(a) The amount of heat released by water when it freezes The amount of heat released by water when it freezes can be calculated using the specific heat capacity and the latent heat of fusion of water.
We know that 1 g of water requires 334 J of energy to change from ice at 0°C to liquid at 0°C. So, 1 kg of water requires 334 kJ of energy to melt from ice to liquid at 0°C.Similarly, 1 kg of water requires 334 kJ of energy to freeze from liquid to ice at 0°C.So, the amount of heat released when 1 kg of water freezes from 0°C to ice at 0°C is 334 kJ/kg of water.At 0°C, 1 kg of water occupies 1 L or 1000 cm³ of volume. Hence, the density of water at 0°C is 1000 kg/m³.
Given, a grower sprays 8.00 kg of water at 0°C onto a fruit tree.So, the amount of heat released by 8.00 kg of water when it freezes can be calculated as follows,
Q = (334 kJ/kg) x (8.00 kg)
Q = 2672 kJ(b) The amount of temperature rise in the tree The amount of temperature rise in the tree can be calculated using the formula,
Q = mcΔT
Where,Q = Heat absorbed by the tree
= Heat released by the water when it freezesm
= Mass of the tree
= 114 kgc
= Specific heat capacity of the tree
= 2.5 x 10³ J/(kg°C)
ΔT = Temperature rise in the tree
So, the amount of temperature rise in the tree can be calculated as follows,ΔT = Q/mcΔT
= (2672 kJ) / (114 kg x 2.5 x 10³ J/(kg°C))
ΔT = 9.37°C
Therefore, the temperature of a 114-kg tree would rise by 9.37°C if it absorbed the heat released in part (a).
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Two pipes of different diameters are joined together in series.
The smaller pipe has a diameter of 0.1m and length of
14m and the larger pipe a diameter of 0.2m and
length of 13m. Oil (density 800kg/m
When two pipes of different diameters are joined in series, the volumetric flow rate remains constant. To find the speed and the volumetric flow rate of the liquid in the two pipes, we use the continuity equation, which is: A1V1=A2V2, where A1 and A2 are the cross-sectional areas of the two pipes, and V1 and V2 are the speeds of the liquids.
The volumetric flow rate can be found using the formula Q=AV, where Q is the volumetric flow rate and V is the speed of the liquid. Assume the speed of the liquid in the smaller pipe is V1, and the speed of the liquid in the larger pipe is V2. Let us take the density of the oil to be 800kg/m³.The cross-sectional area of the smaller pipe is: A1=π(0.1/2)²=0.007854m²
The cross-sectional area of the larger pipe is: A2=π(0.2/2)²=0.031416m²
Using the continuity equation:A1V1=A2V2V2=A1V1/A2V2=0.007854V1/0.031416=0.198V1
The volumetric flow rate is the same in both pipes:Q=AV=0.007854V1=0.031416V2
We can substitute V2 with the expression we derived earlier:
V2=0.198V1Q=0.007854V1=0.031416(0.198V1)Q=0.00493m³/s
The speed of the liquid in the smaller pipe is:
V1=Q/A1=0.00493/0.007854=0.627m/s
The speed of the liquid in the larger pipe is:
V2=Q/A2=0.00493/0.031416=0.157m/s
Therefore, the speed of the liquid in the smaller pipe is 0.627m/s, and the speed of the liquid in the larger pipe is 0.157m/s. The volumetric flow rate of the liquid is 0.00493m³/s. The total length of the two pipes is 14m + 13m = 27m,
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4) A single phase motor draws 37 kW at 0.72 power-factor lagging from a 1.8 kV source. Find the capacitance needed in parallel with the motor to make the combined power factor to 0.95, and determine the current before and after adding that compensator.
A single-phase motor requires a capacitance of 55.7 μF to raise the power factor to 0.95 lagging.
Let’s calculate the reactive power required by the motor using the formula,
P = VI cos (θ)
Reactive power, Q = VI sin (θ)
37,000 = 1,800 I cos (cos⁻¹ 0.72)
⇒ I = 37,000 / (1,800 × 0.72) = 28.87 A
Q = 1,800 × 28.87 × sin (cos⁻¹ 0.72)
Q = 1,800 × 28.87 × 0.69
Q = 36,011.3 VAr (lagging)
Let X be the capacitive reactance that we need to connect in parallel with the motor.
tan (cos⁻¹ 0.95) = (1 - 0.72) / (0.72 + X)
tan (cos⁻¹ 0.95) = 0.94 / (0.72 + X)
X = 55.7 μF
The current before and after adding compensator is:
Before adding compensator:
I = 28.87 A
After adding compensator:
New power factor, cos φ = 0.95
cos⁻¹ 0.95 = 18.19° (leading)
tan 18.19 = (1 - 0.95) / X
X = 48.3 μF
I = 37,000 / (1,800 × cos (cos⁻¹ 0.95))
I = 34.54 A
Therefore, the current before and after adding that compensator is 28.87 A and 34.54 A, respectively.
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If you were to need to move a radioactive source, would you be
better off using tongs, or wearing gloves, if you only had access
to one or the other?
If one needs to move a radioactive source, it is better to use tongs, especially those made of non-metallic and non-conductive materials. If only one of the two items, tongs or gloves, are accessible, the tongs will be a better option than gloves.
If one needs to move a radioactive source, it is better to use tongs, especially those made of non-metallic and non-conductive materials. If only one of the two items, tongs or gloves, are accessible, the tongs will be a better option than gloves. An appropriate pair of tongs can protect the user from the radioactive radiation of the source while they move it. This protection will not be provided by gloves as they are not made to protect against the harmful radiation produced by the radioactive source. This is because gloves are made to provide physical protection to the hands of the user and to shield them from the dangers of chemical substances, which is different from the radiation danger.
The tongs used to move radioactive sources should be non-metallic and non-conductive to protect the user. They should also be heavy-duty and sturdy enough to support the weight of the source being moved. Moreover, one should remember that while moving a radioactive source, one must wear appropriate personal protective equipment such as a lab coat, closed-toe shoes, and safety goggles for extra protection. The radioactive source should also be properly labeled and handled with care, as it has the potential to cause harm if not handled carefully. Furthermore, radioactive materials should be stored properly in a specially designed storage container that minimizes the risk of exposure.
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One of the Milankovitch cycles has to do with changes in the
shape of the Earth’s orbit. Assume for the purposes of this
question that 340.4 W/m2 is the overall average insolation over the
course of one entire cycle of variation in the Earth’s orbit.
Change that average value to a new one that would describe
insolation during a time when the Earth’s orbit is very elliptical
(oval-shaped). Note: Don’t worry about your value being the actual
insolation value, just make it a little different from 340.4 to
describe the expected change in insolation during the elliptical
phase. Would the change you made lead to global warming or
cooling?
A more elliptical orbit would cause changes in the average insolation value, and depending on whether it is increased or decreased, it would lead to global warming or cooling respectively.
The change in the average insolation during a time when the Earth's orbit is very elliptical (oval-shaped) would result in a higher or lower value than 340.4 W/m2. Since the shape of the orbit affects the distance between the Earth and the Sun, a more elliptical orbit would mean that the Earth is closer to the Sun at some points in the orbit and farther away at others. This would lead to variations in the amount of solar radiation reaching the earth's surface.
If the average insolation value is increased, it would lead to global warming as more solar radiation is absorbed by the Earth, increasing the overall temperature. Conversely, if the average insolation value is decreased, it would lead to cooling as less solar radiation is absorbed, resulting in lower temperatures.
To summarize, a more elliptical orbit would cause changes in the average insolation value, and depending on whether it is increased or decreased, it would lead to global warming or cooling respectively.
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Q6. An Alternator rated at 10 kV protected by the balanced circulating current system has its neutral grounded through a resistance of X ohms. The protective relay is set to operate when there is an out of balance current of 1.8 amp in the pilot wires, which are connected to the secondary windings of 1000/5 ratio current transformers. (a) Determine the per cent winding which remains unprotected, (b) Find the minimum value of the Earthing resistance required to protect 75% of the winding. Suppose, X is the last non-zero digit of your student ID. [3*2]
Step (a) involves calculating the percentage of winding that remains unprotected by determining the rated current, actual current, and performing a division and multiplication calculation. Step (b) requires finding the minimum value of the Earthing resistance based on the unprotected winding percentage and using a specific formula, where the last non-zero digit of the student ID is used as a variable.
we need to follow the steps below
(a) Determine the per cent winding which remains unprotected:
- First, calculate the rated current of the alternator by dividing the rated apparent power (10 kV) by the rated voltage (10 kV).
- Then, calculate the actual current flowing through the pilot wires by multiplying the out-of-balance current (1.8 A) with the current transformer ratio (1000/5).
- Finally, determine the percentage of winding remaining unprotected by dividing the actual current by the rated current and multiplying by 100.
(b) Find the minimum value of the Earthing resistance required to protect 75% of the winding:
- Calculate the unprotected winding percentage by subtracting 75% from 100%.
- Use this percentage to determine the minimum value of the Earthing resistance using the formula: R = X / (unprotected winding percentage / 100).
Replace X with the last non-zero digit of your student ID in the above formula to find the specific value.
Note: Please provide your student ID's last non-zero digit for an accurate calculation of the Earthing resistance.
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Problem 1: A spaceship flies directly past you (a negligible distance away) with a speed of 0.5c. You see a clock in the ship through the porthole that reads 12:00. An hour later, as measured by a stationary clock, you look through a telescope at the clock. What time does it read? Give your answer to the nearest second. Caution; this is not an ordinary time dilation problem.
The time observed through a telescope is 13:23:21.
As observed from the earth, the spaceship is moving with a velocity of 0.5c. Thus, the time dilation equation can be applied. This problem is a bit different from regular time dilation problems since the spaceship flies directly past the observer, and a negligible distance away, which means that the perpendicular distance between the observer and spaceship is approximately 0.
Using the time dilation equation; T′=T√1−v2/c2T = 12:00v = 0.5cT′ = 12:00 × √1−(0.5c)2/c2 = 12:00 × 0.866 = 10:23:60 = 10:24Thus, the clock on the spaceship reads 10:24 when it passes the observer. As measured by a stationary clock, an hour later, the time elapsed on the spaceship is given byT′′=T′√1−v2/c2T′′ = 10:24 × √1−(0.5c)2/c2 = 10:24 × 0.866 = 09:00:40After an hour, the elapsed time on the spaceship is 09:00:40.
As measured by the observer's clock, one hour has passed. Therefore, the time elapsed on the observer's clock is 1 hour. Using the formula of elapsed time, we get: Tobs=(T′′−T)Tobs = (09:00:40 − 12:00) = − 02:59:20
Therefore, the time on the spaceship clock that the observer would see through the telescope would be 1 hour and 2:59:20 after the spaceship has passed the observer.
So, the final time would be: 10:24 + 2:59:20 = 13:23:20 ≈ 13:23:21 (to the nearest second)
The time observed through a telescope is 13:23:21.
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3. My hot water system maintains a volume of 130 litres of water, which it heats to a maximum temperature of 60
∘
C in a cylindrical tank 1.5 metres tall. It works by drawing in cold (temperature 10
∘
C ) water at the base of the tank, where the heating element is located. Hot water leaves through a pipe at the top of the system. If the tank is full of water at 60
∘
C, the manufacturer guarantees that it will produce 260 litres of water at or above 50
∘
C in the first hour of use. Temperature diffusion (as per the heat equation) in water has a diffusion coefficient of around 1.5×10
−7
m
2
/s. What is the minimum rate at which the elememt must heat the water (in
∘
C/ litre/minute), to meet the manufacturer's guarantee? Figure 2: Schematic of the hot water system
The heating rate by the volume of water and convert the time to minutes is (Q_total / V) / (1 hour) * (1 litre / 1,000 cm^3) * (60 minutes / 1 hour)
To determine the minimum heating rate required to meet the manufacturer's guarantee, we need to calculate the amount of heat that needs to be supplied to the water in the first hour.
The heat equation for temperature diffusion in water is given by:
∂T/∂t = D * (∂^2T/∂x^2)
In this case, the temperature gradient in the tank is only in the vertical direction, so we can simplify the equation to:
∂T/∂t = D * (∂^2T/∂z^2)
To solve this equation, we assume that the tank is well-mixed, so the temperature is uniform throughout the tank at any given time. This allows us to treat the problem as one-dimensional.
The heat transferred into the water can be expressed as:
Q = m * C * ΔT
The mass of water can be calculated from the volume using the density of water:
m = V * ρ
To meet the manufacturer's guarantee, the system needs to produce 260 liters (260 kg) of water at or above 50°C in the first hour. Therefore, the heat transferred in one hour (Q_total) can be calculated as:
Q_total = m_total * C * ΔT
To calculate the heating rate, we divide the total heat transferred by the time (1 hour or 3,600 seconds):
Heating rate = Q_total / (1 hour)
Finally, to express the heating rate in °C/litre/minute, we divide the heating rate by the volume of water and convert the time to minutes:
Heating rate = (Q_total / V) / (1 hour) * (1 litre / 1,000 cm^3) * (60 minutes / 1 hour)
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Two objects (m 1
=5.25 kg and m 2
=2.70 kg) are connected ty a light string passing over a light, frictionless pulley as in the figure below. The 5.25. kg object is teleased from rest at a point n=4.00 m obove the table. (a) Determine the speed of each object when the two pass each other: Your response differs from the carrect answer by more than 10%. Double check your calculations. m/5 (b) Determine the speed of each object at the moment the 5.25-kg object hits the table. mins (c) How miuch higher does the 2.70−kg object trovel afer the 5.25 kg object hits the toble?
(a) The 2.70 kg object will travel an additional 7.70 m higher than the initial height of the 5.25 kg object when they pass each other.
(b) The speed of the 5.25 kg object at the moment it hits the table is approximately 8.85 m/s.
(c) The 2.70 kg object travels an additional 4.00 m higher after the 5.25 kg object hits the table.
The problem involves two objects, one with a mass of 5.25 kg and the other with a mass of 2.70 kg. These objects are connected by a light string that passes over a light, frictionless pulley. The 5.25 kg object is released from rest at a point 4.00 m above the table.
(a) To determine the speed of each object when they pass each other, we need to consider the conservation of energy. As the 5.25 kg object falls, it gains potential energy which is converted into kinetic energy. At the same time, the 2.70 kg object is being pulled up, gaining potential energy and losing kinetic energy.
Since energy is conserved, the potential energy gained by the 5.25 kg object is equal to the potential energy lost by the 2.70 kg object. Mathematically, we can express this as:
m₁ * g * h₁ = m₂ * g * h₂
where m₁ and m₂ are the masses of the objects, g is the acceleration due to gravity (approximately 9.8 m/s²), h₁ is the initial height of the 5.25 kg object, and h₂ is the final height of the 2.70 kg object.
Substituting the given values, we have:
5.25 kg * 9.8 m/s² * 4.00 m = 2.70 kg * 9.8 m/s² * h₂
Simplifying the equation, we can solve for h₂:
h₂ = (5.25 kg * 9.8 m/s² * 4.00 m) / (2.70 kg * 9.8 m/s²)
h₂ ≈ 7.70 m
This means that the 2.70 kg object will travel an additional 7.70 m higher than the initial height of the 5.25 kg object.
(b) To determine the speed of each object at the moment the 5.25 kg object hits the table, we can use the principle of conservation of mechanical energy. At this point, all the potential energy of the 5.25 kg object is converted into kinetic energy.
The potential energy of the 5.25 kg object is given by:
Potential energy = mass * gravity * height
Potential energy = 5.25 kg * 9.8 m/s² * 4.00 m
The kinetic energy of the 5.25 kg object is given by:
Kinetic energy = (1/2) * mass * velocity²
Setting the potential energy equal to the kinetic energy and solving for the velocity, we get:
(1/2) * 5.25 kg * velocity² = 5.25 kg * 9.8 m/s² * 4.00 m
Simplifying the equation, we can solve for the velocity:
velocity² = 2 * 9.8 m/s² * 4.00 m
velocity² = 78.4 m²/s²
velocity ≈ 8.85 m/s
So, the speed of the 5.25 kg object at the moment it hits the table is approximately 8.85 m/s.
(c) To find out how much higher the 2.70 kg object travels after the 5.25 kg object hits the table, we can subtract the final height of the 5.25 kg object from the initial height of the 2.70 kg object.
Final height of the 5.25 kg object is 0 m (since it hits the table).
Initial height of the 2.70 kg object is 4.00 m.
Therefore, the height difference is:
4.00 m - 0 m = 4.00 m
So, the 2.70 kg object travels an additional 4.00 m higher after the 5.25 kg object hits the table.
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Convert \( 2880^{\circ} \) (a) to revolutions. (c) to radians.
The given angle in degree 2880° is equal to 8 revolutions. The given angle of 2880° is equal to 16π radians.
Given angle in degree: 2880°
(a) Converting 2880° into revolutions.
1 revolution = 360°
Thus, 2880° = 2880/360 revolutions = 8 revolutions
Hence, the given angle in degree 2880° is equal to 8 revolutions.
(c) Converting 2880° into radians.
The conversion between degree and radians is given byπ radians = 180° or 1 radian = 180°/π
Thus, 1° = π/180 radians
Multiplying both sides by 2880, we get
2880° = 2880 × π/180 radians = 16π radians
Therefore, the given angle of 2880° is equal to 16π radians.
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The Observable Universe has a diameter of? 100,000 Light Years 92 Billion Light Years 50 Astronomical Units 14 Billion Light Years
The Observable Universe has a diameter of approximately 92 billion light-years. The correct answer is option : 92 Billion Light Years.
This measurement takes into account the current age of the Universe and the expansion of space over time. It represents the maximum distance that light has had the opportunity to travel since the Big Bang. However, it is important to note that the Observable Universe is not the entire Universe. Due to the expansion of space, there are regions beyond our observable reach. The 92 billion light-year measurement represents the scale of the observable portion, encompassing a vast expanse of galaxies, stars, and other celestial objects that we can potentially observe from Earth. Therefore the correct answer is option : 92 Billion Light Years.
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2. Describe the methods of measuring the ripple contents of a high DC voltage with necessary details. \( [10] \)
In electronics, a power supply delivers electric power to an electrical load. The power supply converts one form of electrical power to another form of electrical power. These electronic power supplies are complex and require careful measurement of the voltage output quality.
Ripple measurement, or the AC voltage that's superimposed on the DC voltage output, is one such quality that must be measured. Here are a few methods of measuring ripple content in a high DC voltage signal:1. Use an oscilloscope:An oscilloscope is used to measure the voltage waveform of an electrical signal. To measure ripple in a DC voltage, connect the oscilloscope probes to the output voltage,
set the scope to AC coupling mode, and check the waveform for any additional AC component superimposed on the DC voltage. If ripple is present, it will be visible on the scope's screen.2. Using a Spectrum Analyzer:A spectrum analyzer is an electronic device that is used to measure the frequency spectrum of an electrical signal. It is used to measure the amplitude and frequency of the ripple in the DC voltage signal. By analyzing the spectrum, the ripple can be measured.
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what is the defining characteristic of a water cycle gizmo answers
The defining characteristic of a water cycle gizmo is its ability to simulate the natural water cycle in a controlled environment.
A water cycle gizmo is a device or model that demonstrates the various processes involved in the water cycle. It typically includes components that represent evaporation, condensation, precipitation, and runoff. The defining characteristic of a water cycle gizmo is its ability to simulate the natural water cycle in a controlled environment.
Water cycle gizmos often use simple mechanisms such as heat sources, condensation chambers, and pumps to mimic the processes that occur in nature. By using a water cycle gizmo, students can gain a hands-on experience and develop a deeper understanding of the water cycle.
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2.17 A transmitter supplies 100 W to a 50 lossless line that is 5.65 wavelength long. The other end of the line is connected to an antenna with a characteristic impedance of 150 + j 25 2. Calculate the: 2.17.1 the normalised impedance in polar form. (2) 2.17.2 the normalised admittance. (2) 2.17.3 the reflection coefficient in polar form.
The answer to the question is:2.17.1 the normalised impedance in polar form: 151.2Ω with an angle of 9.46 degrees.2.17.2 the normalised admittance in polar form: 0.0063346S with an angle of -9.46 degrees.2.17.3 the reflection coefficient in polar form: 77.6Ω with an angle of -18.96 degrees.
The first thing we need to do is to calculate the characteristic impedance of the transmission line. Z0 = sqrt(L/C) where L is the inductance per unit length and C is the capacitance per unit length. If the line is lossless, then the inductance and capacitance will be equal, so
[tex]L = C = 1/(LC)[/tex]
So
[tex]Z0 = sqrt(L/C)[/tex]
= sqrt(1/(LC))
= sqrt(1/1) = 1
Next, we need to calculate the wavelength in the line.
l[tex]amda = c/f[/tex]
= c/2pi
= 3e8/2pi
= 4.77e7 m/s / (2*3.14159*5.65) = 2.67 m
Now we can calculate the normalised impedance.
Z = ZL/Z0
= (150+j25)/(1+j0)
= 150+j25
The normalised impedance in polar form is:
|Z| = sqrt(150^2+25^2)
= 151.2Ω
θ = atan(25/150)
= 9.46 degrees2.17.2 the normalised admittance
The normalised admittance is: Y = 1/Z
= 1/(150+j25)
= 0.0063158-j0.0010526
The normalised admittance in polar form is:|Y| = sqrt(0.0063158^2+0.0010526^2)
= 0.0063346Sθ
= atan(-0.0010526/0.0063158)
= -9.46 degrees
2.17.3 the reflection coefficient in polar form.
The reflection coefficient is:Γ = (ZL-Z0)/(ZL+Z0)
where ZL is the load impedance, which is 150+j25.
Γ = (150+j25-1)/(150+j25+1)
= 74-j24
The reflection coefficient in polar form is:|Γ| = sqrt(74^2+24^2)
= 77.6Ωθ = atan(-24/74)
= -18.96 degrees
Thus, the answer to the question is:2.17.1 the normalised impedance in polar form: 151.2Ω with an angle of 9.46 degrees.2.17.2 the normalised admittance in polar form: 0.0063346S with an angle of -9.46 degrees.2.17.3 the reflection coefficient in polar form: 77.6Ω with an angle of -18.96 degrees.
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Consider the transfer function below:
(a) Identify the poles and zeros of the open-loop system and
discuss the existence of poles and zeros at infinity, as well as
open-loop stability;
(b) Find the
Given Transfer Function:[tex]$$G(s) = \frac{10(s+2)}{(s-1)(s+3)}$$(a)[/tex]Identification of Poles and Zeros and Open-loop StabilityThe numerator and denominator of G(s) can be written as:$$G(s) = \frac{10(s+2)}{(s-1)(s+3)} = 10\frac{(s+2)}{(s-1)}\frac{1}{(s+3)}$$Therefore the poles of the open-loop system are s=1 and s=-3 and the zero is at s=-2. Now, let's discuss the existence of poles and zeros at infinity and the open-loop stability.
In G(s), the degree of numerator is 1 and the degree of the denominator is 2. Thus, we can say that the transfer function approaches 0 as s → ∞. This means there are no poles or zeros at infinity. For the open-loop stability, we need to look at the pole-zero plot. As the poles of the open-loop system lie on the left-hand side of the imaginary axis, the system is stable. So, the open-loop system is stable.
(b) Finding Closed-loop Transfer FunctionLet's find the closed-loop transfer function using feedback loop,Where$$H(s)=1$$$$G(s)=\frac{Y(s)}{X(s)}=\frac{G(s)}{1+G(s)H(s)}$$Substituting H(s) and G(s) in the above equation, we get$$\frac{Y(s)}{X(s)} = \frac{\frac{10(s+2)}{(s-1)(s+3)}}{1+\frac{10(s+2)}{(s-1)(s+3)}(1)}$$$$\frac{Y(s)}{X(s)} = \frac{10(s+2)}{(s+3)(s+12)}$$Hence, the closed-loop transfer function is $$\frac{Y(s)}{X(s)} = \frac{10(s+2)}{(s+3)(s+12)}$$
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The signal to noise ratio of an optical communication system is 45 dB. A pin- photodiode receiver with a quantum efficiency of 60% and operating wavelength of 900 nm is used. The operating bandwidth is 20 MHz, the device dark current is 20 nA, the load resistance is 86 ohm, the amplifier noise figure, Fn = 1 and the operating temperature is 300 K. 2.2.3 Calculate the rms shot noise current. (4) 2.2.4 Calculate the rms thermal noise current. (4)
2.2.3 The rms shot noise current is 0.6928 nA.
2.2.4 The rms thermal noise current is 487.4697 nA.
2.2.3 Calculate the rms shot noise current.
The rms shot noise current is given by the following equation:
i_n = 2qI_dsqrt(BW)
i_n is the rms shot noise current
q is the charge of an electron
I_d is the dark current
BW is the bandwidth
i_n =2 * 1.6 * 10^-19 C * 20 nA * sqrt(20 MHz) = 0.6928203230275509 nA
Therefore, the rms shot noise current is 0.6928203230275509 nA.
2.2.4 Calculate the rms thermal noise current.
The rms thermal noise current is given by the following equation:
i_n = sqrt(4kTBR)
i_n is the rms thermal noise current
k is Boltzmann's constant
T is the temperature
B is the bandwidth
R is the load resistance
i_n = sqrt(4 * 1.38 * 10^-23 J / K * 300 K * 20 MHz * 86 ohm)
i_n = 487.469775059937 nA
Therefore, the rms thermal noise current is 487.469775059937 nA.
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(s+2)² Chapter 14, Problem 18. Draw the Bode plots for G(s)=- s(s+5)(s+10) s = jo
The Bode plots for G(s) = -s(s+5)(s+10) at s = jω consist of a magnitude plot with zero crossings at ω = 5 and ω = 10, and a phase plot with phase shifts of -90° and -180° at ω = 5 and ω = 10, respectively.
The Bode plots for G(s) = -s(s+5)(s+10) at s = jω consist of a magnitude plot and a phase plot.
For the magnitude plot,
At low frequencies (ω → 0), the magnitude is 0 dB (no change).
At ω = 5, there is a zero crossing with a slope of -20 dB/decade.
At ω = 10, there is another zero crossing with a slope of -40 dB/decade.
At high frequencies (ω → ∞), the magnitude approaches 0 dB (no change).
For the phase plot,
At low frequencies (ω → 0), the phase is 0° (no change).
At ω = 5, there is a phase shift of -90°.
At ω = 10, there is an additional phase shift of -180°.
At high frequencies (ω → ∞), the phase approaches -360° (or 0°) due to the double pole.
Since the problem statement mentions s = jo (purely imaginary), the Bode plots are only valid for positive frequencies.
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Which of the following statements correctly describes an object's displacement and distance travelled? (1 Mark) a. The magnitude of displacement is equal to the distance travelled. b. The magnitude of displacement is less than or equal to the distance travelled. c. The magnitude of displacement is greater than or equal to the distance travelled. d. The magnitude of displacement can be less than, equal to, or greater than the distance travelled.
The statement that correctly describes an object's displacement and distance travelled is option d. The magnitude of displacement can be less than, equal to, or greater than the distance travelled.
Displacement and distance are two different quantities used to describe the motion of an object.
Distance refers to the total length of the path covered by an object, regardless of the direction. It is always a positive scalar quantity.
Displacement, on the other hand, refers to the change in position of an object from its initial position to its final position. Displacement takes into account both the distance and direction of the object's motion and is represented as a vector quantity.
In some cases, an object may return to its starting point, resulting in zero displacement but non-zero distance traveled. In other cases, an object may travel a straight path from its initial position to its final position, resulting in the displacement magnitude being equal to the distance traveled. Additionally, displacement can also be greater than the distance traveled if the object takes a non-linear path.
Therefore, the magnitude of displacement can be less than, equal to, or greater than the distance traveled, depending on the specific characteristics of the object's motion (option d).
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A 60Co source is labeled 6.00mCi, but its present activity is found to be 1.93×10 7 Bq. What is the present activity in mCi? (You do not need to enter any units.) 0.522mCi Previous Tries How Ionq aqo did it actually have a 6.00-mCi activity? Submission not graded. Use more significant figures. Tries 4/10 Previous Tries
To calculate the present activity in mCi, we have to use the given formula below:
Activity = λN Where, λ = decay constant
N is the number of radioactive nuclei.
Activity loaded is given in mCi, which is equivalent to 2.22105 disintegrations per second. Thus,
Activity loaded, AL = 2.22×10^5 d/sec
Let the present number of nuclei, N0Thus, the present activity, A0 = λN0
The present activity is given in Bq, which is equivalent to 1 disintegration per second. Thus,Present activity, A0 = 1 disintegration per second
Thus, we can use the following equation, to determine the decay constant, λActivity = λNAL
= λN0
Therefore, λ = AL/N0 Substitute the values in the above equation,AL/N0 = 1.93×10^7 Bq
Substitute the values in the above equation,A0 = λN0
Therefore, A0 = (1.44×10^-3 ) x (0.0115 N0)
= 1.65×10^-5 N0
Activity is generally measured in mCi, so we need to convert it to mCi.Now,1mCi
= 37MBq1Bq
= 2.7×10^-11 CimCi
= 2.7×10^7 disintegration per second
Substitute the values in the above equation, A0 in mCi = 1.65×10^-5 N0 / 2.7×10^7 mCi/Bq
Therefore, A0 in mCi = 0.522 mCiSo, the present activity is 0.522 mCi. Therefore, the answer is 0.522 mCi.
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21. [-/5 Points] The 1 kg standard body is accelerated by only F₁ = (5.0 N) ↑ + (7.0 N) ĵ and F₂ = (−8.0 N)î + (−6.0 N) ĵ. (a) What is the net force in unit-vector notation? F net = DETAILS HRW10 5.P.097. N (b) What is the magnitude and direction of the net force? magnitude direction N ° counterclockwise from the +x-axis (c) What is the magnitude and direction of the acceleration? magnitude m/s² direction ° counterclockwise from the +x-axis MY NOTES ASK YOUR TEACHER
(a) Net force in unit-vector notation The 1 kg standard body is accelerated by F₁ and F₂. Net force is the vector sum of these two forces: [tex]Fnet=F₁+F₂= (5.0 N) ↑ + (7.0 N) ĵ + (−8.0 N)î + (−6.0 N) ĵ = (−3î + N ĵ)N(b)[/tex]
Magnitude and direction of the net force Net force is given as Fnet = −3î + N ĵMagnitude of the net force, Fnet= [tex]√Fnet,x² + Fnet,y²= √(−3 N)² + (1 N)²= √9 + 1= √10 NT[/tex]he direction of the net force in unit-vector notation = tan−1(Fnet,y / Fnet,x)
The direction of the net force in degrees,[tex]θ, = tan−1 (Fnet,y / Fnet,x) = tan−1(1/−3)= −18°[/tex]
Therefore, the magnitude and direction of the net force are √10 N and 18° counterclockwise from the +x-axis, respectively.
(c) Magnitude and direction of the acceleration The acceleration of the 1 kg standard body is given by the Newton's Second Law of motion as:
Fnet = ma,where m is the mass of the body and a is its acceleration.a = Fnet/mThe mass of the body is m = 1 kg, while the net force on it is Fnet = −3î + N ĵ.
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A glass windowpane in a home is 0.620 cm thick and has dimensions of 0.99 m ✕ 1.65 m. On a certain day, the temperature of the interior surface of the glass is 30.0°C and the outdoor temperature is 0°C. Assume the thermal conductivity of the glass is 0.8 W/m · °C.
(a) What is the rate at which energy is transferred by heat through the glass?
W
(b) How much energy is transferred through the window in one day, assuming the temperatures on the surfaces remain constant?
J
(a) The rate at which energy is transferred by heat through the glass is 20.5 watts.
(b) The amount of energy transferred through the window in one day is approximately 1,765,200 joules.
(a) The rate at which energy is transferred by heat through the glass can be determined using the formula for heat transfer:
Rate of heat transfer = (Thermal conductivity) x (Area) x (Temperature difference) / (Thickness)
Thermal conductivity of glass = 0.8 W/m · °C
Area of glass windowpane = 0.99 m x 1.65 m
Temperature difference = (30.0°C - 0°C) = 30.0°C
Thickness of glass windowpane = 0.620 cm = 0.00620 m
Using the given values in the formula, we can calculate the rate at which energy is transferred by heat through the glass:
Rate of heat transfer = (0.8 W/m · °C) x (0.99 m x 1.65 m) x (30.0°C) / (0.00620 m)
Simplifying the equation, we get:
Rate of heat transfer = 20.5 W
Therefore, the rate at which energy is transferred by heat through the glass is 20.5 watts.
(b) To determine the amount of energy transferred through the window in one day, we need to calculate the total energy transferred per unit time and then multiply it by the number of seconds in one day.
The total energy transferred per unit time can be calculated using the formula:
Energy transferred per unit time = Rate of heat transfer x Time
Rate of heat transfer = 20.5 W (from part a)
Time = 1 day = 24 hours = 24 x 60 x 60 seconds
Using the given values in the formula, we can calculate the energy transferred through the window in one day:
Energy transferred per unit time = (20.5 W) x (24 x 60 x 60 seconds)
Simplifying the equation, we get:
Energy transferred per unit time = 1,765,200 J
Therefore, the amount of energy transferred through the window in one day, assuming the temperatures on the surfaces remain constant, is approximately 1,765,200 joules.
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8. [0/1 Points] DETAILS PREVIOUS ANSWERS OSCOLPHYS2016 25.3.WA.013. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER You have enrolled in a scuba diving class and while swimming under water in a nearby lake you look up and note that the Sun appears to be at an angle of 37° from the vertical. At what angle above the horizon does the diving instructor standing on shore see the Sun? Enter a number. vn a figure that represents this situation and shows all of the angles? Can you write Snell's law of refraction for this situation? What is the relationship between the angle at which the sunlight enters the water with respect to the normal and the angle of elevation of the Sun above the horizon as seen by the instructor?° Additional Materials Reading Submit Answer
Given angle of elevation from underwater is 37°. Let's suppose the angle of the Sun from the horizontal is x. So, in right-angled triangle ABD, tan x = AB/BD, If h is the height of the diving instructor, then CD=h, AB = BD x tan x
From Snell's law of refraction, we know that, n₁sin θ₁ = n₂sin θ₂... (i)
As sunlight enters the water, it is refracted. Let us assume that the angle of incidence is i, and the angle of refraction is r, with respect to the normal. For the case in question, the normal is CD and sin r = sin (180 - 37 - i) = sin (143 - i)°
The angle of incidence i and the angle of refraction r are related by Snell's law, i.e. n₁sin i = n₂sin r.... (ii)
From (i) and (ii), n₁sin θ₁ = n₂sin (143 - i)°
The angle of elevation of the Sun is 37° above the horizontal, so it makes an angle of (90 - 37)° = 53° with the vertical. Hence the angle of the Sun from the horizontal is 90 + 53° = 143°. Using the equation, n₁sin θ₁ = n₂sin (143 - i),
n₁sin 53° = n₂sin (143 - i)....(iii)
Again, in right-angled triangle ACD, tan (90 - 37 - i) = h/ACF
rom this equation, we get, AC = h/cos (53 + i)°
Using this in triangle ABC, we get, AB = (h/cos (53 + i)°) tan (143 - i)....(iv)
From (iii) and (iv), we get, n₁sin 53° = n₂(h/cos (53 + i)°) tan (143 - i)
Therefore, the angle above the horizon that the instructor sees the Sun is 90 - i. Putting this in (iii), we get,sin 53° = (n₂/n₁) cos (53 + i)° tan (143 - i)
Therefore, the relationship between the angle at which sunlight enters the water and the angle of elevation of the Sun is given by the above equation. What is the relationship between the angle at which the sunlight enters the water with respect to the normal and the angle of elevation of the Sun above the horizon as seen by the instructor?The relationship between the angle at which sunlight enters the water with respect to the normal and the angle of elevation of the Sun above the horizon as seen by the instructor is given by the following equation:
sin 53° = (n₂/n₁) cos (53 + i)° tan (143 - i)
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4. Given: \( \sigma=35 . \) \( \tau=35.7 \mathrm{lb} \mathrm{ft} \) \( r=0.0240 \mathrm{ft} \) \( F= \) ?
The torque required to twist the shaft is \(2420.57\; lb\; ft\).
The torque \(F\) required to twist the shaft can be calculated by the following formula,
\(F=\dfrac{Tr}{J}\) where, \(T\) is the torque applied to the shaft,\(r\) is the radius of the shaft, \(J\) is the polar moment of inertia.
The polar moment of inertia can be calculated as,
\(J=\dfrac{\pi d^{4}}{32}\) where, \(d\) is the diameter of the shaft.
The polar moment of inertia of the shaft is given by \(J=\dfrac{\pi d^{4}}{32}\)
We know that the radius of the shaft is given by \(r=0.0240\; ft\).
The diameter of the shaft is given by \(d=2r=2\times0.0240=0.0480\; ft\).
Therefore, \(d=0.0480\;ft\).
Substitute the values of \(T\) and \(r\) in the formula \(F=\dfrac{Tr}{J}\),\(\begin{aligned} F&=\dfrac{Tr}{J}\\ &=\dfrac{(35.7)\cdot(0.0240)}{\dfrac{\pi\cdot (0.0480)^{4}}{32}}\\ &=\dfrac{(35.7)\cdot(0.0240)\cdot(32)}{\pi\cdot (0.0480)^{4}}\\ &=2420.57\; lb \; ft \end{aligned}\)
Therefore, the torque required to twist the shaft is \(2420.57\; lb\; ft\).
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Find the work done in lifting the bucket
A 7 lb bucket attached to a rope is lifted from the ground into the air by pulling in 24 ft of rope at a constant speed. If the rope weighs 0.8, how much work is done lifting the bucket and rope?
Assuming the force required to lift the rope is equal to its weight, find the force function, F(x), that acts on the rope when the bucket is at a height of x ft.
F(x)=
The total work done in lifting the bucket and rope is 187.2 foot-pounds (ft-lb).
To find the work done in lifting the bucket and rope, we need to consider two parts:
Part 1: Work done lifting the bucket (without the rope) 24 ft:
The work done in lifting the bucket can be calculated by multiplying the weight of the bucket by the distance it is lifted.
Given:
Weight of the bucket = 7 lb
Distance lifted = 24 ft
Work done lifting the bucket = Weight of the bucket x Distance lifted
Work done lifting the bucket = 7 lb x 24 ft
Please note that the units need to be consistent for the calculation. In this case, we have pounds (lb) and feet (ft).
Part 2: Work done lifting the rope:
Assuming the force required to lift the rope is equal to its weight, we can calculate the work done lifting the rope by multiplying the weight of the rope by the distance it is lifted.
Given:
Weight of the rope = 0.8 lb
Distance lifted = 24 ft
Work done lifting the rope = Weight of the rope x Distance lifted
Work done lifting the rope = 0.8 lb x 24 ft
Now, we can calculate the total work done in lifting the bucket and rope by summing up the work done in both parts:
Total work done = Work done lifting the bucket + Work done lifting the rope
Please note that the units of work are in foot-pounds (ft-lb).
Now, we can calculate the values:
Work done lifting the bucket = 7 lb x 24 ft = 168 ft-lb
Work done lifting the rope = 0.8 lb x 24 ft = 19.2 ft-lb
Total work done = 168 ft-lb + 19.2 ft-lb = 187.2 ft-lb
Therefore, the total work done in lifting the bucket and rope is 187.2 foot-pounds (ft-lb).
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The complete question is:
Find the work done In lifting the bucket A 7 Ib bucket attached to a rope is lifted from the ground Into the air by puling in 24 ft of rope at a constant speed. If the rope weighs 0.8, how much work done lifting the bucket and rope? Part1 -1 Find the work done lifting the bucket (without the rope) 24 ft . ft-Ib Part-2. Assuming the force required to lift the rope is equal to its weight; find the force function, F(x), that acts on the rope when the bucket is at height of x Ft. Part- 3 Setup the Integral that will give the work required to lift the rope 24 ft. Part -4 The total amount of work done lifting the bucket and ft-Ib.
Example 11.12 Find the equivalent parallel resistance and capacitance hat causes a Wien bridge to null with the following component values. R₁ = 3.1 ks2 C₁= 5.2 µF R, = 25 kΩ f=2.5 kHz R₁ - 100 ks2 Gi 2500 15.71 k rad/s
The equivalent parallel resistance and capacitance that cause a Wien bridge to null are 77.91 Ω and 5.2 x 10⁻⁶ F.
R₁ = 3.1 kΩ,
C₁ = 5.2 µF,
R₂ = 25 kΩ,
f = 2.5 kHz, and
R₃ = 100 kΩ.
The bridge is balanced so that,Using a parallel resistance equation and a parallel capacitance equation, we can find the equivalent parallel resistance and capacitance that cause a Wien bridge to null.
The formula for parallel resistance is;
Req = R₁R₂/R₁ + R₂
and the formula for parallel capacitance is;
Ceq = C₁C₂/C₁ + C₂
where C₂ is the equivalent capacitance that causes the Wien bridge to null.
Using the formula for Req,
R₁R₂/R₁ + R₂ = 3.1 x 10³ x 25 x 10³/3.1 x 10³ + 25 x 10³
= 77.91 Ω
Using the formula for Ceq,
C₁C₂/C₁ + C₂ = 5.2 x 10⁻⁶ x C₂/5.2 x 10⁻⁶ + C₂
At null,
C₁/C₂ = 1 and so,
5.2 x 10⁻⁶/C₂
= 1C₂
= 5.2 x 10⁻⁶ F
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You have configured a solar powered electric fence designed to operate 24 hours a day. Your solar panel is rated at 12 nominal volts. When you test the fence, you find it is generating a 2,000 volt electric shock. Which of the following did you need to configure your system? Pick one answer and explain why.
A) Photovoltaic Panel, Inverter, 12 Vdc Battery Bank, Alternating Current Disconnect, Direct Current Voltage Converter
B) Photo Voltaic Panel, Charge Controller, 12 Vdc Battery Bank, Alternating Current Disconnect
C) Photo Voltaic Panel, Charge Controller, 6 Vdc Battery Bank, Direct Current Disconnect, Combiner Box, Inverter
D) Photo Voltaic Panel, Direct Current Disconnect, Charge Controller, 12 Vdc Battery Bank, Direct Current Voltage Converter
The system that you need to configure to have the solar powered electric fence designed to operate 24 hours a day, which generates a 2,000 volt electric shock is B) Photo Voltaic Panel, Charge Controller, 12 Vdc Battery Bank, Alternating Current Disconnect.
A solar-powered electric fence uses a photovoltaic panel to collect energy from the sun and convert it into electrical energy. The voltage of the photovoltaic panel plays a significant role in determining the voltage that the electric fence will generate. Therefore, the photovoltaic panel is the first component you need to configure your system. The charge controller ensures that the 12 Vdc battery bank doesn't overcharge or discharge too much.
The 12 Vdc battery bank provides a stable source of DC power to the fence. The Alternating Current Disconnect is responsible for shutting off the AC power to the fence in case of emergencies. The correct answer is B because it includes the necessary components to configure a solar-powered electric fence designed to operate 24 hours a day.
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short, \( Q \) is the moment of the area about the neutral axis. - Part A - Moment of inertia - Part B - \( Q \) for the given point - Part C - Shear stress
The moment of the area is represented by Q. This represents the moment of inertia about the neutral axis of the element. The moment of inertia is also known as the second moment of area, which is used to define an object's resistance to bending.
The higher the moment of inertia, the more resistant the object is to bending. The shear stress applied on an element is directly proportional to the product of the shear force and the first moment of area.Q for a given point is the moment of area of the element about a given point. This is usually calculated about the centroid of the section.
It is expressed as I/A, where I is the moment of area about the neutral axis and A is the cross-sectional area of the element. Thus,Q = I / Awhere I is the moment of inertia about the neutral axis, and A is the cross-sectional area.The shear stress in an element is determined by dividing the shear force by the area that is perpendicular to the force.
The stress due to the shear force is linearly proportional to the distance from the neutral axis. The maximum shear stress occurs at the neutral axis, where the distance from the neutral axis is zero.
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a) Draw the typical 3-phase induction motor torque-slip characteristics with appropriate labels. (10 marks) b) Draw the two types of rotors for a synchronous machine with appropriate labels.
Three-phase induction motor torque-slip characteristic is a graph of the variation in the torque and slip of an induction motor. The y-axis represents torque and the x-axis represents slip. Slip is defined as the difference between the synchronous speed and the actual rotor speed.
At the beginning of the graph, the torque is zero as the motor is at standstill and the slip is 1. When the motor starts, the rotor speed increases, and the slip decreases. The graph then shows a sharp increase in torque and a decrease in slip as the motor reaches its maximum torque, known as the pullout torque. After the pullout torque, the torque decreases slightly as the slip increases, reaching a point where the motor stalls. This point is known as the breakdown torque. At the breakdown torque, the slip is 1, and the motor stops rotating.
b) The two types of rotors for a synchronous machine are salient-pole rotor and non-salient pole rotor. Salient-pole rotor, also known as a wound rotor, has a large number of poles compared to a non-salient pole rotor. The rotor is a solid steel cylinder with slots to hold the rotor winding. The rotor winding is made up of copper bars, which are placed in the slots and connected by rings at each end. The bars are short-circuited at the ends by end rings to complete the winding.
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Design and simulate a regulated power supply using a bridge rectifier, capacitors, and Zener diode (no Integrated Circuit). The source voltage is 110±10 Vrms, 60 Hz frequency. The output voltage is as follows (+5% ): Type 1:3 V and
Design and simulation of regulated power supply using bridge rectifier, capacitors, and Zener diodeDesign of power supply using Zener diode:Let us begin the design process by setting the parameter values.Source voltage = 110 VFrequency = 60 Hz
The output voltage for Type 1 is 3 VOutput voltage range (+5%) = 0.15 VMinimum output voltage = 2.85 VMaximum output voltage = 3.15 VBridge rectifier:The bridge rectifier is a crucial component of the power supply. It is responsible for converting the incoming AC voltage to DC voltage. We will use a four-diode bridge rectifier for the power supply.Capacitors:The capacitors are connected to the bridge rectifier output and the Zener diode.
The simulation results are shown below:LTSpice simulation resultsThe simulation results show that the output voltage is regulated at 3 V, which is within the desired range. The output voltage is also stable and does not fluctuate despite fluctuations in the input voltage.
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i) Show that the de Broglie wavelength of a particle, of charge e, rest mass mo, moving at relativistic speeds is given as a function of the accelerating potential Vas 2 h 2m,eV (1 + eV 2m,c2 ii) Show how this agrees with 1 = h/p in the nonrelativistic limit.
The de Broglie wavelength of a particle of charge e and rest mass mo moving at relativistic speeds can be given as a function of the accelerating potential as shown below: λ = h / √(2m eV) (1 + eV/2m c²).
The de Broglie wavelength of a particle of charge e and rest mass mo moving at relativistic speeds can be given as a function of the accelerating potential as shown below: λ = h / √(2m eV) (1 + eV/2m c²)
where: λ = de Broglie wavelength of the particle
h = Planck’s constant
e = charge of the particle
V = accelerating potential
m = rest mass of the particle
c = speed of light
This equation was proposed by Schrödinger to give an exact quantum mechanical treatment of electrons inside atoms. In the nonrelativistic limit, the particle speed is much smaller than the speed of light, so we can neglect the term (eV/2mc²) compared to 1. Hence, the equation reduces to: λ = h / p
where: p = momentum of the particle
In conclusion, the above equation is valid only for particles moving at relativistic speeds. In the nonrelativistic limit, the classical equation (λ = h/p) can be used to calculate the de Broglie wavelength of the particle.
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A photon with a wavelength of 5040 nanometers has a frequency of 5.95 e 13 cycles per second. What will be the wavelength (in nanometers) of a photon with a frequency of 3.57 e 14? Select one: A. 5040 nanometers B. 2520 nanometers C. 1260 nanometers D. 10080 nanometers
A photon with a wavelength of 5040 nanometers has a frequency of 5.95 e 13 cycles per second. The wavelength (in nanometers) of a photon with a frequency of 3.57 e 14 is 840 nanometers. There is no correct option.
To find the wavelength of a photon with a given frequency, we can use the equation:
c = λ * f
where c is the speed of light, λ is the wavelength, and f is the frequency.
Wavelength of the first photon ([tex]\lambda_1[/tex]) = 5040 nanometers
Frequency of the first photon (f1) = 5.95 * [tex]10^{13[/tex] cycles per second
We can rearrange the equation to solve for the wavelength:
[tex]\lambda_1 = c / f_1[/tex]
Now we can substitute the known values:
[tex]\lambda_1 = (3.00 * 10^8 m/s) / (5.95 * 10^{13} s^{(-1)})[/tex]
Converting the wavelength to nanometers:
[tex]\lambda_1 = (3.00 * 10^8 m/s) / (5.95 * 10^{13} s^{(-1))} * (10^9 nm / 1 m)[/tex]
Calculating the value of [tex]\lambda_1[/tex]:
[tex]\lambda_1[/tex]≈ 5040 nanometers
So, the wavelength of the first photon is 5040 nanometers.
Now, to find the wavelength of a photon with a frequency of 3.57 * [tex]10^{14[/tex]cycles per second:
[tex]\lambda_2[/tex] = c /[tex]f_2[/tex]
Substituting the known values:
[tex]\lambda_2[/tex] = [tex](3.00 * 10^8 m/s) / (3.57 * 10^{14} s^{(-1)}) * (10^9 nm / 1 m)[/tex]
Calculating the value of [tex]\lambda_2[/tex]:
[tex]\lambda_2[/tex] ≈ 840 nanometers
Therefore, the wavelength of the photon with a frequency of 3.57 *[tex]10^{14[/tex] cycles per second is approximately 840 nanometers.
The correct answer is not among the options provided.
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