When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the buildings. This causes the radio waves to diffract. In this problem, you will see how different wavelengths diffract as they pass through a city and relate this to reception for radios and cell phones. You will use the angle from the center of the central intensity maximum to the first intensity minimum as a measure of the width of the central maximum (where nearly all of the diffracted energy is found).a. Find the angle θ to the first minimum from thecenter of the central maximum (Express your answer in terms λ and a.):b. What is the angle θFM to the first minimum foran FM radio station with a frequency of 101mMHz? (Express your answer numerically indegrees to three significant figures. Note: Do not write youranswer in terms of trignometric functions. Evaluate any suchfunctions in your working.)c. What is the angle θcell for a cellular phonethat uses radiowaves with a frequency of 900MHz? (Express your answer indegrees to three significant figures.)d. What problem do you encounter in tryingto find the angle θAM for an AM radio stationwith frequency 1000kHz?i. The angle becomes zero.ii. The angle can be given only in radians.iii. To find the angle it would be necessary to takethe arcsine of a negative number.iv. To find the angle it would be necessary totake the arcsine of a number greater than one.

Answers

Answer 1

Final answer:

The diffraction of radio waves when passing through city buildings can be represented by single-slit diffraction. The angle to the first minimum can be calculated using the equation θ = sin-1(λ/a), but for AM radio with very large wavelengths, this calculation may not be valid as it could require taking the arcsine of a number greater than one.

Explanation:

When radio waves encounter thin vertical slits such as the spaces between buildings, they diffraction occurs. The property of diffraction can be analyzed using the concept of single-slit diffraction from wave optics. For a single-slit diffraction, the angle θ to the first minimum can be found using the equation θ = sin-1(λ/a), where λ is the wavelength of the wave and a is the width of the slit.

For an FM radio station with a frequency of 101 MHz, we would use the relationship between frequency (f), wavelength (λ), and the speed of light (c) to find the wavelength (λ = c/f) before calculating the angle using the aforementioned equation.

Similarly, for a cellular phone using radio waves with a frequency of 900 MHz, we again find the wavelength using the same relation and then calculate the angle θ to the first minimum.

However, for AM radio, the complication arises because the wavelengths for AM radio are considerably larger. This can lead to a scenario where the slit width is not narrow enough compared to the wavelength, and as a result, the angle θAM calculated using sin-1(λ/a) may result in taking the arcsine of a number greater than one, which is not possible and indicates that the first minimum may not occur.


Related Questions

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. A person can clearly see objects only if they are located between 34 cm and 180 cm away from his eyes. Bifocal lenses are used to correct his vision. What power lens (in diopters) should be used in the top half of the lens to allow him to clearly see distant objects

Answers

Answer:

The power of top half of the lens is 0.55 Diopters.

Explanation:

Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.

The general formula for power of a lens is:

Power = 1/Focal Length in meters

Therefore, for the top half of the lens:

Power = 1/1.8 m

Power = 0.55 Diopters

Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at constant pressure until its volume has doubled. Then it expands adiabatically until the temperature returns to its initial value. Assume that the helium can be treated as an ideal gas.B)What is the total change in internal energy of the helium?C)What is the total work done by the helium?D)What is the final volume of the helium?

Answers

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

Select the correct statement to describe when a sample of liquid water vaporizes into water vapor

Answers

Answer:

This procces is called evaporation.

Explanation:

When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.

Your heart pumps blood at a pressure of 100 mmHg and flow speed of 60 cm/s. At your brain, the blood enters capillaries with such a large total cross-sectional area that the blood velocity is much smaller: about 0.7 mm/s. Ignore viscosity in the capillaries. How long could your neck be so that blood still reaches your brain in m

Answers

Final answer:

The length of the neck that blood can still reach the brain depends on blood pressure, flow speed, and the cross-sectional area of capillaries. Using the given values, the approximate length of the neck is 9.8 meters.

Explanation:

The length of your neck that blood can still reach your brain depends on the blood pressure, flow speed, and the cross-sectional area of the capillaries. The blood enters the capillaries in the brain with a much smaller velocity of about 0.7 mm/s due to the large total cross-sectional area. To determine the length of the neck, we need to consider the time it takes for blood to travel from the heart to the brain. Using the given velocity, we can calculate that the time it takes for blood to reach the brain is approximately 14 seconds. Assuming a constant velocity, the length of the neck can be calculated using the formula: length = velocity x time, which gives us a result of 9.8 meters.

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When he prepares food, Edgar wants to use ingredients that are not high in trans fat. Based on this information, which of the following fats should he include in his recipes?a. Stick margarine made with olive oilb. Partially-hydrogenated peanut oilc. Corn oild. Shortening

Answers

Answer:

Corn oil

Explanation:

Researchers say that corn oil is safer option than olive oil when it comes to reducing sugar levels of the blood and cholesterol levels and that it is often successful in reducing blood pressure.  Corn oil is a polyunsaturated fat with some monounsaturated properties. Also, Corn oil is not high in trans fat responsible for various diseases.

When opening a hinged door, the handle is usually on the side of the door farthest from the hinges. Rank the difficulty of opening the door if you were to push or pull on the following locations on the door:

Hardest To Open

Easiest Open

1. As close to the hinges as possible

2. the center of the door

3. The edge farthest from the hinges

Answers

The order of difficulty from the hardest to the easiest would be subject to the concept we have of Torque.

The Torque principle defines us that

[tex]\tau = Fd[/tex]

Where,

F = Force

d = Distance

As the distance increases, the force applied must be less to make the movement of the object so we have to

[tex]F \propto \frac{1}{d}[/tex]

Hence we have to be the distance inversely proportional to the force to turn the door the order would be:

As close to the hinges as possible (Hardest)> The center of the door> the edge farthest from the hinges (Easies)

1>2>3

The correct ranking of the difficulty of opening the door from hardest to easiest is as follows: 1. As close to the as possible,  2. The center of the door, 3. The edge farthest.

To understand the ranking, consider the physics of opening a door. The difficulty of opening a door is related to the torque required to rotate it around. Torque [tex](\(\tau\))[/tex] is calculated by the equation [tex]\(\tau = r \times F\)[/tex], where [tex]\(r\)[/tex] is the distance from the axis of rotation to the point where the force [tex](\(F\))[/tex] is applied.

1. As close possible: When you apply force close, the value of [tex]\(r\)[/tex] is smallest. This results in the smallest torque, meaning you have to exert more force to achieve the necessary torque to open the door, making it the hardest location to open the door.

2. The center of the door: Applying force at the center of the door increases the distance [tex]\(r\)[/tex]. This results in a larger torque for the same amount of force compared, but it is still not the most efficient point to open the door.

3. The edge farthest: The farthest edge provides the longest lever arm, meaning [tex]\(r\)[/tex] is at its maximum. This allows for the greatest torque for a given force, making it the easiest location to open the door. This is also why door handles are typically placed farthest.

A container made of steel, which has a coefficient of linear expansion 11 ✕ 10−6 (°C)−1, has a volume of 55.0 gallons. The container is filled to the top with turpentine, which has a coefficient of volume expansion of 9.0 ✕ 10−4 (°C)−1, when the temperature is 10.0°C. If the temperature rises to 25.5°C, how much turpentine (in gal) will spill over the edge of the container?

Answers

Final answer:

The amount of spilled turpentine due to temperature rise can be calculated by comparing the changes in volume of the steel container and the contained turpentine, using the principles and mathematical formulas of thermal expansion.

Explanation:

The question asks us to determine how much turpentine, which exhibits a greater rate of thermal expansion compared to the steel container, will spill over when the temperature rises. We need to apply the principles of thermal expansion to calculate this. Both the steel container and the turpentine expand with the increase in temperature, but since turpentine has a higher coefficient of volume expansion than steel, more turpentine will expand than the container can accommodate, resulting in some turpentine spilling over.

To calculate the amount of spilled turpentine, we need to find the change in volume for both the container and turpentine, and subtract the former from the latter. The change in volume due to thermal expansion can be calculated by using the equation ΔV = βV0*(T2 - T1), where β is the coefficient of volume expansion, V0 is the initial volume, and T2 and T1 are the final and initial temperatures respectively.

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Final answer:

To find how much turpentine will overflow, calculate the change in volume due to thermal expansion for both the steel container and the turpentine, and subtract the change in volume of the steel from that of the turpentine.

Explanation:

To solve this problem, we need to consider the thermal expansion of both the steel container and the turpentine. Thermal expansion is the increase in size of a body due to a change in temperature. This is described mathematically by the formula ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the original volume, and ΔT is the change in temperature.

First, calculate the change in volume for the steel container and the turpentine separately. For the steel, β is 11 x 10^-6 (°C)^-1 and for turpentine, β is 9.0 x 10^-4 (°C)^-1. The original volume, V₀, is 55 gallons for both, and the change in temperature, ΔT, is 25.5°C - 10.0°C = 15.5°C.

Performing these calculations will give you the change in volume for the steel and the turpentine. The difference in these two volumes will tell you how much turpentine will overflow as the temperature increases.

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A 100 kg marble slab falls off a skyscraper and falls 200 m to the ground without hitting anyone. Its fall stops within milliseconds, so that there is no loss of thermal energy to its surroundings if its temperature is measured immediately after it stops. By how much has its temperature changed as a result of the fall, if we ignore energy gained or lost as a result of its interaction with the atmosphere? Cmarble = 860 J/(kg oC) 0.57 °C 1.14 °C 2.28 °C 4.56 °C

Answers

Answer:

Δ T = 2.28°C

Explanation:

given,

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C_marble = 860 J/(kg °C)

using conservation of energy

Potential energy = heat energy

  [tex]m g h = m C_{marble}\Delta T[/tex]

  [tex]g h =C_{marble}\Delta T[/tex]

  [tex]\Delta T= \dfrac{g h}{C_{marble}}[/tex]

  [tex]\Delta T= \dfrac{9.8 \times 200}{860}[/tex]

        Δ T = 2.28°C

The temperature change which occurs as a result of the fall, if we ignore energy gained or lost is 2.28°C

What is Temperature?

This is defined as the degree of hotness or coldness of a substance.

Parameters

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C of marble = 860 J/(kg °C)

Using conservation of energy

mgh = mcΔT

ΔT = gh/c

= 9.8 × 200 / 860

 = 2.28°C

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A circuit consists of a coil that has a self-inductance equal to 4.3 mH and an internal resistance equal to 16 Ω, an ideal 9 V battery, and an open switch--all connected in series. At t = 0 the switch is closed. Find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil.

Answers

Answer:

t = 186.2 μs

Explanation:

Current in LR series circuit

[tex]I(t) = I_{s}( 1 - e^{-Rt/L)}[/tex]----(1)

steady current =  I_{s} = V/R

time constant = τ =[tex]L/R =4.3 * 10^{-3} / 16\\[/tex]

                                              = 0.268 ms

magnetic energy stored in coil = [tex]U_{L} = \frac{1}{2}LI^{2}[/tex]

rate at which magnetic energy stored in coil= [tex]\frac{d}{dt}U_{L} =\frac{d}{dt} \frac{1}{2}LI^{2}   \\                            = LI\frac{dI}{dt}\\[/tex]----(2)

rate at which power is dissipated in R:

                                       [tex]P = I^{2}R[/tex]---(3)

To find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil equate (2) and (3)

[tex]I^{2}R=LI \frac{dI}{dt}[/tex]

[/tex]I=\frac{L}{R}\frac{dI}{dt}[/tex]----(4)

differentiating (1) w.r.to t

[tex]I(t)=I_{f} (1-e^{\frac{Rt}{L} })[/tex]

[tex]\frac{dI}{dt} = I_{f}\frac{d}{dt}(1-e^{\frac{-Rt}{L} }   )[/tex]

[tex]\frac{dI}{dt}= I_{f}(-\frac{R}{L} e^{\frac{-Rt}{L} } )\\[/tex]---(5)

substituting (5) in (4)

[tex]I=I_{f}e^{-\frac{Rt}{L} }[/tex]----(6)

equating (1) and (6)

[tex]I_{f}( 1- e^{-\frac{Rt}{L} } ) = I_{f}e^{-\frac{Rt}{L} }[/tex]

[tex]1 - e^{-\frac{Rt}{L} } =  e^{-\frac{Rt}{L} }[/tex]

[tex]\frac{1}{2}= e^{-\frac{Rt}{L} }[/tex]

[tex]t= -\frac{L}{R}ln\frac{1}{2}[/tex]

L= 4.3 mH

R= 16 Ω

t = 186.2 μs

Two long straight wires enter a room through a window. One carries a current of 3.0 ???? into the room while the other carries a current of 5.0 ???? out. The magnitude in Tm of the path integral ∮ ????⃗ ∙ ???????? around the window frame is:

Answers

Answer:

[tex]\begin{equation}\\\oint_LB.dl\\\end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]

Explanation:

If you need calculate

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex]

You can use the Ampere's Law

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = [tex]I_{in}[/tex]μ

   

      Where [tex]I_{in}[/tex]: Current passing through the window

                   μ : Free space’s magnetic permeability

                   μ = 4πx[tex]10^{-7} T.m.A^{-1}[/tex]

Then

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = (3-5)4πx[tex]10^{-7}[/tex]

[tex]\begin{equation}\\\oint_L B.dl\\ \end{equation}[/tex] = -8πx[tex]10^{-7}[/tex]

The magnitude of in the path integral is 2.5*10^-6 T.M

Data;

I1 = 5A

I2 = 3A

μo = 4π * 10^-7 T.m.A^-1

Ampere Law

Ampere law states that the sum of the length of elements multiplied by the magnetic field in the path of the length elements is equal to the permeability multiplied by the electric current enclosed in the loop.

Mathematically;

∮B.dl = I*μo

I = current passing through the windowμo = free space magnetic permeability

[tex]\int B.\delta s = \mu I\\\int B.\delta s = \mu (5A - 3A)\\\int B.\delta s = 4\pi * 10^-^7 * 2A\\\int B.\delta s = 2.5*10^-^6 T.M[/tex]

The magnitude of in the path integral is 2.5*10^-6 T.M

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You are playing a speed-based card game with your 64-year-old grandfather. The object of the game is to get rid of your cards as fast as you can. Once the first card is turned over, each player tries to play by deciding on which pile to play his or her card. When you were younger, your grandfather always beat you in this game. Now, you always beat him. Your grandfather is likely experiencing a slight decline in_____________.

Answers

Now, you always beat him. Your grandfather is likely experiencing a slight decline in perceptual speed.

Explanation:

The speed of perception refers to the capacity to accurately (and completely) compare words letter, digits, objects, images, etc. When testing, these objects can be displayed simultaneously or one after the other. This type of test can be included in the proficiency test.

For example, we have also seen all the puzzles that ask the reader to notice the differences between the two pictures. The time it takes to recognize these differences is a measure of the speed of perception. Likewise, in getting rid of cards at the given situation, grandfather experiences a less decline in his perceptual speed.

Your grandfather is likely experiencing a slight decline in; Perceptual speed.

The grandfather is playing a speed based card game.

Now we are told that the object of the game is to get rid of the cards as fast as possible.

We are told that when you were younger your grandfather used to beat you always in the game. This means that his speed in comparing the cards to know which one to get rid of was fast before but has declined now since he can't beat you again.

Finally, we can say that his perceptual speed has declined because perceptual speed is defined as the ability to compare letters, numbers, objects, patterns e.t.c

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In the fastest measured tennis serve, the ball left the racquet at 73.14 m/s. A serve tennis ball is typically in contact with the racquet for 30.0 ms and starts from rest. Assume constant acceleration.(a) what was the ball's acceleration during this serve??(b) how far did the ball travel during the serve???

Answers

Answer:

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

Explanation:

a) since

v = vo + a*t ,

where v= velocity at time t , vo= velocity at time t=0 and a= acceleration

,then

a= (v-vo)/t

replacing values

a= (v-vo)/t = (73.14 m/s - 0 m/s)/( 30* 10⁻³ s) = 2438 m/s²

b) the distance travelled d is

v² = vo² + 2*a*d  

then

d = (v² - vo²) /(2*a) = (73.14 m/s)² - 0²)/(2*2438 m/s²)= 1.0971 m

a)  the acceleration is a= 2438 m/s²

b) the distance travelled during serve is d = 1.0971 m

What is acceleration?

Acceleration represents the rate at which velocity should be changed with time, with respect to both speed and direction. Since acceleration contains both a magnitude and a direction, it is a vector quantity.

Calculation of acceleration & distance:

a) since

[tex]v = vo + a\times t[/tex]

Here

v= velocity at time t ,

vo= velocity at time t=0

and a= acceleration

Now

[tex]a= (v-vo)\div t\\\\ =(73.14 m/s - 0 m/s)/( 30\times 10^{-3} s)[/tex]

= 2438 m/s²

b) Now the distance traveled d is

[tex]v^2 = vo^2 + 2\times a\times d \\\\d = (v^2 - vo^2) \div (2\timesa) \\\\=(73.14 m/s)^2 - 0^2)\div (2\times 2438 m/s^2)[/tex]

= 1.0971 m

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

[tex]q_{1} +q_{2} =-q_{3}[/tex] -----eqution 1

where,

[tex]q_{1}[/tex] is the heat absorbed by the solid at 0⁰C

[tex]q_{2}[/tex] is the heat absorbed by the liquid at 0⁰C

[tex]q_{3}[/tex] the heat lost by the warmer water sample

Important equations to be used in solving this problem

[tex]q=m *c*\delta {T}[/tex], where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

[tex]\delta {T}[/tex] is change in temperature

Again,

[tex]q=n*\delta {_f_u_s}[/tex] -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

[tex]=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O[/tex]

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

[tex]q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ[/tex]

This means that equation (1) becomes

79.13 KJ + [tex]q_{2} = -q_{3}[/tex]

Step 4: calculate the final temperature of the water

[tex]79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}[/tex]

Substitute in the values; we will have,

[tex]79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})[/tex]

79.13 kJ + 990.66J* [tex](T_{f}-218})[/tex] = -1463J*[tex](T_{f}-100})[/tex]

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* [tex](T_{f}-218})[/tex] = -1.463KJ*[tex](T_{f}-100})[/tex]

[tex]79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3[/tex]

collect like terms,

2.45366[tex]T_{f}[/tex] = 283.133

∴[tex]T_{f} =[/tex] = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the magnitude of the tension in the string is F. The boy then speeds up the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions every second. What happens to the tension in the sting?

(A) The magnitude of the tension increases to four times its original value, 4F.
(B) The magnitude of the tension reduces to half of its original value, F/2.
(C) The magnitude of the tension is unchanged.
(D) The magnitude of the tension reduces to one-fourth of its original value, F/4.
(E) The magnitude of the tension increases to twice its original value, 2F.

Answers

Final answer:

When a stone is whirled at double the speed, the tension in the string increases to four times its original value, assuming the radius of the whirl remains the same.

Explanation:

The tension in a string whirling a stone in a circle at a constant speed is directly proportional to the square of the speed. If the boy doubles the speed in the scenario you gave, keeping the radius of the circle unchanged, the tension in the string would increase as the square of that factor. So, between the options given, if the boy increases the speed of the stone so that it makes two complete revolutions every second instead of one, the magnitude of the tension in the string increases to four times its original value. Thus, the correct answer is (A) the magnitude of the tension increases to four times its original value, 4F.

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Final answer:

The tension in the string of a whirling stone increases by a factor of four when the speed of rotation doubles and the radius remains the same. It is because the tension is directly proportional to the square of the speed of the stone.

Explanation:

The tension in the string of a whirling stone is related to the centripetal force, which is directly proportional to the square of the speed of rotation and the mass of the stone, and inversely proportional to the radius of the circle. If the speed of rotation doubles (from one revolution per second to two revolutions per second) and the radius of the circle remains the same, the resulting tension in the string (centripetal force) increases by a factor of four.

Hence, the answer is (A) The magnitude of the tension increases to four times its original value, 4F.

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10 kg of liquid water is in a container maintained at atmospheric pressure, 101325 Pa. The water is initially at 373.15 K, the boiling point at that pressure.The latent heat of water -> water vapor is 2230 J/g. The molecular weight of water is 18 g.103 J of heat is added to the water.1)How much of the water turns to vapor?mass(vapor)=

Answers

Answer:

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

Explanation:

Given:

mass of water, [tex]m_w=10\ kg[/tex]pressure conditions, [tex]P=101325\ Pa[/tex]temperature conditions, [tex]T=373.15\ K[/tex]latent heat of vapourization of water, [tex]L=2230\ J.g^{-1}[/tex]amount of heat supplied to the water, [tex]103\ J[/tex]

Now using the equation of heat considering latent heat only:

(since water already at boiling point at atmospheric temperature)

[tex]Q=m.L[/tex]

[tex]103=m\times 2230[/tex]

[tex]m=0.0462\ g[/tex] of water is converted into vapour.

Final answer:

In the situation provided, about 46.2 grams of the water will have transitioned from a liquid to a vapor after being supplied with 103 kJ of energy, given the specified latent heat of vaporization.

Explanation:

Given that the latent heat of water's vaporization is 2230 J/g and 103 J of energy was provided to the water, we first convert all our units to be consistent. Remember that the latent heat of vaporization is the amount of heat energy required to change one gram of a substetance from a liquid to a gas at constant mperature and pressure. In this case, we're transitioning water to water vapor.

The input energy is 103 kJ, and the latent heat of vaporization is 2.23 kJ/g, so we can calculate the mass of the water that was vaporized using the equation: mass (g) = energy input (kJ) / latent heat of vaporization (kJ/g). By plugging in the values we get: mass = 103 / 2.23 = 46.2 grams.

So, approximately 46.2 grams of the water will have transitioned from a liquid to a vapor given the provided energy input of 103 kJ.

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The circuit below shows some of the circuity in a small toy robot. When the circuit is on, the robot moves its arms (the motor) and blinks a light on its head (the light bulb).

A. How much current is in the circuit when these things are happening? Hint: To find the total resistance of the circuit, add the two resistances together.

Answers

Answer:

dear can you provide the circuit of the robot

2 eagles hit each other at a 90 degree angle they grab each other, what is their velocity and direction?

Answers

Answer:

This is an example of completely inelastic collision, since they grab each other and move as a single object after the collision.

The conservation of momentum requires

[tex]\vec{P_1} + \vec{P_2} = \vec{P}_{final}[/tex]

where

[tex]\vec{P_1} = m_1v_1 \^x\\\vec{P_2} = m_2v_2 \^y[/tex]

Their final momentum is

[tex]\vec{P}_{final} = m_1v_1 \^x + m_2 v_2 \^y[/tex]

Their final velocity and direction is

[tex]\vec{v}_{final} = \frac{m_1v_1}{m_1 + m_2}\^x + \frac{m_2v_2}{m_1 + m_2}\^y[/tex]

Although not states in the question, let's consider the case that the masses and speeds of the eagles are the same:

[tex]\vec{v}_{final} =  \frac{v}{2}\^x + \frac{v}{2}\^y[/tex]

You are driving west in your car on a clear, warm day. Up ahead, you see a dark band of ominous clouds. Soon, you drive beneath the towering bank of clouds, and things get very dark. For twenty minutes, you are pelted with rain and even some hail. You keep driving west, even though it seems a little sketchy at times. On the radio, you hear that the next county to the north is under a tornado warning. Then the sky lightens, and there is only a little rain. You notice that the temperature outside has dropped by 15°F (8°C). What have you just driven across?

Answers

Answer:

You have just driven across the Cold front.

Explanation:

Weather Front:

In Meteorology, weather fronts are simply boundaries between two air masses of different densities. There are four weather fronts and one of them is cold front and other fronts are warm front, stationary front and occluded front.

Cold Front:

In Meteorology, cold front is a boundary of cold air mass that is advancing under the warm air mass.

In our scenario, as the driver was moving in the west and he listened on radio, the next country on north is under tornado warning. So, when he passed the cold front, he experienced the conditions described.

Because sunspots are a little cooler than the average temperature of the photosphere, they prevent some energy from being released from the part of the surface they occupy, but this energy is usually released from hotter and brighter than average areas nearby. Please answer the question: What could happen to the Sun if this energy release did not happen

Answers

Answer: Corona Mass Ejections(CME) and Solar flares

Explanation: Corona Mass Ejections and Solar flares are eruptions that occur in the sun due to the instability in the magnetic field of the sun. This Corona Mass Ejections and Solar flares are prevented by sun spots.Corona mas Ejections are Large and massive eruptions,solar flares are somewhat small eruptions,both of them are prevented from occuring by Sunspots which help to dissipate cooler temperatures in the sun.

Students connect a spring scale to a block on a rough horizontal surface. The students use the spring scale to measure the magnitude of the horizontal force needed to pull the block at a constant speed. Which of the following statements explains why two forces exerted between objects ar equal in magnitude?
a. The gravitational and normal forces exerted on the block, because they are a Newton's third-law pair.
b. The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair.
c. The normal force and the frictional force exerted on the block, because objects always exert forces of equal magnitude on each other.
d. The frictional forces that the block and the surface exert on each other, because objects always exert forces of equal magnitude on each other.

Answers

For there to be a reaction there must also be action. In the horizontal movement there is a balance in which the magnitude of the Forces in opposite directions must be in total 0. The only horizontal forces are friction and the force exerted by the spring scale. For this reason the correct answer is B:

The frictional force and force exerted by the spring scale on the block, because they are a Newton's third-law pair.

A swimming pool heater has to be able to raise the temperature of the 40 000 gallons of water in the pool by 10.0 C°.

How many kilowatt-hours of energy are required?

(One gallon of water has a mass of approximately 3.8 kg and the specific heat of water is 4 186 J/kg⋅°C.)
a. 1 960 kWh
b. 1 770 kWh
c. 330 kWh
d. 216 kWh

Answers

Answer:

b. 1 770 kWh

Explanation:

The heat needed to change the temperature of a certain amount of a substance is given by:

[tex]Q=mC\Delta T[/tex]

Here m is the mass of the susbtance, C is the specific heat of the substance and [tex]\Delta T[/tex] is the temperature change

[tex]Q=(40000*3.8kg)(4186\frac{J}{kg\cdot ^\circ C})(10^\circ C)\\Q=6.36*10^9J[/tex]

Recall that one watt hour is equivalent to 1 watt (1 W) of power sustained for 1 hour. One watt is equal to 1 J/s. So, one watt hour is equal to 3600 J and one kilowatt hour is equal to [tex]3600*10^3 J[/tex]

[tex]Q=6.36*10^9J*\frac{1kW\cdot h}{3600*10^3J}\\Q=1766.66kW\cdot h[/tex]

Final answer:

To heat 40,000 gallons of water by 10.0 C° in a swimming pool, 1,767 kilowatt-hours of energy are required, rounding to the nearest so, option gives (b) 1,770 kWh as the answer.

Explanation:

The question asks: How many kilowatt-hours of energy are required to raise the temperature of 40,000 gallons of water in a pool by 10.0 C°? To solve this, we need to calculate the energy needed using the formula for heat energy: Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

Firstly, convert the volume of water from gallons to kilograms. 40,000 gallons is approximately 40,000 x 3.8 kg = 152,000 kg. Next, use the specific heat of water (4,186 J/kg°C) and the temperature change (10.0 C°) to find the energy in joules: Q = 152,000 kg x 4,186 J/kg°C x 10.0 C° = 6,362,720,000 J.

To convert joules to kilowatt-hours, divide the total joules by 3,600,000 (the number of joules in one kilowatt-hour): 6,362,720,000 J / 3,600,000 J/kWh = 1,767 kWh. Therefore, the energy required is 1,767 kWh, making option (b) 1,770 kWh the nearest correct answer.

A certain transverse wave is described by y(x,t)=Bcos[2π(xL−tτ)],where B = 6.40 mm , L = 26.0 cm , and τ = 3.90×10^−2 s .a. Determine the wave's amplitude.b. Determine the wave's wavelength.c. Determine the wave's frequency.d. Determine the wave's speed of propagation. e. Determine the wave's direction of propagation.

Answers

Answer

given,

[tex]y(x,t)=B cos[2\pi (\dfrac{x}{L} - \dfrac{t}{\tau})][/tex]

B = 6.40 mm ,  L = 26 cm ,    τ = 3.90 × 10⁻² s

general wave equation

  y = A cos (k x - ωt)

where A is the amplitude of the

a) Amplitude of the given wave

      B = 6.40 mm

b) Wavelength of the given wave

       λ = L

       λ = 26 cm

c)  wave frequency

     [tex]f = \dfrac{1}{\tau}[/tex]

     [tex]f = \dfrac{1}{3.9 \times 10^{-2}}[/tex]

            f = 25.64 Hz

d) speed of wave will be equal to

     v = f λ

     v = 25.64 x 0.26

     v = 6.67 m/s

e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.

A rigid, insulated tank that is initially evacuated is connected though a valve to a supply line that carries steam at 1 MPa and 300∘C. Now the valve is opened, and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed.

Determine the final temperature of the steam in the tank, in ∘C.

Answers

Answer:

Final Temperature of the steam tank = 456.4°C

Explanation:

Assuming it to be a uniform flow process, kinetic and potential energy to be zero, and work done and heat input to be zero also.  We can conclude that,

Enthalpy of the steam in pipe = Internal Energy of the steam in tank

Using the Property tables and Charts - Steam tables,  

At Pressure= 1 MPa and Temperature= 300°C,

Enthalpy = 3051.2 kJ/kg

At Pressure= 1 MPa and Internal Energy= 3051.2 kJ/kg,

Temperature = 456.4°C.

Cart 111 of mass mmm is traveling with speed v_0v 0 ​ v, start subscript, 0, end subscript in the + x+xplus, x-direction when it has an elastic collision with cart 222 of mass 3m3m3, m that is at rest. What are the velocities of the carts after the collision?

Answers

Answer:

In an elastic collision, the momentum and the kinetic energies are conserved.

Momentum:

[tex]\vec{P_i} = \vec{P_f}\\\vec{P}_1 + \vec{P}_2 = \vec{P}_1' + \vec{P}_2'\\m\vec{v_0} + 0 = m\vec{v_1}' + 3m\vec{v_2}}' \\v_0 = v_1 + 3v_2[/tex]

Kinetic energy:

[tex]K_i = K_f\\K_1 + K_2 = K_1' + K_2'\\\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}m{v_1'}^2 + \frac{1}{2}3m{v_2'}^2\\v_0^2 = {v_1'}^2 + 3{v_2'}^2[/tex]

We have two equations and two unknowns:

[tex]v_0 = v_1' + 3v_2'\\v_0^2 = {v_1'}^2 + 3{v_2'}^2\\\\3v_2' = v_0 - v_1'\\3{v_2'}^2 = {v_0}^2 - {v_1'}^2\\\\3{v_2'}^2 = (v_0 - v_1')(v_0 + v_1') = 3{v_2}'(v_1' + v_0)\\\\v_2' = v_1' + v_0\\3v_2' = v_0 - v_1'\\\\4v_2' = 2v_0\\\\v_2' = v_0/2\\v_1' = -v_0/2[/tex]

Explanation:

The first cart hits the second cart at rest and turns back with half its speed.

The second cart starts moving to the right with half the initial speed of the first cart.

Answer:

v1 = -v0/3 ,v2 = 2v0/3

Explanation:

An 80.0-g piece of copper, initially at 295°C, is dropped into 250 g of water contained in a 300-g aluminum calorimeter; the water and calorimeter are initially at 10.0°C.

What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 0 and 0.215 cal/g⋅°C, respectively. cw = 1.00 cal/g°C)
a. 12.8°C
b. 16.5°C
c. 28.4°C
d. 32.1°C

Answers

Answer:

b. 16.5°C

Explanation:

[tex]m_{c}[/tex] = mass of piece of copper = 80 g

[tex]c_{c}[/tex] = specific heat of piece of copper = 0.0920 cal/g°C

[tex]T_{ci}[/tex] = Initial temperature of piece of copper = 295 °C

[tex]m_{w}[/tex] = mass of water = 250 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/g°C

[tex]T_{wi}[/tex] = Initial temperature of piece of copper = 10 °C

[tex]m_{al}[/tex] = mass of calorimeter  = 300

[tex]c_{al}[/tex] = specific heat of calorimeter = 0.215 cal/g°C

[tex]T_{ali}[/tex] = Initial temperature of calorimeter = 10 °C

[tex]T[/tex] = Final equilibrium temperature

Using conservation of heat

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

[tex]m_{c} c_{c} (T_{ci} - T) = m_{w} c_{w} (T - T_{wi})+ m_{al} c_{al} (T - T_{ali})\\(80) (0.092) (295 - T) = (250) (1) (T - 10) + (300) (0.215) (T - 10)\\T = 16.5 C[/tex]

The final temperature of the system is 16.4 ⁰C.

Conservation of energy

The final temperature of the system is determined by applying the principle of conservation of energy as shown below;

Heat lost by piece of copper = heat gained by water + heat gained by calorimeter

mCc(Tc - T) = mCw(T - Tw) + mCl(T - Tl)

80 x 0.09 x (295 - T) = 250 x 1 x (T - 10) + 300 x 0.215 x (T - 10)

2124 - 7.2T = 250T - 2500 + 64.5T - 645

5269 = 321.7T

T = 5269/321.7

T = 16.4 ⁰C

Thus, the final temperature of the system is 16.4 ⁰C.

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A solar heating system has a 25.0% conversion efficiency; the solar radiation incident on the panels is 1 000 W/m2.

What is the increase in temperature of 30.0 kg of water in a 1.00-h period by a 4.00-m2 -area collector? (cw = 4 186 J/kg⋅°C)

a. 14.3°Cb. 22.4°Cc. 28.7°Cd. 44.3°C

Answers

Answer:

c. 28.7C

Explanation:

Since the solar radiation incident on the panel is 1000W/m2 and the collector has an area of 4m2. We can conclude that the solar power generated is

1000 * 4 = 4000 W or J/s

Within 1 hour, or 3600 seconds this solar power generator should generate (with 25 % efficiency):

E = 0.25 * 4000 * 3600 = 3600000 J or 3.6 MJ

This energy will be converted to heat to heat up water. Using water heat specific of 4186 J/kg C we can find out how much temperature has raised:

[tex]\Delta T = \frac{E}{c*m} = \frac{3600000}{30 * 4186} = 28.7^oC[/tex]

So C. is the correct answer

If a 6-m cylindrical bar has a constant density of p= 5 g/cm for its left half and a constant density p = 6 g/cm for its right half, what is its mass?

Answers

Final answer:

To find the mass of the cylindrical bar, calculate the volumes of the two halves using the formula V = πr²h, where r is the radius and h is the height, and multiply them by their respective densities. Then divide the total mass by the total length of the bar.

Explanation:

To find the mass of the cylindrical bar, we need to consider the densities of its left and right halves. The left half has a density of 5 g/cm, while the right half has a density of 6 g/cm. The total length of the bar is 6 m, so we can calculate the volumes of the two halves using the formula V = πr²h, where r is the radius and h is the height.

Let's assume the left half has a radius of r1 and a height of 6 m, and the right half has a radius of r2 and a height of 6 m. The total mass can then be calculated by multiplying the volume of each half by its respective density and summing the results. Finally, we divide the mass by the total length of the bar to get the mass per meter.

An observer sits in a boat watching wave fronts move past the boat. The distance between successive wave crests is 0.80 m, and they are moving at 2.2 m / s.

What is the wavelength of these waves?
a. 1.6 m
b. 2.2 m
c. 0.80 m

What is the frequency of these waves?
a. 0.36 Hz
b. 2.8 Hz
c. 0.80 Hz

What is the period of these waves?
a. 0.80 s
b. 0.36 s
c. 2.8 s

Answers

To solve this problem we will use the three requested concepts: Wavelength, frequency and period.

The wavelength is the distance between each crest, therefore it is already given and is 0.8m

The correct answer is C.

The frequency can be described as a relationship between wave speed and wavelength therefore

[tex]f = \frac{v}{\lambda}[/tex]

[tex]f = \frac{2.2}{0.8}[/tex]

[tex]f = 2.75Hz \approx 2.8Hz[/tex]

The correct answer is B.

The period is the inverse of the frequency therefore

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{2.8}[/tex]

[tex]T = 0.35s[/tex]

The correct answer is B.

(a) The wavelength of the wave is 0.80m and the right option is c.

(b) The frequency of the wave is 2.8 Hz and the right option is b.

(c) The period of the wave is 0.36 s and the right option is b

(a) The distance between successive wave crests = wavelength of the wave

From the question,

(a) Wavelength = 0.80 m

Hence the wavelength = 0.80 m

(b) Using,

     V = λf.............. Equation 1

Where V = Velocity of the wave, λ = wavelength of the wave, f = frequency of the wave.

f = V/λ.................... Equation 2

Given: V = 2.2 m/s, λ = 0.80 m

Substitute these values into equation 2

f = 2.2/0.8

f = 2.75 Hz.

f ≈ 2.8 Hz

Hence the frequency of the wave is 2.8 Hz

(c) f = 1/T.............. Equation 3

Where T = period.

Therefore,

T = 1/f .................. 4

Given: f = 2.8 Hz,

T = 1/2.8

T = 0.357

T ≈ 0.36 s

Hence the period of the wave = 0.36 s

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Spaceships A and B are traveling directly toward each other at a speed 0.5c relative to the Earth, and each has a headlight aimed toward the other ship.


What value do technicians on ship B get by measuring the speed of the light emitted by ship A's headlight?


Answer choices


1).75c


2)1.0c


3)1.5c


4) .5c

Answers

The answer is 2) 1.0c. Light will always propagate through a vacuum at the speed of light “c”; even when moving at a significant fraction of the speed of light, observers will still measure this as the speed of light and the difference is resultant of time dilation.

Which of the following statements correctly describes the law of conservation of energy? Group of answer choicesa. Mass cannot be created but it can be destroyed under extreme pressures.b. Mass cannot be conserved during a chemical reaction; a little bit of mass is always lost.c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed.d. When added to a system, energy can destroy mass.

Answers

To solve this problem we will also apply the concept related to the conservation of the mass, which announces that: "In an isolated system, during any ordinary chemical reaction, the total mass in the system remains constant, that is, the mass consumed by the reagents is equal to the mass of the products obtained. "

If the mass is in a closed system, it cannot change. This assessment should not be confused with the transformation of the matter within it, for which it is possible that over time the matter will change from one form to another. For example during a chemical reaction, there is a rupture of links to reorganize into another, but said mass in the closed system is maintained.

The correct answer is:

C. "The mass of a closed system cannot change over time; mass cannot be created or destroyed."

The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed

The law of conservation of mass states that the mass is an isolated system that can not be created nor destroyed.

conserved means saved, so according to the law of conservation of mass refers to the "saving" of mass.

Thus, The following statements correctly describe the law of conservation of energy - c. The mass of a closed system cannot change over time; mass cannot be created nor destroyed

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