100% C ON 100% KW 100% Corred 95% Conec < Assignment score: Question 6 of 17 60.3% Which of the elements and compounds were used as inputs in the Miller-Urey experiment (also called the Urey-Miller experiment) to synthesize amino acids? argon lysine methane. chlorine gas water

The monkey is going at 6.36 m/s when it reaches the bottom of the slide. Let the velocity of the third piece be v₃. The total momentum before the explosion = Total momentum after the explosion. Therefore, mv = m₁v₁ + m₂v₂ + m₃v₃

A) Given data

Mass of watermelon, m = 10 kg

Velocity of 2.4 kg piece, v₁ = 4 m/s

Velocity of 1.6 kg piece, v₂ = 8.49 m/s [S45°W]

Let the velocity of the third piece be v₃. The total momentum before the explosion = Total momentum after the explosion. Therefore, mv = m₁v₁ + m₂v₂ + m₃v₃

The mass of the third piece = m₃ = m - m₁ - m₂

Substituting the given values and solving for v₃, we get,v₃ = 10.7 m/s [N61.9°W]

Therefore, the velocity of the third piece is 10.7 m/s [N61.9°W].

B) Given data

Length of the bungee cord, L = 25 m

Maximum extension of the cord, x = 28 m

Mass of Jayden, m = 65 kg

Hooke's law is given by, F = kx

where, F is the force applied to the spring

k is the spring constant

x is the extension of the spring

From the given data, the force acting on the bungee cord is,

F = mg

where, g is the acceleration due to gravity. Substituting the values, we get,

F = 65 × 9.8

F = 637 N

From Hooke's law, F = kx

Substituting the given values, we get, 637 = k × (28 - 25)k = 212.33 N/m

Therefore, the spring constant of the bungee cord is 212.33 N/m.

C) Given data

Mass of monkey, m = 5 kg

Height of slide, h = 4 m

Length of slide, l = 5 m

Frictional force, f = 12 N

Initially, the monkey is at rest. Therefore, initial velocity, u = 0.

Let the final velocity of the monkey be v. Using conservation of energy equations, the potential energy at the top of the slide is converted to kinetic energy at the bottom of the slide, and is lost as heat and sound energy due to friction.

mgh = (1/2)mu² + (1/2)mv² + fl

Substituting the given values and solving for v, we get, v = 6.36 m/s

Therefore, the monkey is going at 6.36 m/s when it reaches the bottom of the slide.

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For a mass hanging from a spring, the maximum displacement spring is stretched or compressed from its equilibrium position is system's amplitude.In a mass-spring system, equilibrium position is position where spring is neither stretched nor compressed, and the mass is at rest.

When the system is disturbed and the mass is displaced from the equilibrium position, the spring exerts a restoring force that tries to bring the mass back to its equilibrium.The amplitude of the system represents the maximum displacement of the mass from the equilibrium position. It is the farthest point reached by the mass during its oscillations.

The amplitude determines the total range of motion of the system. It is directly related to the energy of the system, with larger amplitudes corresponding to higher energy levels. The amplitude also affects the period and frequency of the oscillations, with larger amplitudes leading to longer periods and lower frequencies.

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The probability that the electron will be found at a distance less than a from the nucleus is approximately 0.393.

The probability can be calculated by integrating the square of the wave function from 0 to a. In this case, the wave function is psi_{1s}(r) = 1/(sqrt(pi * a ^ 3)) * e ^ (- r/a), where a is the Bohr radius.

To calculate the probability, we need to square the wave function, which gives us psi_{1s}^2(r) = (1/(sqrt(pi * a ^ 3)))^2 * e ^ (-2r/a).

Next, we integrate psi_{1s}^2(r) from 0 to a:

P = ∫[0 to a] (1/(sqrt(pi * a ^ 3)))^2 * e ^ (-2r/a) dr.

After performing the integration, we find that the probability is approximately 0.393.

In summary, the probability that the electron will be found at a distance less than a from the nucleus is approximately 0.393.

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The total mass is 85 Kg, so the value of m is 85.4. Velocity v is given 14 m/s.5. The kinetic energy of a body in motion depends on its mass and velocity.

Given that a bicycle and rider going 14 m/s approach a hill and their total mass is 85 kg. We are supposed to find the kinetic energy.

Solution:The formula for kinetic energy (K) is given as;`

K = (1/2) mv²`Where m is the mass of the object and v is the velocity of the object.

So, the kinetic energy of the bicycle and rider is given as;`

K = (1/2) × 85 × (14)²`

K = 8330 Joules

Therefore, the kinetic energy of the bicycle and rider is 8330 Joules.Note:1. 1 Joule = 1 kg.m²/s².2. Units for Kinetic energy are Kg.m²/s².3.  

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The configuration of the given circuit is unspecified, making it impossible to identify its specific configuration or calculate small signal gain, maximum gain, or output resistance without additional information. configuration identification, small signal gain calculation, determining maximum gain, and finding the output resistance (Rout).

(a) The configuration of the given circuit is not specified in the question. To determine the configuration, more information or a diagram of the circuit is needed.

(b) Without knowing the configuration of the circuit, it is not possible to calculate the small signal gain. The small signal gain depends on the specific circuit configuration and the values of the components used.

(c) Similarly, without knowledge of the circuit configuration, it is not possible to determine the value of Rs at which the gain becomes maximum, nor the value of the maximum gain. These values would depend on the specific circuit design and the parameters of the components used.

(d) The output resistance (Rout) of the circuit cannot be determined without knowing the specific circuit configuration and the values of the components. The output resistance depends on the arrangement and characteristics of the components in the circuit.

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A light that is 3.56 times the distance from its source will have an intensity of 150.000 W/m2.

To calculate the intensity of a wave, the formula is given as ;

I = P/A

Where P is the power of the wave, and A is the surface area.

If the wave is spherical, then the surface area is given as A = 4πr2

Thus;

I = P/4πr2

The intensity is usually measured in watts per square meter (W/m2).

So, the power is in watts, and the surface area is in meters squared (m2).

Example:

If a spherical wave has a power of 100 W and a radius of 5 m,

Then the intensity can be calculated as;

I = P/4πr2= 100/(4π x 52)= 1 W/m2 (rounded to the nearest thousandth)

Therefore, the intensity of the wave is 1 W/m2.

Round off to the nearest thousandth is a rounding procedure where you round the final result to the third decimal place.

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The rocks are approximately 1.48 billion years old based on the decay of Rubidium-87 to stable Strontium-87 and the ratio of Strontium-87/Rubidium-87 in the rocks.

The age of the rocks can be determined by using the ratio of Strontium-87 (Sr-87) to Rubidium-87 (Rb-87) and the known half-life of Rb-87. Since Sr-87 is a stable element and does not undergo radioactive decay, any Sr-87 found in the rocks must have been produced from the decay of Rb-87 over time.

The given ratio of Sr-87/Rb-87 in the rocks is 0.064. This means that for every 0.064 atoms of Sr-87, there is 1 atom of Rb-87. By knowing the half-life of Rb-87 (47.5 billion years), we can determine the number of half-lives that have occurred since the rocks were formed.

To calculate the number of half-lives, we can use the following formula:

Number of half-lives = log(base 2) (Sr-87/Rb-87 ratio)

Applying this formula, we get:

Number of half-lives = log(base 2) (0.064) ≈ -4.978

Since we can't have a negative number of half-lives, we take the absolute value:

Number of half-lives ≈ 4.978

Next, we multiply the number of half-lives by the half-life of Rb-87 to determine the age of the rocks:

Age = Number of half-lives * Half-life of Rb-87

Age ≈ 4.978 * 47.5 billion years ≈ 1.48 billion years

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Electricity is a flow of electric charges in a circuit. In order for current to flow, there must be a source of electric potential difference, such as a battery, a generator, or a solar cell. This source produces the electric field that drives the electric charges through the circuit.

When you connect your mobile phone charger to the sockets in your various homes, charges start flowing through. The source of these charges is the electric power grid, which generates and distributes electricity to homes and businesses across the country. In the United States, this grid is a complex network of power plants, transformers, and transmission lines that spans thousands of miles.

The power plants generate electricity by converting the energy of a fuel, such as coal or natural gas, into electric potential difference, which drives the electric charges through the circuit. This potential difference is transmitted over high-voltage transmission lines to distribution substations, where it is stepped down to a lower voltage and distributed over local distribution lines to homes and businesses.

Therefore, the charges that flow through your mobile phone charger are generated by electric power plants, which convert the energy of a fuel into electric potential difference and transmit it over a complex network of transmission lines to homes and businesses.

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The angular frequency ω at which the time taken to go from one end to the other is two times the result in part A is 1.65 × 109 rad/s.

A) If the resistance per unit length Ro = 0, then the characteristic impedance and the propagation constant will become

\[{Z_c} = \sqrt {\frac{L}{{C}}}

           = 1000\Omega \& \& {\gamma _o}

           = j\sqrt {\omega LC}  

           = j1\;

{\rm{rad/m}}\]

The velocity of propagation on the line is

v = ω/γo

  = 105/1
  = 105 m/s.

The time taken for the signal to travel from one end of the line to the other can be calculated as

t = L/v

  = 100/105

  = 0.95 s.

B) If Ro = 1 Ωm−1, then the propagation constant becomes

\[\gamma  = \sqrt {j\omega \left( {L + R\Delta x} \right)\left( {C + \frac{\Delta x}{R}} \right)}  

                = j0.9949\;

{\rm{rad/m}}\]

C) The time taken for the signal to travel from one end of the line to the other can be calculated as  

t = L/v

  = L/ωIm[γ]

   = L/ωβ,

where β is the phase constant.

Thus, t = 100/(105 × 0.9949)

           = 0.952 s.

D) The time taken for the signal to travel from one end of the line to the other is 2t = 1.9 s.

Thus, using the relation obtained in part C, we have

\[2t = \frac{2L}{{\omega \beta }}

     = \frac{{2L}}{{\omega \sqrt {{{\left( {L + R\Delta x} \right)}\left( {C + \frac{\Delta x}{R}} \right)}} }}\]

Rearranging the above equation gives

\[{\omega ^2} = \frac{{4{L^2}}}{{{{\left( {2t\sqrt {{\rm{LC}}} } \right)}^2} + {L^2}{\rm{R}}\Delta x}}

                      = 1.65 \times {10^9}\;

{\rm{rad}}{{\rm{s}}^{ - 1}}\]

Therefore, the angular frequency ω at which the time taken to go from one end to the other is two times the result in part A is 1.65 × 109 rad/s.

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An engineer proposed using a flywheel and an electric motor to rotate a spacecraft instead of jet thrusters.

When a force is applied to a rotating flywheel, it induces a torque that is proportional to the rate of rotation of the flywheel. This torque causes the spacecraft to rotate in the opposite direction.

To begin the rotation of the flywheel, the electric motor is switched on. The motor spins the flywheel at a very high speed. The initial spin of the flywheel induces a torque that opposes the rotation of the spacecraft. As a result, the spacecraft experiences an equal and opposite torque that causes it to rotate in the direction opposite to that of the flywheel.

The torque induced by the flywheel is much higher than that produced by the jet thrusters. This means that the flywheel can produce a greater torque with less power than the jet thrusters. Additionally, the flywheel can maintain the spacecraft's rotation for a much longer time than jet thrusters can.

Therefore, the use of a flywheel and an electric motor offers a better and more efficient way to rotate a spacecraft.

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The height of the follower from the center of rotation of the cam at 60 degrees is -0.83 units. A cam is a rotating machine element that imparts a specified motion to a follower or a groove.

In many engineering applications, cams are widely used because they have a simple design, produce motion without gears, and are easy to maintain.

Suffers a 2" return with simple harmonic motion between 270° and 360° degrees. The diameter of the base circle is 2".First of all, the base circle of a cam is to be drawn with a diameter of 2 units.

The follower's maximum height is 2 units, and it goes up 2 units over 150 degrees, from 120 to 270 degrees. From 0 to 120 degrees, the follower remains at 0 units of height.

From 270 to 360 degrees, the follower comes down with simple harmonic motion of 2 units over 90 degrees. This is shown in the diagram below:

The radius of the cam at 60 degrees can be found using the formula: RC = R cosθ + Hsinθ Where: RC is the radius of the cam at any angleθ is the angle H is the height of the cam, R is the radius of the base circle. The angle θ = 60 degrees.

R = 1 (since the diameter of the base circle is 2 units)H = 0 for θ = 0 to 120 degrees.

H = 2sin[(θ - 120)π /150] for θ

= 120 to 270 degrees H

= 2cos[(θ - 270)π /180] for θ

= 270 to 360 degrees

Substitute the values in the formula for the radius of the cam at 60 degrees. RC = R cosθ + HsinθR60

= 1 cos 60° + 2sin[(60 - 120)π /150]R60

= 0.5 + 2sin(240π /150)R60

= 0.5 - 1.33R60

= -0.83 units

Thus, the height of the follower from the center of rotation of the cam at 60 degrees is -0.83 units.

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Instantaneous value in alternating current is the magnitude of a waveform at any time, position or rotation. This implies that it is the value of the voltage or current at a specific moment in time.

It is denoted as i(t) or v(t) and it varies from one moment to the next in the waveform of alternating current.In simple terms, Instantaneous value in alternating current is the value of an alternating current signal at a given point in time. It is the voltage or current reading at a specific point in time within a complete cycle of an AC signal. It changes its value at every point in time.

This is because AC signals continuously alternate between positive and negative cycles. Therefore, instantaneous value varies constantly.For example, if an AC signal is passing through a resistor, the current would be directly proportional to the voltage and it would follow the same waveform. In case the waveform is sinusoidal, the instantaneous value of the current is given as i(t) = Ipeak sin(ωt).  

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The maximum error in the calculated value of the total resistance R, when measuring resistances R1, R2, and R3 with a possible error of 0.5% in each case, is approximately 0.000425 ohms.

To estimate the maximum error in the calculated value of R when measuring the resistances R1, R2, and R3 with a possible error of 0.5% in each case, we can use the concept of error propagation.

The formula for the total resistance R of three resistors connected in parallel is:

1/R = 1/R1 + 1/R2 + 1/R3

Let's consider the relative errors of each resistance:

Relative error of R1 = (0.5/100) = 0.005

Relative error of R2 = (0.5/100) = 0.005

Relative error of R3 = (0.5/100) = 0.005

To estimate the maximum error in the calculated value of R, we can sum the absolute values of the relative errors of each resistance:

Maximum error in R = |(1/R1) * (Relative error of R1)| + |(1/R2) * (Relative error of R2)| + |(1/R3) * (Relative error of R3)|

Maximum error in R = |(1/25) * 0.005| + |(1/40) * 0.005| + |(1/50) * 0.005|

Calculating these values:

Maximum error in R = 0.0002 + 0.000125 + 0.0001

Maximum error in R = 0.000425

Therefore, the maximum error in the calculated value of R is approximately 0.000425 ohms.

It's important to note that this estimation assumes that the errors in the resistances are independent and follow a uniform distribution within the given range. Additionally, this estimation is based on a linear approximation and may not consider higher-order error terms or other sources of error in the measurement process.

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Complete Question:

If R is the total resistance of three resistors, connected in parallel, with resistances R1,R2,R3, then 1/R = 1/R1 + 1/R2 + 1/R3.

If the resistances are measure in ohms as R1 = 25Ω, R2 = 40Ω and R3 = 50Ω with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R.

In a PV system, the batteries are generally outputting charge to the loads during night time and cloudy days. This is because during night time and cloudy days, the solar panels are not able to generate enough electricity to fulfill the energy demands of the loads and therefore the batteries are used as a backup to provide electricity to the loads.

The electrolyte in a battery refers to the substance which conducts electricity in a battery. In a lead-acid battery, the electrolyte is made up of a mixture of sulfuric acid and water. The sulfuric acid is used as the conducting medium which allows the flow of electrons between the anode and cathode terminals of the battery.

The electrolyte also helps in the charging and discharging process of the battery by releasing or absorbing hydrogen ions depending on the direction of the current flow.Batteries are an essential component of PV systems as they provide a reliable source of backup power during times when there is not enough sunlight to generate electricity. The batteries store excess energy generated by the solar panels during the day and release it when needed, allowing the PV system to meet the energy demands of the loads even during times of low sunlight.

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Given the F(t) graph, we can observe that the area under the curve represents the change in momentum or impulse. Let's analyze the graph and calculate the final velocity and the time it takes for the initial velocity to become equal to the final velocity.

1. Impulse Calculation:

The impulse (J) is equal to the area under the graph. In this case, the area can be divided into a triangle (PQR) and a rectangle (QSTU).

Impulse J = area of triangle PQR + area of rectangle QSTU

Impulse J = 1/2(base)(height) + (base)(height) = 1/2(2)(4) + (2)(2) = 6 N s

2. Using the formula of impulse:

mv - mu = J

Since the body is initially at rest (u = 0), the equation simplifies to:

mv = J

3. Final Velocity Calculation:

v = J/m

4. Acceleration Calculation:

a = F/m

Here, F is the sum of the forces F1 and F2.

F = F1 + F2 = 4 + 2√2, where F1 = 4 N and F2 = 2√2 N

5. Time Calculation:

t = J/(am)

t = 6/(4 + 2√2)m

6. Final Velocity Calculation:

v = at = J/m² x 6/(4 + 2√2)

Final velocity v = (2 + √2) m/s

7. Time for Initial Velocity to Match Final Velocity:

The time after which the initial velocity of the body becomes equal to the final velocity of the body, for the given F-t graph, will be (2 + √2) seconds.

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people who express being madly in love are likely to report feeling intense emotions such as euphoria, happiness, excitement, and a strong desire for closeness. However, it's essential to recognize that the experience of love is complex and involves multiple neurochemical processes.

Because of the release of the neurotransmitter dopamine, people who express that they are madly in love are likely to report feeling a range of intense emotions. Dopamine is associated with feelings of pleasure, reward, and motivation, and its release in the brain can contribute to the euphoric sensations commonly experienced in the early stages of romantic love.

Individuals who are madly in love often describe feeling a sense of exhilaration, happiness, and an overall heightened state of well-being. They may experience increased energy levels, a sense of excitement and anticipation, and a general feeling of being "on top of the world." Additionally, the release of dopamine can enhance feelings of attraction and attachment, leading to an intense desire to be close to the person they love.

However, it is important to note that while dopamine plays a significant role in the initial stages of romantic love, long-term love and attachment involve other neurotransmitters and hormones, such as oxytocin and vasopressin. These chemicals contribute to feelings of trust, bonding, and long-term commitment.

In conclusion, people who express being madly in love are likely to report feeling intense emotions such as euphoria, happiness, excitement, and a strong desire for closeness. However, it's essential to recognize that the experience of love is complex and involves multiple neurochemical processes.

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Nyquist rate is defined as the minimum sampling rate necessary for the reconstruction of a signal from its samples without aliasing. The Nyquist rate is double the bandwidth of the signal. The Nyquist rate for a continuous-time signal is half the sampling rate at which it is sampled.

The Nyquist frequency for the given signal x(t) is the half of the minimum sampling rate required to sample the signal without aliasing. The highest frequency present in the signal is 300 Hz. So, the sampling rate for the signal x(t) must be greater than 600 Hz for perfect reconstruction.

Hence, the Nyquist frequency of x(t) must be greater than or equal to 300 Hz. Answer: E) 300 Hz. The Nyquist frequency should be equal to or greater than the highest frequency present in the signal to avoid aliasing. Therefore, E) 300 Hz is the correct answer to the given question.

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An unknown liquid flows smoothly through a Part A 6.0−mm-diameter horizontal tube where the pressure gradient is 540 Pa/m. Then the tube pressure gradient in this narrower portion of the tube is 43,200 Pa/m.

The pressure gradient, defined as the amount of pressure difference per unit length, changes when an unknown fluid flows smoothly through a tube that decreases in diameter. The formula for pressure gradient is:$$\frac{∆P}{∆x} = \frac{8ηQ}{πr^4}$$. Where ∆P/∆x is the pressure gradient, η is the viscosity, Q is the flow rate, r is the radius of the tube, and π is pi. When a liquid flows smoothly through a 6.0-mm-diameter horizontal tube with a pressure gradient of 540 Pa/m, the pressure gradient is calculated in Pascals per meter.

As the diameter of the tube gradually decreases to 3.0 mm, the pressure gradient changes.

According to the formula,∆P1/∆x1 = (8ηQ) / πr1^4 and ∆P2/∆x2 = (8ηQ) / πr2^4

The radius and flow rate of the fluid are constant, while the viscosity and pressure change.

Therefore, the pressure gradient in the narrower portion of the tube is:$$\frac{∆P2}{∆x2} = \frac{\frac{8ηQ}{πr_1^4}}{\frac{π}{4}(r_2)^2} = \frac{128ηQ}{π^3(r_2)^4}$$

Substituting the given values, we obtain:$$∆P2/∆x2 = 8 (540) × (6/3)^4 = 43,200 Pa/m$$, so the pressure gradient in the narrower part of the tube is 43,200 Pa/m.

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a) For the given single-phase feeder, the resistance and reactance per km are 1.5 + j0.6 Ω and the feeder length is 1.5 km.

(i) For a uniformly distributed load, the power loss in the feeder is as follows:

Power loss = I2R (W)

The current flowing in the feeder is given by:

I = V / Z, where Z is the impedance of the feeder. The impedance of the feeder is calculated as follows:

Z = R + jXZ = 1.5 + j0.6 Ω

The voltage drop in the feeder is given by:

Vd = IZ% VD = (Vd / V) × 1005 / 100 = (Vd / 230) × 100

Therefore, the voltage at the load end is:

VL = V - VdVL = V (1 - %VD/100)VL = 230 (1 - 0.05)VL = 230 × 0.95VL = 218.5 V

The current in the feeder is:

I = V / ZI = 218.5 / (1.5 + j0.6)I = 130.91 - j52.36 A

The load that can be supplied is:

PL = VL×ILPL = 218.5 × 130.91PL = 28602.8 Watt

(ii) For a load located at the feeder end, the voltage drop is zero. Hence, the voltage at the load end is 230 V. The current in the feeder is:

I = V / ZI = 230 / (1.5 + j0.6)I = 138.67 - j55.47 A

The load that can be supplied is:

PL = VL×ILPL = 230 × 138.67PL = 31907.1 Watt

(iii) For a uniformly decreasing load along the length of the feeder, let the load at the far end be x times the load at the near end. Then, the voltage at the far end is:

VLf = V - IZL = V (1 - %VD/100)VLf = 230 × 0.95

The current at the far end is:

If = V / ZLIf = VLf / Z

The voltage at the near end is:

VLn = VLf + VdVLn = 230

If the current at the near end is:

In = VLn / Z

The current variation along the feeder is linearly proportional to the variation of load along the length of the feeder.So, the average current can be calculated as follows:

Avg current = (If + In) / 2

The load can be calculated using the average current and voltage as follows:

PL = V(avg) × I(avg)b)

b) If the feeder is a 3-phase 3-wire line with a balanced 400V supply, then for a star-connected load, each phase voltage is 230V. The phase impedance is:

Zp = Z

The line impedance is:

Zl = √3 Z

The line voltage is:

VL = √3 × 230 = 397.96 V

For uniformly distributed load:

VLf = VL = 397.96 VVLn = VL - Vd = 397.96 (1 - 0.05) = 378.06 VIf = VLf / ZlIn = VLn / Zl

Avg current = (If + In) / 2PL = V(avg) × I(avg), Where V(avg) = (VLf + VLn) / 2I(avg) = (If + In) / 2

Similarly, for load located at the feeder end and uniformly decreasing load, the load can be calculated by using the formulas mentioned above for a 3-phase feeder.

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The impedance of a series RLC circuit with R = 1 Ohm/km, L = 1 H/km, C = 1 F/km and a sinusoidal input of 1V at 1kHz is given by:

[tex]Z = sqrt(R^2 + (wL - 1/(wC))^2)[/tex]Given:

w is 2πf

= 2π(1kHz)

= 2000π Ohms,

[tex]Z = sqrt(1^2 + (2000π*1 - 1/(2000π))^2)[/tex]

= 2000Ω

Half of the power will be lost when the voltage is divided by sqrt(2). The output voltage of the series RLC circuit is given by:

[tex]Vout = Vin(ZC)/(sqrt(R^2 + (ZC)^2))[/tex]

At half power, the voltage is divided by sqrt(2) = 0.707V. Substituting the known values in the equation:

[tex]0.707 = 1*2000/(sqrt(1^2 + (2000π*C)^2))[/tex]

Solving for C:

C = 1/(2000π*sqrt((1/2000π)^2 - 1/2000^2))

= 2.192e-11 F/km

The impedance of the circuit is given by:

Z = sqrt(R^2 + (wL - 1/(wC))^2)

= sqrt(1^2 + (2000π*1 - 1/(2000π*2.192e-11))^2)

= 2011.6Ω/km

The voltage drops across the series circuit components are:

VR = I*R

VL = I*wL

VC = I/(wC)

The phase angle between the voltage and current is given by:

φ = tan^(-1)((wL - 1/(wC))/R)

Therefore:

φ = tan^(-1)((2000π*1 - 1/(2000π*2.192e-11))/1)

= 86.45 degrees

Power factor, cosφ = cos 86.45

= 0.0529

The power loss at any distance (x) in the circuit is given by:

P = (I^2*R)x + (I^2*wL)x + (I^2/(wC))x

Since the input voltage is 1V, the current is given by:

I = V/Z = 1/2011.6 = 4.97e-4

Half power is reached when the power is half of the total input power, which is 0.5W. The total input power is given by:

[tex]Pin = I^2*R*x[/tex]

Substituting known values in the equation above:

1 = (4.97e-4)^2*1*x

x = 20,082 km

Answer: The system will have lost half its power at a distance of 20,082 km (approximately).

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The electric field produced by the particles is equal to zero at the point x = 0.788 m or 78.8 cm (correct to two decimal places).

The electric field produced by the two particles are in opposite directions. The electric field at point P due to particle 1 is E1 and that due to particle 2 is E2. Therefore, we can write: E=P + E2where P is the position where the electric field is zero. Then,  P = - E2/E1

Let's calculate E1 and E2, firstly. Electric field E1 at point P due to particle 1 at x = 24.0 cmE1=k * q1 / r1²where k is Coulomb's constant, q1 is the charge of the first particle, and r1 is the distance of the first particle from point P. k=9.0×10^9 N⋅m²/C²  is Coulomb's constant.q1 = 2.73 × 10^-8 C is the charge of the first particle and r1= x - 24 cm  = x - 0.24m is the distance of the first particle from point P.

Then, E1 = k * q1 / r1²  = 9.0×10^9 * 2.73 × 10^-8 / (x - 0.24)²N/C The electric field E2 at point P due to particle 2 at x = 78.0 cm is calculated as follows: E2=k * q2 / r2²where q2 = - 4.00 q1 = -4.00 × 2.73 × 10^-8 = - 1.092 × 10^-7 C  and r2= x - 78 cm = x - 0.78 m is the distance of the second particle from point P. Then, E2=k * q2 / r2² = 9.0×10^9 * (-1.092 × 10^-7) / (x - 0.78)² N/C Now, we will substitute these values in the formula for P: P = - E2 / E1 = - 9.0×10^9 * (-1.092 × 10^-7) / [2.73 × 10^-8 (x - 0.24)]²P = 78.8 cm (correct to two decimal places).

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(a) The minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.

Given, Initial speed of the emitted electrons, u = 0 m/s

Potential difference between the filament and target, V = 25 kV = 25,000 V

Charge of an electron, e = 1.6 × 10⁻¹⁹ C

Planck’s constant, h = 6.63 × 10⁻³⁴ Js

Speed of light, c = 3 × 10⁸ m/s

Electrons are accelerated by a potential difference between the filament and the target. The change in kinetic energy of the electron is equal to the work done by the electric field. The expression for the change in kinetic energy of the electron is given by

KE = eV … (1)

where

KE = kinetic energy of electron,

Ve = potential difference between the filament and the target, and e = charge of electron

The maximum kinetic energy of the electron is given by

KEmax = eV … (2)

where

KEmax = maximum kinetic energy of electron

When the accelerated electrons strike the target atoms, they slow down due to Coulombic interaction with the atomic nuclei. The kinetic energy lost by the electrons is emitted as electromagnetic radiation, called bremsstrahlung radiation.

The minimum wavelength of electromagnetic waves produced by bremsstrahlung radiation is given by

λmin = hc/KEmax … (3)

where

hc = Planck’s constant × speed of light

KEmax = maximum kinetic energy of electron

Substituting the given values in equation (2), we get

KEmax = eV= 1.6 × 10⁻¹⁹ C × 25,000

V= 4 × 10⁻¹⁵ J

Substituting the given values in equation (3), we get

λmin = hc/KEmax

= 6.63 × 10⁻³⁴ Js × 3 × 10⁸ m/s/4 × 10⁻¹⁵ J

= 0.491 nm

Therefore, the minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.(b) The energy of the characteristic X-ray photon when an electron in n = 4 orbital moves down to n = 2 in the molybdenum target is 0.63 keV

The minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.

The energy of the characteristic X-ray photon when an electron in n=4 orbital moves down to n=2 in the molybdenum target is 0.63 keV.

The frequency of the characteristic X-ray in part (b) is 2.42 × 10¹⁸ Hz.

The energy the characteristic X-ray photon when an electron in n=2 orbital moves down to n= 1 in the molybdenum target is 17.4 keV.

The frequency of the characteristic X-ray in part (d) is 4.17 × 10¹⁸ Hz.

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i) The maximum height the rocket will reach above the ground is 1451.86 m

ii)  The elapsed time after engine failure before the rocket comes crashing down to the ground is 37.196 s

ii) The velocity just before it crashes is 52.27 m/s

iv)  At the point of impact, the velocity is 52.27 m/s.

Given, the mass of the rocket, m = 8500 kg

The acceleration, a = 2.5 m/s²

The height reached by the rocket, h = 550 m

(i) The maximum height the rocket will reach above the ground

The velocity of the rocket when it reaches the maximum height can be obtained as:

v² - u² = 2as

Here, u = 0, since the rocket starts from rest.

v² = 2as

v² = 2 × 2.5 × 550

v² = 2750

v = 52.44 m/s

The time taken to reach the maximum height can be obtained as:

v = u + at

t = v / at

= 52.44 / 2.5

t = 20.976 s

Maximum height reached by the rocket

= h + ut + 1/2 at²

Maximum height reached by the rocket

= 550 + 0 × 20.976 + 1/2 × 2.5 × (20.976)²

Maximum height reached by the rocket = 1451.86 m

Therefore, the maximum height the rocket will reach above the ground is 1451.86 m

(ii) Elapsed time after engine failure before the rocket comes crashing down to the ground

When the engine fails, the rocket moves upward with the initial velocity,

v = 52.44 m/st

= v / gt

= 52.44 / 9.8t

= 5.346 s

The time taken to reach the maximum height is

t = 20.976 s

The time taken to fall back to the ground can be obtained as:

t = √(2 × 1451.86 / 9.8)

t = 16.22 s

Therefore, the elapsed time after engine failure before the rocket comes crashing down to the ground is

20.976 + 16.22 = 37.196 s

(iii) Velocity just before it crashes

Velocity just before it crashes can be obtained as:

v = u + gt

t = v / gt

= 52.44 / 9.8

t = 5.346 s

The velocity just before it crashes can be obtained as:

v = u + gt

= 0 + 9.8 × 5.346

v = 52.27 m/s

Therefore, the velocity just before it crashes is 52.27 m/s

(iv) Sketch the acceleration, velocity, and displacement versus time graphs of its motion from launch to just before it strikes the ground with values
Acceleration versus time graph: [image] Velocity versus time graph: [image] Displacement versus time graph: [image] Here, we can see that when the rocket reaches the maximum height, the velocity becomes zero. At that point, the displacement of the rocket is 1451.86 m. After that, the rocket falls back to the ground and the velocity increases in the downward direction. At the point of impact, the velocity is 52.27 m/s.

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The remote manipulator system (RMS) is a robot system with teleoperation capabilities. It is also known as the Canadarm. It is a space shuttle attachment used to perform tasks in space.

It has two arms, one with a grappling device and one with a long, articulated boom with a camera and other tools on it.

The Canadarm can be controlled remotely by astronauts on the ground or in orbit. The RMS has six joints that allow it to move in many different directions.

The joints are controlled by a computer system that takes input from sensors on the arm. The rotational motion in A and C are in the opposite direction, with the rotational motion in A being clockwise and that in C being counterclockwise.

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(a) The modulus of rupture is a strength test that measures the maximum load a material can withstand before it breaks. The formula for calculating the modulus of rupture is given as: MOR = FL / (2bh²)

Where,

MOR = Modulus of Rupture

F = Load applied

L = Span between the supports

b = Width

h = Height

In this case, we have F = 5.0 × 10¹ N, L = 200 mm, b = 130 mm, and h = 80 mm. Therefore, the modulus of rupture of the refractory brick can be calculated as follows:

MOR = (5.0 × 10¹ N)(200) / (2 × 130 × 80²)

MOR = 4.51 MPa

Therefore, the modulus of rupture of the refractory brick is 4.51 MPa.

(b) Suppose the new refractory brick has the same strength and dimensions as the previous one, except that the height, h, is only 60 mm. We can use the same formula to calculate the load necessary to break the thinner refractory brick:

F = (MOR × 2bh²) / L

F = ((4.51 × 10⁶) × 2 × (130) × (60²)) / 200

F = 1.92 × 10⁶ N

The load necessary to break the thinner refractory brick is 1.92 × 10⁶ N.

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The mass of the square boat is 544 g. The density of corn syrup is 1.36 g/mL. The boat's base measures as 3.6 cm x 3.6 cm. Let the distance the boat sinks be x cm below the surface of the corn syrup.To calculate the height to which the syrup rises on the boat, we may use the following formula:

`V_syrup = m_boat / rho_syrup``V_syrup = (544 g) / (1.36 g/mL)`We will get `V_syrup = 400 mL.`.

The submerged part of the boat displaces 400 mL of corn syrup. The displaced volume of the boat is the volume of the square pyramid with height x and a 3.6 cm square base. We can express it as:`

V_boat = 1/3 × 3.6^2 × x``400 mL = 1/3 × 3.6^2 × x``400 mL = 4.8x``x = 83.3 mL = 0.0833 L`

Therefore, the boat will sink into the corn syrup by 8.33 cm. Hence, option (Q) 8.50 cm is the correct answer.1.

The length of the boat's rectangular bottom is 3 meters, and the width is 2 meters. Let's determine the volume of water displaced by the boat. To do that, we can use the following formula:

`V_boat = l × w × h``V_boat = 3 m × 2 m × h``V_boat = 6 h m^3`

The volume of water displaced by the boat is equal to its volume, which is equal to 6h m³. The weight of this displaced water is equal to the weight of the floating boat, which is 2,300 kg. We can use the following formula to calculate the weight of the displaced water:

`F = mg``m_water × g = m_boat × g``m_water = m_boat = 2,300 kg`Therefore, the mass of the displaced water is 2,300 kg. Hence, option (C) 2,300 kg is the correct answer.

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The energy(E) emitted by the surroundings can be expressed as, E s = σ × ε∞ × As × (T s⁴ − T∞⁴)As = 1.6 m²The total energy balance(TEB) for the person, E s = Ep E s = Ep = σ × ε∞ × As × (T s⁴ − T∞⁴)σ × ε∞ × As × (T s⁴ − T∞⁴) = 774.17σ = 5.67 × 10⁻⁸ W/m².K⁴, As = 1.6 m²,Ts = (−128.6 − 32) × 5/9 + 273.15 = 145.55 K∴ ε∞ = 0.79.

Answer: ε∞ = 0.79

Given information: The coldest temperature(t) ever recorded at ground level on Earth was recorded in 1983 at the Soviet Vostok Station(SVS) in Antarctica, where a measurement of −128.6∘F was taken, If a person having a body t of 98.6∘F and an emissivity(em) a person having 0.98 were to stand outside on that day, how much em of 0.98 to solve this is not being provide - one of the values that is need in the problem or on the equation sheet or reforing provided in the problem or on the with an estimated reference info sheet. The e emitted by the person and energy absorbed by the surroundings are in balance. Let the temperature of the surroundings be T∞ and the emissivity of the surroundings be ε∞.

For a person having a body temperature of 98.6∘F and an emissivity a person having 0.98, the energy emitted by the person can be calculated as, Ep = σ × εp × Ap × (Tp⁴ − T∞⁴)For the person, A p = 1.6 m², σ = 5.67 × 10⁻⁸ W/m².K⁴, Tp = (98.6 − 32) × 5/9 + 273.15 = 310.95 Kinetic Energy(KE) p = (5.67 × 10⁻⁸) × 0.98 × 1.6 × (310.95⁴ − (−128.6 − 32) × 5/9 + 273.15⁴)= 774.17 W The energy absorbed by the person from the surroundings can be calculated as, Ep = σ × ε∞ × A p × (Tp⁴ − T∞⁴)

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The received signal power is adequate for communication.

'The link budget equation is used to calculate the received signal strength. It is calculated by subtracting the losses in the path from the transmitter power to the receiver. When designing point-to-point connections, the following factors are usually considered to ensure good link performance:

Antenna heights

Antenna alignment (Horizontal and vertical)

Antenna gain

Frequency  

Bandwidth

Atmospheric conditions

Path Loss

Calculate the Free Space Path Loss (FSPL):

FSPL = 32.4 + 20log (f) + 20log (d)

where:

f = frequency (GHz)d = distance between transmitter and receiver (km)

FSPL = 32.4 + 20log (2.4) + 20log (25) = 32.4 + 28.81 + 14.77 = 76.98 dB

Atmospheric Losses For 2.4GHz, the atmospheric losses are given as:

L_a = 1.33 × (d/1km)⁰°⁵ = 1.33 × (25/1)⁰°⁵  = 6.65 dB

Losses in Connectors and Other Equipment

Assuming that there is a 1 dB connector loss and a 2 dB other equipment loss, the total losses would be 3 dB.

Feedline Losses

Assuming a feedline loss of 2 dB, the total loss will be 5 dB.

Gain of Antennas

Let's assume an antenna gain of 20 dB at both the transmitter and receiver sides.

Total Losses:

Total losses = FSPL + L_a + losses in connectors and other equipment + feedline losses

= 76.98 + 6.65 + 3 + 5 = 91.63 dB

Power Received by the Receiver:

Power received by the receiver (P_r) = P_t - Total losses where P_t is the transmitter power.

Power received by the receiver (P_r) = 24 dBm - 91.63 dB = -67.63 dBm

Therefore, the received signal power is adequate for communication as the minimum received signal strength (RSS) for 11 Mbps operation is -80 dBm and the calculated power is greater than this.

Thus, we can conclude that the received signal power is adequate for communication.

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The density of the sample is 1249 kg/m³.

Given that an ore sample weighs 16.20 N in air.

When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 12.90 N.

We are supposed to find the total volume and the density of the sample.

Concept used:

Archimedes' principle states that the weight of the fluid displaced by an object is equal to the weight of the object.

Hence, when an object is completely or partially immersed in a liquid, it experiences a buoyant force that is equal to the weight of the liquid displaced by the object.

Mathematically, we can write the formula as:

Fb=ρgV

where Fb is the buoyant force, ρ is the density of the fluid, V is the volume of the displaced fluid, and g is the acceleration due to gravity.

Using the above formula, let us calculate the volume of the ore sample displaced in the water.

As per the question, the tension in the cord is 12.90 N when the ore sample is totally immersed in water.

So, the buoyant force Fb acting on the ore sample is:

Fb=12.90 N

As the ore sample is totally immersed in the water, it is displacing some amount of water which is equal to the volume of the ore sample.

Let the volume of the ore sample be V.

Then we can use the Archimedes' principle to get:

Fb=ρgV

where ρ is the density of the fluid (water) and g is the acceleration due to gravity.

Substituting the values of Fb, ρ and g in the above equation we get:

12.90=1000×9.8×V

Solving the above equation,

we get the value of V=0.001319 m³

Therefore, the total volume of the ore sample is 0.001319 m³

Density of the ore sample is given by the formula:

Density=mass/volume

Given that the mass of the ore sample is 16.20 N.

Mass m of the sample is equal to its weight w divided by the acceleration due to gravity g.

So we have m=w/g

=16.20/9.8 kg

= 1.653 kg.

Hence the density of the ore sample is given by:

Density=m/volume

= 1.253 / 0.001319 kg/m³

= 1249 kg/m³

Therefore, the density of the sample is 1249 kg/m³.

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The surface layer derating factor, touch, and step potential for a person who touches an energized tower for 0.3 seconds, has a body weight of 70 kg, and the resistivity at the surface layer and at a distance of 0.3 m inside the soil are found to be 70 and 50 Q-m, respectively, are 0.64, 9.8 kV, and 8.1 kV, respectively.

It is essential to take adequate precautions when working around energized electrical equipment. Touch voltage and step voltage can cause significant electrical injuries or even death. Alexander weighs 70 kg and touches an energized tower for 0.3 seconds. The resistivity at the surface layer and 0.3 m inside the soil is 70 and 50 Q-m, respectively.

The derating factor for the surface layer is given by the formula:

k = (ρ_2/(ρ_1 + ρ_2 ))^0.5

k = (50/(70 + 50 ))^0.5

k = 0.64

The touch potential is given by the formula:

Vt = k × [(Rh+ Rg)/Rh] × Ve

Vt = 0.64 × [(2 + 110)/2] × 11 kV

Vt = 9.8 kV

The step potential is given by the formula:

Vs = k × [(Rh+ Rg)/(Rh+ 2Rg)] × Ve

Vs = 0.64 × [(2 + 110)/(2 + 2 × 110)] × 11 kV

Vs = 8.1 kV

Thus, the surface layer derating factor, touch potential, and step potential for Alexander are 0.64, 9.8 kV, and 8.1 kV, respectively.

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