by the following formula:

D = (Eba - V)/Eba,

where Eba is the motor's back emf at rated speed. If we assume that the speed is halved, or 750 rpm, we can calculate the new back emf as follows:

Eba' = (N'/N) × Eba

       = (750/1500) × 212.5

       = 106.25 V

The duty ratio can now be calculated as follows:

D = (Eba' - V)/Eba'

   = (106.25 - 220)/106.25

   = -1.067

The duty ratio of a chopper cannot be negative, so we will have to flip the switch and convert it to a first-quadrant chopper with a duty ratio of 0.067.

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1. If the meals total production energy of 16.1 MJ/portion, the production energy of the meal is 4.4818 kWh/portion. 2. he energy used to produce your meal would power a 60W incandescent light bulb for is 74.6967 hours. 3. The nearest calorie the energy of your meal is 3857 calories/portion.

1. To convert the meal's total production energy of 16.1 MJ/portion to kWh/portion, we can use the conversion factor 1 Megajoule = 0.278 kWh.

16.1 MJ/portion * 0.278 kWh/1 MJ = 4.4818 kWh/portion

So, the production energy of the meal is 4.4818 kWh/portion.

2. To determine the energy used to produce the meal in terms of powering a 60W incandescent light bulb, we need to convert the energy from Megajoules to kWh. Then, we can divide this value by the power of the light bulb (60W) to find the duration in hours.

16.1 MJ/portion * 0.278 kWh/1 MJ = 4.4818 kWh/portion

4.4818 kWh/portion / 0.06 kW (60W = 0.06 kW) = 74.6967 hours

Rounding to the nearest hour, the energy used to produce the meal would power a 60W incandescent light bulb for approximately 75 hours.

3. The meal's total energy of 16.1 MJ/portion, we can convert it to calories using the conversion factor 1 Megajoule = 239.01 calories.

16.1 MJ/portion * 239.01 calories/1 MJ = 3856.961 calories/portion

Rounding to the nearest calorie, the energy of the meal is approximately 3857 calories/portion.

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The given information states that sound absorbing materials like acoustic foam are utilized to lessen background noise. If an absorbing material lessens the sound level by 30 dB, the sound intensity decreases by a factor of 10¹⁵.

We can use the following formula to determine the ratio between two sound intensities:I₁ / I₂ = (d₁ / d₂)²where I₁ and I₂ are the sound intensities, and d₁ and d₂ are the distances between the sound source and the listener. Since the question is about the attenuation of sound by an absorbing material, we can assume that the distance between the sound source and the listener is constant.

Therefore, we can use the following formula to calculate the attenuation in decibels:

dB = 10 log (I₀ / I)

where I₀ is the reference sound intensity

(0 = 10⁻¹² W/m²), and I is the actual sound intensity.

In this case, the absorbing material reduces the sound level by 30 dB.

Therefore, we can write:

30 dB = 10 log (I₀ / I)

⇒ log (I₀ / I) = 3

⇒ I₀ / I = 10³

= 1000

This means that the sound intensity is reduced by a factor of 1000, or 10¹⁵ in power units (since intensity is proportional to the square of the sound pressure).

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Higher mass stars are able to use a higher fraction of their mass for fusion due to the increased gravitational pressure within their cores. The gravitational force in massive stars is stronger, causing a greater compression of the core. This compression results in higher temperatures and pressures, enabling fusion reactions to occur more efficiently.

The higher temperature and pressure facilitate the fusion of heavier elements, such as carbon, nitrogen, and oxygen, which require more energy to overcome their stronger electrostatic repulsion. In contrast, lower mass stars primarily undergo fusion of lighter elements like hydrogen and helium.

Additionally, higher mass stars have longer lifetimes, allowing them to sustain fusion for a more extended period. This extended duration provides more time for the fusion reactions to proceed, effectively utilizing a larger fraction of their mass for energy production.

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the angular acceleration must be greater than amax / r for the blade tip to fracture.

To determine the angular speed ω at which the propeller blade's tip will fracture, we need to consider the relationship between angular acceleration, angular speed, and radius.

The angular acceleration α is related to the angular speed ω and time t through the equation:

α = ω / t

We can rearrange this equation to solve for time:

t = ω / α

Now, let's consider the linear acceleration a at the blade tip, which can be related to angular acceleration α and radius r through the equation:

a = α * r

If the magnitude of the acceleration at the blade tip exceeds a certain value amax, the blade tip will fracture. Therefore, we can set up the following inequality:

a > amax

Substituting the expression for a, we have:

α * r > amax

Solving for α, we get:

α > amax / r

Now, we can substitute the expression for α in terms of ω and t:

ω / t > amax / r

Substituting t = ω / α:

ω / (ω / α) > amax / r

ωα / ω > amax / r

α > amax / r

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In polar form, the phasor of voltage is 55 ∠240°. In rectangular form, the phasor of voltage is (-1/2) + j(√3/2).

Instantaneous voltage/current phasors for the given equations:

a) i(t) = 2/2 cos(wt + 45°)A  

Instantaneous current = 2/2 cos(wt + 45°)

Acos(wt+45) = cos w t cos 45 + sin w t sin 45

= 1/√2 cos w t + 1/√2 sin w t

We know that,

I = Irms cos (wt +θ)Therefore, here

Irms = 2/2 = 1A

Now, the phasor of current can be represented as

I = 1 ∠45°

In polar form, the phasor of current is 1 ∠45°.In rectangular form, the phasor of current is (1/√2) + j(1/√2). b) v(t) = 110V2 cos(wt - 120°)

Instantaneous voltage = 110V2 cos(wt - 120°)cos(wt - 120) = cos w t cos 120 + sin w t sin 120= -1/2 cos w t + √3/2 sin w tWe know that,

V = Vrms cos (wt +θ)

Therefore, here

Vrms = 110V/2 = 55V

Now, the phasor of voltage can be represented as

V = 55 ∠240°

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The downlink CNR is 84.08 dB.

The operating frequency of a satellite is 6 GHz, distance of 35,780km above the earth station, Effective isotropic radiated power =80 dBW, Atmospheric absorption loss of 2 dB, satellite antenna with physical area of 0.5 m² and aperture efficiency of 80%, effective noise temperature of 190°K, noise bandwidth of 20 MHz and the threshold CNR for this satellite is 25 dB.

To determine whether the transmitted signal shall be received with satisfactory quality at the satellite or not, we have to calculate the received signal power and noise power. For this, we have to use the Friis transmission formula: Pr = Pt + Gt + Gr - L - 20 log f - 147.55

Where, Pr = received power at satellite in dBm Pt = transmitted power at earth station in dBm Gt = gain of transmitting antenna in dBi Gr = gain of receiving antenna in dBi L = system losses in dB f = operating frequency in MHz

Using the above formula, the received power can be calculated as follows:

Pr = 80 + 2 + 10 log [(0.8 x 0.5) / (4 x π x (35,780 x 1000)² x 6 x 10⁹)] - 20 log 6 - 147.55Pr = -107.67 dBm

Now, we can calculate the carrier-to-noise ratio (CNR) as follows:

CNR = Pr - Ls - PnCNR = -107.67 - 2 - 228.6

CNR = -338.27 dBi.e. CNR is less than the threshold CNR of 25 dB, hence the transmitted signal shall not be received with satisfactory quality at the satellite.

To calculate the downlink CNR, we can use the same formula. The noise power in this case is given by kTB, where k is the Boltzmann constant, B is the noise bandwidth and T is the effective noise temperature.

Pn = kTB = 1.38 x 10⁻²³ x 190 x 20 x 10⁶Pn = -213.52 dBm

Now, the received power at earth station is given by Pt = Pr + Ls + Lp - Gt - GrPt = -107.67 - 2 - 0.8 + 10 log [(0.8 x 0.5) / (4 x π x (35,780 x 1000)² x 6 x 10⁹)] - 20 log 6Pt = -129.44 dBm

Now, the CNR can be calculated as before:

CNR = Pt - PnCNR = -129.44 + 213.52CNR = 84.08 dB

Since the CNR of the satellite link is greater than the threshold CNR of 25 dB, the transmitted signal shall be received with satisfactory quality at the earth station.

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In the given nuclear reaction sRa226 – X + 2He", we are asked to determine the daughter element "X" produced.

To identify the daughter element in the nuclear reaction, we need to understand the notation used. The notation sRa226 represents the parent nuclide, which is radium-226.
The notation 2He" represents the particle emitted, which is a helium nucleus (alpha particle) with a charge of +2.

In a nuclear reaction, the daughter element is formed when the parent nuclide undergoes decay by emitting particles.
In this case, the emission of a helium nucleus indicates that the parent nuclide loses two protons and two neutrons.

By subtracting two protons and two neutrons from the atomic number and mass number of the parent nuclide, respectively, we can determine the atomic number and mass number of the daughter element.

Radium-226 (sRa226) has an atomic number of 88 and a mass number of 226. Subtracting two protons (atomic number) and two neutrons (mass number), we get an atomic number of 86 and a mass number of 222.

The element with atomic number 86 is radon (Rn), so the correct answer is b) 86Rn222.
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In terms of minutes, it would take light about 160 minutes or 2 hours and 40 minutes to travel from Earth to Uranus. It would take light approximately 100,000 years to cross the diameter of the Milky Way galaxy.

The distance between Earth and Uranus and the time it takes for light to cross the diameter of the Milky Way galaxy are as follows:

Earth to Uranus: The average distance from Earth to Uranus varies depending on their positions in their respective orbits around the Sun. On average, the distance between Earth and Uranus is approximately 2.871 billion kilometers. In terms of minutes, it would take light about 160 minutes or 2 hours and 40 minutes to travel from Earth to Uranus.

Light crossing the diameter of the Milky Way: The Milky Way galaxy has a diameter of about 100,000 light-years. Since light travels at a speed of approximately 299,792 kilometers per second, we can calculate the time it takes for light to cross the diameter of the Milky Way.

Using the formula: Time = Distance / Speed

Distance = 100,000 light-years * 9.461 trillion kilometers (conversion factor)

Distance ≈ 946,100,000,000,000 kilometers

Time = 946,100,000,000,000 kilometers / 299,792 kilometers per second

Time ≈ 3,157,815,750 seconds

Converting seconds to years:

Time ≈ 100,000 years

Therefore, it would take light approximately 100,000 years to cross the diameter of the Milky Way galaxy.

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The time constant of the circuit is 0.68 s. The capacitance of the circuit is 0.000047 F. The charge on the capacitor when it is fully charged is 0.00963 C.

(a) Calculation of Time constant

The expression for the charging of a capacitor through a resistor is given as, q(t) = Q(1 - e^(-t/T))

Where, T = Time constant = RC

The given expression is, p = 75% = 0.75

It means that the capacitor is 75% charged.

The expression for the percentage of charge on the capacitor is given as, p = q(t)/QWhere, q(t) = Charge on the capacitor at time t, Q = Charge on the capacitor at time t = ∆V × C

Where, ∆V = Voltage applied = AV = 85 V, C = Capacitance

The expression for the charging of a capacitor can be written as,

q(t) = ∆V × C(1 - e^(-t/T))

Putting the given values,

0.75 = ∆V × C(1 - e^(-0.68/T))0.75 = 85 C × (1 - e^(-0.68/T))0.0088235

= 1 - e^(-0.68/T)e^(-0.68/T)

= 1 - 0.0088235e^(-0.68/T)

= 0.9911765T

= -0.68 / ln(0.9911765)T

= 0.68 s

Therefore, the time constant of the circuit is 0.68 s.

(b) Calculation of Capacitance

The expression for the charging of a capacitor through a resistor is given as, q(t) = Q(1 - e^(-t/T))

Where, T = Time constant = RC

The expression for the percentage of charge on the capacitor is given as, p = q(t)/Q

Where,

q(t) = Charge on the capacitor at time t,

Q = Charge on the capacitor at time t = ∆V × C

Where ∆V = Voltage applied = AV = 85 V, C = Capacitance

Putting these values,

p = q(t)/Q= q(t) / (∆V × C)0.75 = (q(t) / 85C)q(t) = 63.75 C

Substituting these values in the expression of the charging of the capacitor,

q(t) = Q(1 - e^(-t/T))63.75 C = 85 C (1 - e^(-0.68/T))0.75

= 1 - e^(-0.68/T)e^(-0.68/T)

= 0.25T = -0.68 / ln(0.25)T

= 1.386 s

Also, T = RC = 1.386 s = R × C

On substituting the given value, we get

6.52 C = 1.386 sC = 0.000047 F

Therefore, the capacitance of the circuit is 0.000047 F.

(c) Calculation of Charge

The expression for the charging of a capacitor through a resistor is given as, q(t) = Q(1 - e^(-t/T))

At steady-state, the capacitor gets fully charged.

It means that q(t) = Q

Substituting this in the expression of charging of a capacitor,

q(t) = Q(1 - e^(-t/T))Q

= Q(1 - e^(-t/T))e^(-t/T)

= 0e^(-t/T) = 1T = -t / ln(1)T

= t = 0.68 C

also, T = RC = 0.68 s = R × C

On substituting the given value, we get

6.52 C = 0.68 sC = 0.00010461 C

Now, the expression for the charge on the capacitor is given as,

Q = ∆V × C = 85 V × 0.00010461 CQ = 0.00963 C

Therefore, the charge on the capacitor when it is fully charged is 0.00963 C.

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The X-rays have a wavelength of 154.2 pm (picometers) and they produce reflections from the 200 planes and the 111 plane of Cu, which has an FCC (face-centered cubic) structure.

To calculate the diffraction angles, we can use Bragg's law: n * λ = 2 * d * sin(θ), where n is the order of the reflection, λ is the wavelength, d is the spacing between the planes, and θ is the angle of diffraction.

For the 200 planes, we have d = a / sqrt(200), where a is the lattice parameter. For the FCC structure, a = 4 * r / sqrt(2), where r is the atomic radius of Cu.
Similarly, for the 111 plane, we have d = a / sqrt(3)
The density of Cu is given as 8.935 g/cm³. From the density, we can calculate the atomic mass of Cu.

The diffraction of X-rays from crystal planes can be described using Bragg's law, which states that the angle at which diffraction occurs depends on the wavelength of the X-rays and the spacing between the crystal planes.
Using these values, we can substitute them into Bragg's law to calculate the diffraction angles for the 200 planes and the 111 plane.

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Complete Question:

The X - rays of wavelength 154.2 pm produce reflections from the 200 planes and the 111 plane of Cu which has FCC structure and density of 8.935 g /cm3 . At what angles will the diffracted intensity be maximum?

By substituting the values of a, d, λ, and solving for θ in Bragg's law, we can find the angles at which the diffracted intensities will occur for the (200) and (111) planes of Cu.

To determine the angles at which the diffracted intensities will occur, we can use Bragg's law, which relates the angle of incidence, the wavelength of X-rays, and the spacing between crystal planes:

nλ = 2d sin(θ)

where n is the order of diffraction, λ is the wavelength of X-rays (154.2 pm = 1.542 Å), d is the spacing between crystal planes, and θ is the angle of incidence.

For the (200) planes of Cu in an FCC crystal structure, the spacing between planes can be calculated using the formula:

d = a / √(h^2 + k^2 + l^2)

where a is the lattice constant and (hkl) represents the Miller indices for the planes. In the case of (200) planes, the Miller indices are (2, 0, 0).

Similarly, for the (111) planes, the Miller indices are (1, 1, 1).

To calculate the lattice constant (a) for Cu, we can use the relation between the density (ρ), Avogadro's number (Nₐ), and the atomic mass (M):

ρ = (Nₐ * M) / (a^3 * Z)

where Z is the number of atoms in the unit cell of the crystal structure. For FCC, Z = 4.

By rearranging the equation, we can solve for a:

a = (Nₐ * M / (ρ * Z))^(1/3)

Using the known values, we can calculate the lattice constant a for Cu.

Substituting the values of a, d, λ, and solving for θ in Bragg's law, we can find the angles at which the diffracted intensities will occur for the (200) and (111) planes of Cu.

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Transcranial magnetic stimulation (TMS) is a non-invasive brain stimulation technique that involves the use of magnetic pulses to stimulate specific areas of the brain. The resulting magnetic field that is uniform over a circular area of diameter 2.34 cm inside the patient's brain during this procedure can be calculated using the given parameters.

First, calculate the rate of change of magnetic field by using the formula;emf = -N dφ/dtWhere N is the number of turns, dφ is the change in magnetic flux, and dt is the change in time. The negative sign shows that the induced emf opposes the change in magnetic flux.φ = Bπr²where B is the magnetic field, π is a constant, and r is the radius of the circle. Here, B = 6.00 T and r = 1.17 cm = 0.0117 m.φ = 6.00 T × π (0.0117 m)²= 2.34 × 10⁻⁴ WbWhen the magnetic field rises from zero to its peak in 83.0 μs, the rate of change of magnetic flux is given by;dφ/dt = φ/t = (2.34 × 10⁻⁴ Wb) / (83.0 × 10⁻⁶ s)= 2.82 VThe number of turns is not given, so the induced emf cannot be determined. Therefore, the answer is 2.82 V.

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(a)i(t) = 8 cos(2πft) / 10.909 ∠-5.7107°= 0.732 cos(2πft + 5.7107°) A, (b) the voltage leads the current by 5.7107°

(a)To calculate the current supplied by the source as a function of time, we need to determine the total impedance of the circuit. We can use the following equation to calculate the impedance of the parallel combination of the resistor and inductor:

Z = R || XL= R || jXL= R || j(2πfL)

where R is the resistance (12 Ω), XL is the inductive reactance (j2 Ω), and f is the frequency (100 Hz).

Substituting the given values, we get:

Z = 12 || j2(2π × 100 × 0.002)= 12 || j1.2566= 10.909 ∠-5.7107°V/I

Let us now calculate the current supplied by the source as a function of time:

i(t) = v(t) / Z

where v(t) = 8 cos(2πft) is the voltage supplied by the source.

Substituting the value of Z, we get:

i(t) = 8 cos(2πft) / 10.909 ∠-5.7107°= 0.732 cos(2πft + 5.7107°) A

(b)The phase relationship between voltage and current is given by the phase angle between the two waveforms. Since the voltage and current waveforms are sinusoidal, we can use the following formula to calculate the phase angle:φ = θv - θi

where θv and θi are the phase angles of the voltage and current waveforms, respectively.

Substituting the values, we get:φ = 0° - (-5.7107°)= 5.7107°

Therefore, the voltage leads the current by 5.7107°.

This means that the current waveform lags behind the voltage waveform.

In other words, the current does not instantaneously follow the voltage, but instead takes some time to respond. This is due to the presence of the inductor in the circuit, which causes the current to lag behind the voltage.

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We have a 1 m² cross-section of a river, and the water is flowing at 2 m/s, then the power potential from the river is 2000 W.

The power potential from a river per unit cross-sectional area can be calculated using the following formula:

Power potential = (1/2)*density of water*velocity of water^3 * area

where:

density of water is the density of water, in kg/m³

velocity of water is the velocity of water, in m/s

cross-sectional area is the cross-sectional area of the river, in m²

In this case, we have:

density of water = 1000 kg/m³

velocity of water = 2 m/s

cross-sectional area = 1 m²

Substituting these values into the formula, we get:

Power potential = (1/2) * 1000 kg/m³ * 2 m/s^3 * 1 m² = 2000 W

Therefore, the power potential from a river per unit cross-sectional area is 2000 W.

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To measure the free-fall acceleration on the distant planet, we can make use of the period of the pendulum's swing. The formula for the period of a simple pendulum.

The ability of carbon to form four covalent bonds: Carbon has four valence electrons, allowing it to form up to four covalent bonds with other atoms. This versatility in bonding allows for the formation of complex and diverse carbon-based molecules.The orientation of those bonds in the form of a tetrahedron: Carbon atoms bonded to four different groups tend to adopt a tetrahedral geometry. This arrangement contributes to the three-dimensional shape and structural diversity of carbon-based molecules.

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The value of lambda and Cp for a 3-blade wind turbine with a rotor diameter of 20 meters can be determined using the Betz limit formula. According to the Betz limit, the maximum possible Cp for a wind turbine is 0.59.

The value of lambda is given by the ratio of the actual power extracted by the turbine to the maximum power that could be extracted according to the Betz limit. The value of Cp is given by the ratio of the actual power extracted by the turbine to the power available in the wind.
The Betz limit formula is expressed as:
P = 0.5 × rho ×A ×v³ × Cp
Where,
P = power output
rho = air density
A = area swept by the blades
v = wind speed
Cp = coefficient of power
Thus, the value of lambda is given by:
lambda = P / (0.5 × rho × A × v³ × 0.59)
The value of lambda will vary with wind speed because the power output of the turbine depends on wind speed. As wind speed increases, the power output of the turbine increases, which affects the value of lambda. The value of Cp will also vary with wind speed because it depends on the power available in the wind.
In conclusion, the values of lambda and Cp for a 3-blade wind turbine with a rotor diameter of 20 meters can be calculated using the Betz limit formula. The values of lambda and Cp will vary with wind speed because they depend on the power output and power available in the wind, respectively.

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Lighting systems operating at 30 volts or less shall consist of a 600-volt power supply, low-voltage luminaires, and associated equipment that are all identified for use.

These systems may be used in wet locations and other hazardous locations because the voltage is low enough to prevent any serious hazards.

The low voltage wiring shall have a minimum 90° C rating and a minimum 600-volt insulation rating. Transformers, wiring, and other equipment that produce or handle low-voltage circuits shall comply with the National Electrical Code (NEC).

The use of low-voltage systems provides energy savings, and they are more durable than high-voltage alternatives. In addition, they provide enhanced safety, making them an excellent choice for various applications, including residential, commercial, and industrial facilities.

In conclusion, lighting systems operating at 30 volts or less shall consist of a power supply, low-voltage luminaires, and associated equipment that are all identified for use.

These systems are designed for safety, durability, and energy savings, making them ideal for a wide range of applications.

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Therefore, the nature of the object is a black hole.

The nature of the object is a black hole. The signals from the probe became more and more redshifted, and eventually, at a distance of 40 km from the object, the probe and the signals get ‘frozen'. This indicates that the probe has reached the event horizon of the object.

Therefore, the nature of the object is a black hole.

(b) The mass of the object can be calculated using the formula

Rsch = 2GM/c²

The Schwarzschild radius can be given as follows:

Rsch = 2GM/c²

where G is the gravitational constant,

M is the mass of the object,

and c is the speed of light.

Rearranging the formula for mass, we get:

M = Rsch * c²/2G

Now,

we can Calculate the mass of the object using the values of

Rsch and G.Rsch = 40 km = 40,000 m (as the units of Rsch should be in meters)

G = 6.674 × 10^-11 m³/kg s²c

= 3.00 × 10^8 m/s

Substituting the values of Rsch,G, and c in the equation for M,

we get:

M = (40,000 * 3.00 × 10^8 * 3.00 × 10^8) / (2 * 6.674 × 10^-11)M

= 2.26 × 10^30 kg

= 1.13 solar masses

Therefore, the mass of the object is 2.26 × 10^30 kg or 1.13 solar masses. This value of mass confirms that the object is a black hole, as it is more than three times the mass of the sun.

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a) The volume of the glass sphere is equal to the volume of water displaced, so the volume of the glass sphere is 25.04 cm^3.

b) The density of the glass is 2.20 g/cm^3.

c) The density of ethanol is 1.41 g/cm^3.

(a) To find the volume of the glass sphere, we need to use the principle of buoyancy. The weight of the sphere in air minus the weight of the sphere in water gives us the buoyant force, which is equal to the weight of the water displaced by the sphere.

Buoyant force = Weight in air - Weight in water

Buoyant force = 55.1032 g - 30.1082 g = 24.995 g

Since the density of water is given as 0.9982 g/cm^3, we can use the equation density = mass/volume to find the volume of the water displaced by the sphere.

Volume of water displaced = Mass of water displaced / Density of water

Volume of water displaced = 24.995 g / 0.9982 g/cm^3 = 25.04 cm^3

The volume of the glass sphere is equal to the volume of water displaced, so the volume of the glass sphere is 25.04 cm^3.

(b) To find the density of the glass, we can use the equation density = mass/volume. Since we know the mass of the glass sphere from the weight in air measurement, we can divide it by the volume we just calculated.

Density of glass = Mass of glass sphere / Volume of glass sphere

Density of glass = 55.1032 g / 25.04 cm^3 = 2.20 g/cm^3

So, the density of the glass is 2.20 g/cm^3.

(c) To find the density of ethanol, we can use a similar approach as in part (b). Since we know the mass of the ethanol displaced by the glass sphere, we can divide it by the volume of the glass sphere.

Density of ethanol = Mass of ethanol displaced / Volume of glass sphere

Density of ethanol = 35.3353 g / 25.04 cm^3 = 1.41 g/cm^3

Therefore, the density of ethanol is 1.41 g/cm^3.

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