1). The hoop will go up the incline approximately 0.656 m when rolling with a speed of 3.6 m/s. 2). It will take approximately 0.322 s for the hoop to arrive back at the bottom of the incline.
To determine how far up the incline the hoop will go, we can analyze the energy conservation in the system. When the hoop reaches the incline, its initial kinetic energy is converted into potential energy as it moves up the incline. The total mechanical energy of the system is conserved, neglecting any energy losses due to friction.
Initial speed, v = 3.6 m/s
Incline angle, θ = 17°
The height the hoop will reach on the incline, we need to equate the initial kinetic energy to the potential energy at the highest point:
1/2 * I * ω² = m * g * h
The moment of inertia (I) for a thin hoop of mass m and radius r is I = m * r².
The linear velocity v of the hoop is related to the angular velocity ω by v = r * ω.
Plugging these values into the equation, we have:
1/2 * m * r² * (v / r)² = m * g * h
Simplifying the equation, we get:
1/2 * v² = g * h
Solving for h, we have:
h = (1/2 * v²) / g
Substituting the given values:
h = (1/2 * 3.6²) / g
The acceleration due to gravity, g, is approximately 9.8 m/s².
h = (1/2 * 3.6²) / 9.8
Calculating the value, we find:
h ≈ 0.656 m (rounded to three significant figures)
Therefore, the hoop will go up the incline approximately 0.656 m.
Now, let's move on to Part B, which asks for the time it takes for the hoop to arrive back at the bottom of the incline.
We can find the time using the kinematic equation:
s = ut + (1/2)at²
where:
s = displacement (height of the incline)
u = initial velocity (0 since the hoop starts from rest at the top)
a = acceleration (due to gravity, -9.8 m/s²)
t = time
Rearranging the equation, we have:
t = [tex]\sqrt{(2s)/a}[/tex]
Substituting the known values:
t = sqrt([tex]\sqrt{(2 * 0.656) / 9.8}[/tex])
Calculating the value, we find:
t ≈ 0.322 s (rounded to three significant figures)
Therefore, the hoop will take approximately 0.322 s to arrive back at the bottom of the incline.
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Question 1 At the high velocity, drag force is proportional to the squared velocity of a particle as kv². Find its acceleration in the unit of m/s² when a falling speed becomes 0.89 times its terminal velocity. Use the gravitationalacceleration, g = 9.8m/s². Answer: Question 2. A roller-coaster car with a mass of 470 kg moves at the bottom of a circular dip of radius, R= 18.5 m, with a speed of v = 42.7 m/s. Find the normal force of the track on the car at the bottom of the dip in the unit of kN. Use the gravitational acceleration, g = 9.81 m/s². R Answer:
The terminal velocity of an object is the maximum velocity attainable by an object as it falls through a fluid (air is the most common example). The normal force of the track on the car at the bottom of the dip is given by:N = mv² / R + mgN = 470 × 42.7² / 18.5 + 4614.7N = 27660 N or 27.7 kN
In simpler words, it is the constant speed that an object reaches when the force of gravity is balanced by the force of drag. At terminal velocity, there is no acceleration since the net force acting on the object is zero. In the case when a falling speed becomes 0.89 times its terminal velocity, the velocity can be expressed as:u = 0.89vTWe know that the drag force, Fd, is proportional to the squared velocity of a particle, kv², where k is a constant.
The force required to keep an object moving in a circular path of radius R with a speed of v is given by:F = mv² / RWe are required to find the normal force of the track on the car at the bottom of the dip. At the bottom of the dip, the car is in contact with the track. Hence, the normal force provides the centripetal force. Thus, we can write:N = mv² / R + mgHere,m = 470 kgv = 42.7 m/sR = 18.5 mg = 470 × 9.81 = 4614.7 N
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why does the volume of water increase when it freezes
The volume of water increases when it freezes due to a unique property known as the anomalous expansion of water. Most substances contract and become denser as they transition from a liquid to a solid state. However, water defies this trend.
When water molecules cool down, they start to form stable hydrogen bonds with neighboring molecules. In the liquid state, these hydrogen bonds are constantly forming and breaking.
However, as the temperature drops below 4 degrees Celsius, the water molecules slow down, allowing more stable hydrogen bonding to occur.
In the process of freezing, the water molecules arrange themselves in a hexagonal lattice structure, with each molecule bonded to four neighboring molecules.
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2) How amplitude of Wien Bridge Oscillator can be stabilized against temperature variation? References:
To stabilize the amplitude of a Wien Bridge Oscillator against temperature variation, techniques such as thermistors, temperature compensation networks, and thermal design are employed.
The amplitude of a Wien Bridge Oscillator can be stabilized against temperature variation by employing temperature compensation techniques. One common method is the use of a temperature-sensitive resistor (thermistor) in the feedback network of the oscillator. The thermistor's resistance changes with temperature, and by appropriately selecting its characteristics, it can counteract the temperature-induced variations in the gain of the amplifier.Additionally, a temperature compensation network can be incorporated into the oscillator circuit. This network typically includes components such as resistors, capacitors, or diodes that exhibit temperature-dependent characteristics. By carefully selecting and arranging these components, the effects of temperature changes on the oscillator's gain and frequency response can be minimized.Furthermore, proper thermal design and component selection are crucial to reduce the impact of temperature variations. This includes using components with low-temperature coefficients, providing proper heat sinking, and ensuring the thermal stability of critical components.In conclusion, stabilizing the amplitude of a Wien Bridge Oscillator against temperature variation can be achieved through techniques such as using temperature-sensitive resistors, employing temperature compensation networks, and implementing effective thermal design practices.References:1. A. Sedra and K. Smith, "Microelectronic Circuits," 7th edition, Oxford University Press, 2014.2. J. G. Webster, "Encyclopedia of Medical Devices and Instrumentation," John Wiley & Sons, 2006.For more questions on temperature
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Aspherical cavity of radius 5.00 cm at the center of a metus sphere of radius 180 cm. A point charge 4 10 JC rests at the very center of the cavity wheas the metal conductor cames no net charge
The spherical cavity of radius 5.00 cm, where a point charge of 4.00 × 10⁻⁶ C is placed, is 1.01 × 10⁷ N/C.
Given data: Radius of the spherical metal shell, R = 180 cm Radius of the spherical cavity, r = 5 cm Charge enclosed by the spherical cavity, q = 4×10⁻⁶ C The net charge on the spherical metal shell is zero.
Therefore, the electric field inside the metal shell is zero. As the cavity is present inside the metal shell, the electric field inside the cavity will also be zero. Now, using Gauss's law, the electric field at a point inside the cavity at a distance r from the center is given as:E = q/4πε₀r²
where ε₀ is the permittivity of free space.ε₀ = 8.85 × 10⁻¹² C²/Nm²Putting the given values, we get: E = 4×10⁻⁶ / (4π × 8.85 × 10⁻¹² × (5 × 10⁻²)²)= 1.01 × 10⁷ N/C To be more accurate, you can state that the electric field at a point inside the spherical cavity of radius 5.00 cm, where a point charge of 4.00 × 10⁻⁶ C is placed, is 1.01 × 10⁷ N/C.
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Under constant-volume conditions, 2700 J of heat is added to 1.5 moles of an ideal gas. As a result, the temperature of the gas increases by 86.6 K. How much heat would be required to cause the same temperature change under constant-pressure conditions? Do not assume anything about whether the gas is monatomic, diatomic, etc. QP=
The amount of heat required to cause the same temperature change under constant-pressure conditions is 3779.986 JOULE.
At constant volume, the conditions are:
heat = 2700 J
number of mole (gas) n = 1.5 moles
change in temperature ΔT = 86.6 k
Now according to the rules of thermodynamic Change in internal energy at constant volume is ΔU =2700 J and change of entropy in a constant pressure will be equal to the transfer heat.
At constant volume :
[tex]Q=mc_v\Delta T\\\\ 2700\ \text{Joule}=1.5\ \text{mole}\times c_v \times\ 86.6\ K\\\\ c_v=20.79 \dfrac{\text{Joule}}{\text{mole}\cdot{K}}[/tex]
since gas undergoes the same temperature change in both process change in internal energy is same.
By Mayors equation :
[tex]c_p-c_v=R[/tex]
[tex]c_p-20.79=8.314\\\\c_p=29.099 \dfrac{\text{Joule}}{\text{mole}\cdot{K}}[/tex]
Heat would be required at constant pressure condition:
[tex]Q=mc_p \Delta T\\\\Q=1.5 \times29.099\times 86.6\\\\Q=3779.988 \rm J[/tex]
hence, the heat at constant pressure is 3779.988 J
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Find the resultant force on the screw eye. One rope is horizontal, the other rope is vertical. The force in the rope is 175lb.
The resultant force on the screw eye is approximately 247.55 lb.
The resultant force on the screw eye can be found by analyzing the two ropes separately and then combining their effects.
To start, let's consider the horizontal rope. Since the rope is horizontal, the force it applies on the screw eye will act purely in the horizontal direction. This means that the vertical component of this force is zero. Therefore, the only force to consider from this rope is its horizontal force, which is 175 lb.
Now, let's focus on the vertical rope. Since the rope is vertical, the force it applies on the screw eye will act purely in the vertical direction. This means that the horizontal component of this force is zero. Therefore, the only force to consider from this rope is its vertical force, which is also 175 lb.
To find the resultant force, we need to combine the horizontal and vertical forces. Since these forces are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the resultant force.
By using the Pythagorean theorem, we can calculate the magnitude of the resultant force as follows:
Resultant force = √((175 lb)² + (175 lb)²)
= √(30625 lb² + 30625 lb²)
= √(61250 lb²)
= 247.55 lb (rounded to two decimal places)
Therefore, the resultant force on the screw eye is approximately 247.55 lb.
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If a person looks at himself on a bright Christmas tree sphere, which has a diameter of 9 cm, when his face is 30 cm away from it.
a. Find the place where the image is located (mathematically and perform the ray tracing)
b. Describes the nature of the image (real or virtual, right or inverted, larger or smaller than the object.
Place where the image is locatedThe position of the image can be calculated mathematically.Using the mirror equation, (1/u) + (1/v) = (1/f), whereu is the object distance from the mirror,v is the image distance from the mirror, andf is the focal length of the mirror.
Using the data given in the question, we can obtain the value of f:Focal length, f = R/2Where R is the radius of curvature of the mirror.R = 2 × 4.5 cm = 9 cm (Radius of the mirror is half of the diameter)Focal length, f = 4.5 cmNow, we need to find the object distance, u. The question states that the person is 30 cm away from the mirror.Object distance, u = -30 cm (negative sign because the person is on the other side of the mirror).
Let us substitute the values into the mirror equation:1/-30 + 1/v = 1/4.5Simplifying this equation, we get:v = -90 cmThis negative value for the image distance indicates that the image is virtual and located on the same side of the mirror as the person. Using the ray-tracing diagram, we can represent the formation of the image. b) Nature of the imageThe image formed by the mirror is virtual, upright, and enlarged compared to the object.
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please help me with answering those questions thanks
Question 1
Radiation exposure decreases with exposure time.
True
False
Question 2
Radiation exposure decreases with increasing distance from the source.
True
False
Question 3
Radiation exposure increases with increasing intervening material.
True
False
Radiation exposure decreases with exposure time is true.
Radiation exposure decreases with exposure time. This means that the amount of radiation exposure that a person is exposed to decreases as the duration of exposure decreases. The shorter the time of exposure, the less radiation exposure there is, and the lower the risk of harmful effects.
Question 2: Radiation exposure decreases with increasing distance from the source is true
Radiation exposure decreases with increasing distance from the source. This means that the farther away someone is from the source of radiation, the less radiation exposure they will experience. This is because radiation spreads out as it travels, so the intensity of the radiation decreases as the distance from the source increases.
Question 3: Radiation exposure increases with increasing intervening material is false
Radiation exposure decreases with increasing intervening material. This means that any material that comes between the source of radiation and a person can help to reduce the amount of radiation exposure that the person receives. This is why lead and other dense materials are often used in radiation shielding.
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Cuestion 7 Not yet antwered Mathed oul of 300 In a pn junction, under forward bias, the built-in electric field stops the diffusion current Select one True Fation
In a p-n junction, under forward bias, the built-in electric field stops the diffusion current. This statement is True. The built-in electric field in a p-n junction opposes the movement of charge carriers and works to prevent current from flowing. When the forward bias voltage is applied, it reduces the potential barrier.
The positive terminal of the battery is connected to the p-type material, and the negative terminal is connected to the n-type material. The holes in the p-type region are pushed toward the n-type region, while the electrons in the n-type region are pushed toward the p-type region by the electric field generated by the battery.
The amount of bias voltage applied determines the amount of electric field, which in turn determines the number of holes and electrons that diffuse across the junction. The current flowing through the circuit is proportional to the number of charge carriers that diffuse across the junction. The flow of current in a p-n junction under forward bias is referred to as forward current.
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For the circuit shown below, find the complex power on inductor \( L_{2} \), Assume \( v_{s}(t)= \) \( 160 \cos (2 \pi 60 t)(\mathrm{rms}) \)
The complex power on the inductor \(L_2\) is 7.88 + j 10.65 VA.
Complex power is defined as the complex conjugate of voltage multiplied by the complex conjugate of current. It is a complex number and its real part is the actual power consumed by the circuit and the imaginary part is the reactive power. The formula for complex power is:S = VI*
For inductive circuits, the current lags the voltage.
So, the current is given by the expression:i = Imax sin(ωt - φ)where Imax = Vmax/XL and XL is the inductive reactance given by the formula:XL = 2πfL
Given the circuit shown below, we can obtain the value of inductive reactance of \(L_2\) as follows:
XL = 2πfL = 2π(60)(0.35) = 131.95 Ω
The voltage across the inductor is the same as the voltage of the source, that is:V = Vmax cos(ωt) = 160 cos(2π60t) = 80 V
To find the current, we need to find the phase angle φ. To do this, we first need to find the impedance Z of the inductor. We can use the following formula:Z = jXL = j131.95 Ω
So, the current is given by:i = Imax sin(ωt - φ)i = Vmax/XL sin(ωt - φ)i = 80/131.95 sin(2π60t - φ)
The power factor is defined as the ratio of the real power to the apparent power.
The real power is given by P = Vrms Irms cosφ, while the apparent power is given by S = Vrms Irms.
Therefore, the power factor is cosφ = P/S.
Let's start by finding the rms current, which is given by:Irms = Imax/√2Irms = Vmax/(XL√2)Irms = 80/(131.95√2)Irms = 0.4405 A
Now, we can use this value to find the real power consumed by the circuit:P = Vrms Irms cosφ
But, we still need to find the phase angle φ to obtain the power factor.
To do this, we can use the impedance of the inductor as follows:Z = R + jXL
So, the phase angle φ is given by:tanφ = XL/Rφ = atan(XL/R)φ = atan(131.95/50)φ = 1.22 rad
Now we can find the real power consumed by the circuit:P = Vrms Irms cosφP = (Vmax/√2)(Imax/√2)cosφP = (80/√2)(0.4405/√2)cos(1.22)P = 17.76 W
Finally, we can find the apparent power consumed by the circuit as:S = Vrms IrmsS = (Vmax/√2)(Imax/√2)S = (80/√2)(0.4405/√2)S = 19.8 VA
The power factor is cosφ = P/S. So, the power factor is:cosφ = 17.76/19.8cosφ = 0.895
We can now find the complex power on the inductor using the formula:S = VI*S = Vrms Irms cosφ + jVrms Irms sinφS = (Vmax/√2)(Imax/√2)cosφ + j(Vmax/√2)(Imax/√2)sinφS = (80/√2)(0.4405/√2)(0.895 + j sin(1.22))S = 7.88 + j 10.65 VA
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Monochromatic light with wavelength 538 nm is incident on aslit with width 0.025 mm. The distance from the slit to a screen is3.5 m. Consider a point on the screen 1.1 cm from the centralmaximum. Calculate (a) θ for that point, (b) α and (c)the ratio of the intensity at that point to the intensity at thecentral maximum.
The given wavelength is λ = 538 nm = 538 × 10⁻⁹ m
Width of the slit is a = 0.025 mm = 0.025 × 10⁻³ m
Distance between the slit and the screen is D = 3.5 m
Position of the point on the screen is y = 1.1 cm = 1.1 × 10⁻² m
(a) To find θ, we can use the formulaθ = y/D
For the given values,θ = y/D= (1.1 × 10⁻²)/(3.5)= 3.14 × 10⁻³ rad
(b) To find α, we can use the formulaα = λ/a
For the given values,α = λ/a= (538 × 10⁻⁹)/(0.025 × 10⁻³)= 2.152 × 10⁻⁵ rad
(c) To find the ratio of intensity at the given point to the intensity at the central maximum, we can use the formulaI
/I₀ = [sin(πa/λ) / (πa/λ)]² × [sin(πy/λD) / (πy/λD)]²
For the central maximum, y = 0.
So,I/I₀ = [sin(πa/λ) / (πa/λ)]²
For the given point, we have already found θ.
So,I/I₀ = [sin(πaθ/λ) / (πaθ/λ)]² = [sin(π(0.025 × 3.14 × 10⁻³)/(538 × 10⁻⁹)) / (π(0.025 × 3.14 × 10⁻³)/(538 × 10⁻⁹))]²
I/I₀ = 0.0386
So, the ratio of intensity at the given point to the intensity at the central maximum is 0.0386.
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The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV.
(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?
(a) The kinetic energy of the electron in the first excited state of the hydrogen atom is -6.8 eV.
(b) The potential energy of the electron in the first excited state of the hydrogen atom is 3.4 eV.
(c) The choice of the zero of potential energy does not affect the values of kinetic and potential energy, only the overall reference point.
(a) To find the kinetic energy of the electron in the first excited state of the hydrogen atom, we need to subtract the potential energy from the total energy. The total energy is given as -3.4 eV, which includes both kinetic and potential energy components. Since the electron is in a bound state, the total energy is negative.
The kinetic energy is equal to the total energy minus the potential energy:
Kinetic energy = Total energy - Potential energy
In this case, the total energy is -3.4 eV, and the potential energy is the negative of the total energy:
Potential energy = -(-3.4 eV) = 3.4 eV
Therefore, the kinetic energy can be calculated as:
Kinetic energy = -3.4 eV - 3.4 eV = -6.8 eV
(b) The potential energy of the electron in the first excited state of the hydrogen atom is given as 3.4 eV. This represents the energy associated with the attraction between the electron and the proton in the hydrogen atom. Since the total energy is negative, the potential energy is positive, indicating a stable bound state.
(c) None of the answers above would change if the choice of the zero of potential energy is changed. The choice of the zero of potential energy is arbitrary and does not affect the relative values of the kinetic and potential energy components. It only affects the overall reference point for potential energy calculations. In this case, if the zero of potential energy were shifted, both the kinetic and potential energy values would change by the same amount, but their relative difference and the total energy would remain unchanged.
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Find the magnitude of the projected component of the force
acting along the pole. The pole is a 3.00 m tall vertical pole. The
force is 4.00 kN and acts along a cable between the top of the pole
and a
In this problem, we are asked to find the magnitude of the projected component of the force acting along the pole. The pole is a 3.00 m tall vertical pole. The force is 4.00 kN and acts along a cable between the top of the pole and a point on the ground that is 6.00 m from the bottom of the pole.
We can solve this problem by using trigonometry.Let's start by drawing a diagram to represent the situation. Let θ be the angle between the force vector and the horizontal axis, and let F be the force vector acting along the cable. Then, the projected component of the force acting along the pole is given by Fcos(θ). [tex]F_{\parallel}=F \cdot cos(\theta)[/tex]We can use the Pythagorean theorem to find the length of the cable. Since the pole is vertical, the length of the cable is equal to the hypotenuse of a right triangle whose legs are 3.00 m and 6.00 m.
Therefore, the length of the cable is[tex]L=\sqrt{3^2+6^2}=6.71m[/tex]Next, we need to find θ. We know that the tangent of θ is equal to the opposite side over the adjacent side (in this case, the opposite side is 3.00 m and the adjacent side is 6.00 m). Therefore,[tex]tan(\theta)=\frac{3.00}{6.00}=0.5[/tex]Taking the arctangent of both sides, we find that [tex]\theta=tan^{-1}(0.5)=26.6^\circ[/tex]
Now we can use the formula we derived earlier to find the magnitude of the projected component of the force acting along the pole:[tex]F_{\parallel}=F\cdot cos(\theta)=4.00\ kN\cdot cos(26.6^\circ)=3.63\ kN[/tex]Therefore, the magnitude of the projected component of the force acting along the pole is 3.63 kN.
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Its not 4 A hydrogen atom in an excited state can be ionized with less energy than when it is in its ground state. What is the minimum energy level n of an electron in a a hydrogen atom if 0.84eV of energy can ionize it?
The minimum energy level of an electron in a hydrogen atom can be determined by calculating the energy difference between the ionized state and the ground state.
Given that 0.84 eV of energy is required to ionize the hydrogen atom, we can find the corresponding energy level using the equation for the energy of a hydrogen atom.
The energy levels of electrons in a hydrogen atom are determined by the equation E = -13.6 eV/n^2,
where E is the energy of the electron, n is the principal quantum number representing the energy level, and -13.6 eV is the ionization energy of a hydrogen atom in its ground state.
To find the minimum energy level required for ionization, we can rearrange the equation as n^2 = -13.6 eV / E and substitute the given ionization energy:
n^2 = -13.6 eV / 0.84 eV
Simplifying the equation, we get:
n^2 ≈ 16.19
Taking the square root of both sides, we find:
n ≈ 4.03
Therefore, the minimum energy level of an electron in a hydrogen atom that requires 0.84 eV of energy for ionization is approximately n = 4.
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quickly please
7. If the retort temperature was 121 C and the highest temperature reached on cold point was value will be: a. 117 b.6 c. 1.8 d. 121 e. 4
The correct option is (c) 1.8.
Given: Retort temperature, t1 = 121°CCold point temperature, t2 =?
The value of the highest temperature reached on the cold point will be 117 °C.
Given t1 = 121°C and t2 = 117°C, the processing time and lethality are calculated by using the following formula: T = F0 / [((121 - Fo) / Z) + 1]Where T is the processing time, F0 is the lethality, Z is the temperature sensitivity valueThe temperature sensitivity value, Z is given as 10.
The lethality F0 is calculated by using the following formula:F0 = ((t1 - t2) / Z) × 10
Putting all the given values into the equation for F0:F0 = ((121°C - 117°C) / 10) × 10F0 = 4
The value of F0 obtained is 4.
Putting this value in the first equation: T = F0 / [((121°C - 4) / 10) + 1]T = 4 / [11.7]T = 0.34 minutes = 20.4 seconds
Hence, the correct option is (c) 1.8.
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What inductance must be connected to a 20 pF capacitor in an
oscillator capable of generating 600 nm (i.e., visible)
electromagnetic waves?
The inductance must be connected to a 20 pF capacitor in an oscillator capable of generating 600 nm (i.e., visible) electromagnetic waves is 21 nH.
In order to generate electromagnetic waves of 600 nm, the required frequency would be 5 x 10^14 Hz (c = λν,
where c is the speed of light,
λ is the wavelength, and
ν is the frequency).
Formula of resonance frequency:
f = 1 / 2π√LC
Where
f is the frequency,
L is the inductance, and
C is the capacitance.
Replacing the values:
f = 5 x 10^14 Hz and
C = 20 pF.
The required value of L would be approximately 21 nH (nanohenries).
Therefore, the inductance must be connected to a 20 pF capacitor in an oscillator capable of generating 600 nm (i.e., visible) electromagnetic waves is 21 nH.
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1.
2.Enumerate and explain briefly using a suitable
diagrams various methods of starting a polyphase induction
motor
9-4. How is induced torque developed in a single-phase induction motor (a) according to the double revolving-field theory and \( (b) \) according to the cross-field theory?
1. Various methods of starting a polyphase induction motorThe polyphase induction motors are generally started in any of the following ways:Direct-on-line startingStar-delta startingRotor resistance starting Autotransformer startingSoft-startingDirect-on-line starting
The most simple and economical method of starting a three-phase induction motor is DOL starting. This method is also known as full-voltage starting. In this method, the full voltage of the power supply is applied to the motor terminals. Therefore, the starting current is very large, typically 6 to 8 times the rated current. It is only used for small motors.Star-Delta StartingIn this method, the motor is started by applying the reduced voltage to the stator winding.
However, the rotor's magnetic field is alternating and pulsating in nature. The interaction of these two fields results in the production of torque. The alternating flux induces the current in the rotor. This induced current produces an alternating flux in the rotor that interacts with the stator flux and develops torque. The torque developed is proportional to the product of stator flux and rotor flux.
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Should the leakage inductance of an inductor be in parallel or in series with the magnetizing inductance?
a. In parallel
b. In series
c. It depends
The leakage inductance of an inductor should be in series with the magnetizing inductance. The leakage inductance in an inductor results from the incomplete magnetic linkage between the primary and secondary winding of the transformer caused by the leakage flux.
Leakage flux or magnetic flux is generated in the inductor as a result of the inductor's current. When the current in the inductor changes, the magnetic field also changes, causing the magnetic flux in the inductor to change.In parallel, the leakage inductance should not be used with the magnetizing inductance.
The leakage inductance generates an unwanted voltage drop and distorts the current flowing in the primary winding.
The magnetizing inductance, on the other hand, is utilized for energy storage and is the inductance necessary to maintain the magnetic field in the inductor.
As a result, the magnetizing inductance must be in series with the leakage inductance to prevent the leakage inductance from impeding the flow of current and causing unnecessary energy loss.
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5. Discuss the limitations of the "super diode" precision half-wave rectifier circuit and also explain a suitable circuit to overcome the same. [CO3] 10 Marks
The "super diode" precision half-wave rectifier circuit has limitations in terms of accuracy, bandwidth, and the need for a negative supply. A suitable circuit to overcome these limitations is the precision full-wave rectifier.
The "super diode" precision half-wave rectifier circuit is a modification of the conventional half-wave rectifier, which is designed to minimize the voltage drop that occurs across the diode. However, this circuit has limitations in terms of accuracy, bandwidth, and the need for a negative supply. The accuracy of the circuit is limited by the forward voltage drop of the diode, which can cause errors in the output voltage.
The bandwidth of the circuit is also limited by the time constant of the RC circuit. To overcome these limitations, a suitable circuit is the precision full-wave rectifier. This circuit is designed to produce a full-wave rectified output without the need for a negative supply. The precision full-wave rectifier uses a differential amplifier to compare the input voltage to a reference voltage, and switches the output to the positive or negative rail depending on the polarity of the input signal. This circuit is more accurate and has a wider bandwidth than the "super diode" precision half-wave rectifier circuit.
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3. On my way home one night, I am driving at a speed of 19.0: As I approach a stoplight, I see it turn yellow and speed up to make it through. 1 cover the next 36 meters in 1.65 seconds. Assume the acceleration during this 1.65 s is constant a. What is my noceleration while I speed up? b. What is my final speed? 4. You and your roommate are doing physics problems while in your bunk beds. You make a mistake and ask your roommate to toss up an craser. You are 1.40 m above your friend 1. What speed must your roommate throw the craser at in order for it to just barely reach you? (Remember that velocity is equal to zero at the highest point) b. How long does it take the craser to travel from your friend's hand to your hand? c. You like to snack while you study, so your fingers are covered in Cheeto dust. Your gross fingers cause you to drop the eraser from your top bunk, a height 2.50 m above the floor. How fast is the craser moving just before it hits the floor? Assume it is not moving before you drop it (an initial velocity of zero)
The acceleration while you speed up is 2.122 m/s². The final speed of the car is 48.1 m/s. The required speed at which your roommate must throw the eraser is 4.19 m/s. The speed of the eraser just before it hits the floor is 7.02 m/s.
a. The acceleration while you speed up is 2.122 m/s².
We can use the kinematic equation below to find the acceleration: Δx = vit + 1/2 at²
Here, Δx is the displacement (36 m), vi is the initial velocity (19.0 m/s), t is the time interval (1.65 s), and a is the acceleration.
Rearranging this equation, we get:
a = 2(Δx - vit)/t²
= 2(36 - 19.0 × 1.65)/1.65²
= 2.122 m/s²
b. The final speed of the car is 48.1 m/s. We can use the kinematic equation below to find the final velocity:
v² = vi² + 2aΔx
Here, vi is the initial velocity (19.0 m/s), a is the acceleration (2.122 m/s²), and Δx is the displacement (36 m). Rearranging this equation,
we get:
v = √(vi² + 2aΔx)= √(19.0² + 2 × 2.122 × 36)= 48.1 m/s
b. The required speed at which your roommate must throw the eraser is 4.19 m/s. We can use the kinematic equation below to find the initial velocity:
Δy = viyt - 1/2 gt²
Here, Δy is the displacement (1.40 m), t is the time taken to reach the highest point (when the velocity is zero), viy is the initial velocity in the y-direction, and g is the acceleration due to gravity (9.81 m/s²).
Since the velocity is zero at the highest point, we can use the following equation:
viy = gt.
Rearranging this equation, we get:
t = viy/g.
Substituting this value of t in the first equation, we get:
1.40 = viy(viy/g) - 1/2 g(viy/g)²= viy²/2gviy = √(2gΔy)= √(2 × 9.81 × 1.40)= 4.19 m/s
c. The speed of the eraser just before it hits the floor is 7.02 m/s. We can use the kinematic equation below to find the final velocity:
vf² = vi² + 2gΔy
Here, vi is the initial velocity (zero), g is the acceleration due to gravity (9.81 m/s²), and Δy is the displacement (2.50 m). Rearranging this equation, we get:
vf = √(vi² + 2gΔy)= √(2 × 9.81 × 2.50)= 7.02 m/s
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A 10-kW toaster roughly takes 6 minutes to heat four slices of bread. Find the cost of operating the toaster, in cents, once per day for 1 month (30 days). Assume energy costs of \( 0.74 \) cents/kWh.
Therefore, the cost of operating the toaster, in cents, once per day for 1 month (30 days) is 22.2 cents.
Given information: The power of toaster, P = 10 kW
Number of slices of bread, n = 4Time taken to heat four slices of bread, t = 6 minutes = 0.1 hour
Energy cost per kWh, C = 0.74 cents
To find: Cost of operating the toaster for once per day for a month (30 days)We know that the energy consumed by the toaster in terms of kWh is:
Energy consumed,
E = P × t
= 10 kW × 0.1 hour = 1 kWh
For 4 slices of bread, energy consumed = 1 kWh
Cost of operating the toaster for once
= Energy consumed × Cost per kWh = 1 kWh × 0.74 cents/kWh = 0.74 cents
For a day, the cost of operating the toaster
= 0.74 cents
For 30 days, the cost of operating the toaster = 0.74 cents/day × 30 days
= 22.2 cents
Therefore, the cost of operating the toaster, in cents, once per day for 1 month (30 days) is 22.2 cents.
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A baseball (m = 154 g) approaches a bat horizontally at a speed of 43.6 m/s (97.6 mi/h) and is hit straight back at a speed of 54.4 m/s (122 mi/h). If the ball is in contact with the bat for a time of 1.83 ms, what is the average force exerted on the ball by the bat? Neglect the weight of the bat, since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction. Number i 8247 Units N Vf
We can use the principle of impulse and momentum to solve the given problem. In order to do that, we need to find the initial momentum (p1) and final momentum (p2) of the baseball.
Then, we can find the change in momentum (Δp = p2 - p1) and use it to calculate the average force (F = Δp / Δt) exerted on the ball by the bat. Let's start by finding the initial and final momenta. Initial momentum: The baseball is approaching the bat horizontally with a speed of 43.6 m/s.
Therefore, its initial momentum is given by:p1 = m × v1where m is the mass of the baseball and v1 is its initial velocity.p1 [tex]= 154 g × (43.6 m/s) = 6718.4 g·m/s = 6.7184 kg·m/s[/tex]Final momentum: The baseball is hit straight back by the bat at a speed of 54.4 m/s. Therefore, its final momentum is given by:
p2 = m × v2where v2 is its final velocity.p2 = 154 g × (54.4 m/s) = 8369.6 g·m/s = 8.3696 kg·m/sChange in momentum: The change in momentum of the baseball is given by:[tex]Δp = p2 - p1Δp = 8.3696 kg·m/s - 6.7184 kg·m/s = 1.6512 kg·m/s[/tex]
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Two wave pulses of the same magnitude amplitude exist at the same location in space, and the result is sown in the figure. What kind of interference would this be considered? This is constructive interference. This is destructive interference. The interference is neither constructive or destructive. The interference is both constructive and destructive.
The interference is an example of constructive interference. Constructive interference occurs when two waves meet and their amplitudes add up, resulting in a larger wave. In this case, the two wave pulses have the same magnitude amplitude and are at the same location in space. As a result, when the waves overlap, they reinforce each other and create a larger wave.
To explain further, when two waves have the same amplitude and align perfectly, their crests and troughs coincide, causing the wave amplitudes to add up. This results in a wave with a higher amplitude and energy. In the figure, we can see that the overlapping waves create a wave with a greater magnitude compared to the individual waves.
In constructive interference, the phase difference between the waves is either zero or a whole number multiple of the wavelength. This means that the two waves are in sync and reinforce each other, leading to an increase in amplitude.
On the other hand, destructive interference occurs when two waves meet and their amplitudes cancel each other out. This happens when the waves have opposite phases or a phase difference of half a wavelength. In this case, the resulting wave would have a smaller or even zero amplitude.
In conclusion, the interference is considered to be constructive interference because the overlapping waves reinforce each other, resulting in a larger wave.
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Calculate the current ia and the vc for all values oft (time), the initial voltage of the capacitor is 0 V
The impedance of the circuit can be calculated using the formula, Z = RAs no values are given for the inductance, capacitance and resistance of the circuit, the calculation of i and vc cannot be done. Hence, the final answer is, there is insufficient information to calculate the current ia and the vc for all values of time (t). The given information is inadequate.
Given that the initial voltage of the capacitor is 0V, to calculate the current ia and the vc for all values of time (t), the circuit diagram of a series RLC circuit is required:
RLC Circuit Diagram
The equation for current in the circuit is given by, i = [V0 / Z] * sin (ωt - φ)
Where,
Z = Impedance of the circuit
ω = Angular frequency = 2πf (where f is the frequency of the AC source)
V0 = Amplitude of the AC voltage
φ = Phase angle
At resonance, the impedance of the circuit is minimum. Hence, the current in the circuit will be maximum at resonance. The resonant frequency of the circuit is given by, f = 1 / (2π√LC)
Where,L = Inductance of the circuit C = Capacitance of the circuit
At resonance, the phase angle φ is 0°.
Therefore, the current in the circuit can be calculated using the formula,i = V0 / R
Since the values of the RLC circuit are not provided, the calculation of i and vc cannot be done.
However, the formulae for the same are, i = [V0 / Z] * sin (ωt - φ)
vc = V0 sin (ωt - φ)
Here, V0 is the voltage of the AC source.In order to calculate the value of Z, the formulae for inductive reactance and capacitive reactance is required.
XL = 2πfLXC = 1 / 2πfC
Calculating the impedances of the inductor and the capacitor, respectively,
ZL = jXLZC
= 1 / jXC
At resonance, the impedances of the inductor and capacitor will be equal and opposite, hence they will cancel out each other. Thus, the only impedance that will remain in the circuit is the resistance R.
Therefore, the impedance of the circuit can be calculated using the formula, Z = RAs no values are given for the inductance, capacitance and resistance of the circuit, the calculation of i and vc cannot be done.
Hence, the final answer is, there is insufficient information to calculate the current ia and the vc for all values of time (t). The given information is inadequate.
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I would appreciate a small description or showing which formulas were used. 2.A load absorbs 10-j4 kVA of power from a 60-Hz source with a peak voltage of 440 V a.(3 pts Find the peak current drawn by the load b.2 pts Find the power factor of the load.Include whether it is leading or lagging. C. 4 pts Sketch and label the power triangle
a) The formula to calculate the peak current (Ip) drawn by the load is given as: Where θ is the phase angle between the voltage and current vectors. Since the load absorbs power, the power factor will be lagging. Therefore, the power factor of the load is given by:
Ip = P / (√2 * Vp)
Where:
P = Power in Watts
Vp = Peak voltage
So, the peak current (Ip) drawn by the load is given by:
Ip = 10000 / (√2 * 440) = 31.57 A
Hence, the peak current drawn by the load is 31.57 A.
b) The formula to calculate the power factor is given as:
PF = cos(θ)
Where θ is the phase angle between the voltage and current vectors. Since the load absorbs power, the power factor will be lagging. Therefore, the power factor of the load is given by:
PF = cos(θ) = cos(arccos(10 / √(116^2 + 10^2))) = cos(0.0874) = 0.996
Hence, the power factor of the load is 0.996 leading.
c) The sketch of the power triangle is as follows:
The magnitude of the impedance is given by:
|Z| = √(R^2 + X^2) = √(0^2 + 4^2) = 4 Ω
The phase angle between the voltage and current vectors is given by:
θ = arctan(-4/0) = -90°
The apparent power is given by:
S = Vrms * Irms = (440 / √2) * (10 / √2) = 2200 VA
The reactive power is given by:
Q = S * sin(θ) = 2200 * sin(-90°) = -2200 VAR
The real power is given by:
P = S * cos(θ) = 2200 * cos(-90°) = 0 W
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About how many half-lives have elapsed when a radioactive substance has decayed to less than 1% of its original amount? 05 O 50 07 O 10 32 99
About 7 half-lives have elapsed when a radioactive substance has decayed to less than 1% of its original amount.
Radioactive decay is a process where a nucleus of an unstable atom loses energy by emitting radiation. The amount of time it takes for half of a sample to decay is called the half-life of the substance. If we want to know the amount of time it takes for a radioactive substance to decay to less than 1% of its original amount, then it would require a minimum of 7 half-lives to pass by.
This is because, after each half-life, the amount of radioactive substance will be reduced by 50%. So, if we take 50% for 7 times (7 half-lives), it will give us a value that is less than 1%. Therefore, about 7 half-lives have elapsed when a radioactive substance has decayed to less than 1% of its original amount.
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A 0.6 specific gravity gas flows from a 2-in pipe through a 1-in nozzle-type choke. The upstream pressure and temperature are 120psi and 70∘F, respectively. The downstream pressure is 90psi (measured 2ft from the nozzle). The gas-specific heat ratio is 1.3. The gas viscosity is 0.0125cp. (1) What is the expected pressure at the nozzle outlet? (5 Mark) (2) What is the expected daily flow rate? (10 Marks) (3) Assuming the compressibility ratio of the upstream fluid to downstream fluid is approximately 1.0, is icing a potential
The pressure at the nozzle outlet can be determined by taking the downstream pressure and multiplying it by 0.916. P2 (at nozzle outlet) = 90 psi * 0.916 = 82.44 psi Pressure at the nozzle outlet = 82.44 psi.
Formula:
Volumetric flow rate, V = m * R / Pm = mass flow rateR = Universal gas constant / Molecular weight of gas = 10.732 * 10⁶ / 22.4 * 0.6 = 0.2142P = Pressure of gas in psiaP = 90 psi * 144 / 14.696 = 922.08 psiaV = 1.24 * 10³ * m / PQ = V * 24 * 3600 = 33.696 * m
Daily flow rate, Q = 33.696 * mQ = 33.696 * 0.2125 * 60 * 60 * 24 = 553.93 bbl/dayThe expected daily flow rate is 553.93 bbl/day.
Assuming the compressibility ratio of the upstream fluid to downstream fluid is approximately 1.0, If the compressibility ratio of the upstream fluid to downstream fluid is about 1.0, icing is not likely to be a problem since there will be no sharp changes in pressure or temperature that could lead to water or other substances condensing and then freezing.
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write the answers as VECTORS! with XY coordinates! other answer on
here incorrect...
Given: A sphere, having a mass of \( W \), is supported by two smooth, inclined surfaces. A horizontal force \( F \) acts at the center of the sphere, as shown. Find: Determine the reaction forces act
The reaction forces acting on the sphere are (W/2sin(θ)) (cos(θ), sin(θ)). The XY coordinate values of the required vector are (W/2sin(θ)) cos(θ) and (W/2sin(θ)) sin(θ).
Given: A sphere, having a mass of W, is supported by two smooth, inclined surfaces. A horizontal force F acts at the center of the sphere, as shown. We need to determine the reaction forces acting on the sphere. Let us consider the figure below for the derivation of the required solution:
The forces acting on the sphere are its weight W and the horizontal force F. Let R1 and R2 be the reaction forces on the inclined planes 1 and 2, respectively.
The reaction forces can be resolved into their components as shown:
R1 cos(α) - R2 cos(β)
= 0 (i)R1 sin(α) + R2 sin(β)
= W
(ii)The horizontal force F acts at the center of the sphere, which is at the midpoint of the lines joining the points of contact between the sphere and the inclined planes.
Therefore, the reaction forces R1 and R2 are equal.
Hence,R1 = R2 = R
From equations (i) and (ii), we get:
R cos(α) = R cos(β)
Therefore, α = β
Let the angle α = β = θ.
Therefore, equation (ii) becomes:
R sin(θ) = W/2
Hence, R = W/2sin(θ)
The XY coordinate values of R are (R cos(θ), R sin(θ))
Therefore, R = (W/2sin(θ)) (cos(θ), sin(θ))
The reaction forces R1 and R2 can be obtained as follows:
R1 = R2 = R = (W/2sin(θ)) (cos(θ), sin(θ))
Hence, the reaction forces acting on the sphere are (W/2sin(θ)) (cos(θ), sin(θ)). The XY coordinate values of the required vector are (W/2sin(θ)) cos(θ) and (W/2sin(θ)) sin(θ).
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A coaxial cable is being used to transmit a signal with frequencies between 20MHz and 50MHz. The line has a propagation velocity of 200Mm/s. At what physical line length (in meters) would you need to begin worrying about transmission line theory? (Use the λ/16 rule of thumb)
The physical line length is 160m
Given:
Frequency range: 20MHz to 50MHz
Velocity of propagation: 200Mm/s
Calculation:
The formula for wavelength (λ) is given by: λ = c/f
Substituting the given values: λ = 3 × 10^8 m/s ÷ (20 × 10^6 Hz)
Calculating: λ = 15 m
Using the λ/16 rule of thumb:
λ/16 = 15/16 = 0.9375 m
Determining the line length at which transmission line theory is significant:
Dividing 150 by 0.9375: 150 ÷ 0.9375 = 160
Conclusion:
The physical line length at which we need to start worrying about transmission line theory is approximately 160 meters.
Therefore, the answer is 160 meters
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A horizontal aluminum rod 4.3 cm in diameter projects 4.4 cm from a wall. A 1300 kg object is suspended from the end of the rod. The shear modulus of aluminum is 3.0-1010 N/m². Neglecting the rod's mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.
Shear stress on the rod: The shear stress, τ, on a solid cylindrical rod is given by:
τ = (F/A) [1 + (r/h)]
where, F = Load applied to the rod
A = Cross-sectional area of the rod r = Radius of the rod h = Height of the rod
The cross-sectional area of the rod,
[tex]A = (π/4) × d² = (π/4) × (4.3 cm)² = 14.45 cm²[/tex] where d is the diameter of the rod.
Substituting the given values:
[tex]F = 1300 kg g = 9.8 m/s²= 1.274 × 10⁴[/tex]N(here g is the acceleration due to gravity)
A = 14.45 cm²r = 2.15 cm
= 0.0215 m h = 4.4 cm
= 0.044 mτ = (F/A) [1 + (r/h)]
= (1.274 × 10⁴ N)/(14.45×10⁻⁴ m²) [1 + (0.0215 m)/(0.044 m)]
= 6.727 × 10⁸ N/m²
(b) Vertical deflection of the end of the rod:
y = (FL)/(Ah²) [3L/h - 4r/πh]
Substituting the given values:
[tex]L = 4.4 cm = 0.044 mF = 1300 kgg = 9.8 m/s²= 1.274 × 10⁴[/tex]N
(here g is the acceleration due to gravity)
[tex]A = 14.45 cm²r = 2.15 cm = 0.0215 mh = 4.4 cm = 0.044[/tex]my = (FL)/(Ah²)
[3L/h - 4r/πh]
[tex]= (1.274 × 10⁴ N × 0.044 m)/(14.45×10⁻⁴ m²[/tex] ×[tex](0.044 m)²) [3 × 0.044 m/0.044 m - (4 × 0.0215 m)/(π × 0.044 m)][/tex]
= 0.0138 m
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