Answer:
a. 1.0 eV
Explanation:
Given that
Voltage difference ,ΔV = 1 V
From work power energy
Work =Change in the kinetic energy
We know that work on the charge W= q ΔV
For electron ,e= 1.6 x 10⁻¹⁹ C
q=e= 1 x 1.6 x 10⁻¹⁹ C
Change in the kinetic energy
ΔKE= q ΔV
Now by putting the values
ΔKE= 1 x 1.6 x 10⁻¹⁹ x 1 C.V
We can also say that
ΔKE= 1 e.V
Therefore the answer will be a.
a). 1.0 eV
what consistent physiological pattern is more common in men's' teeth? A. Men have more room in their mouths for their molars. B. Men generally have stronger enamel. C. Men have more teeth. D. Men generally have larger teeth.
Answer:
B.
Explanation:
Men generally have stronger enamel than the women.
There was an accident at a uranium processing plant in Tokai, Japan, on September 30, 1999. Using your favorite search engine, find out what happened and compare this accident with the ones at Chernobyl and Three Mile Island. How could these accidents have been avoided? Do these incidents prove that nuclear power plants should be phased out? Why or why not?
For the three accidents mentioned there were human factors that caused the destabilization of the reactors and generated catastrophies.
In the case of Tokai, there was an excess of Uranium due to the fact that workers (Not qualified, since such work did not merit it) added to the containers, which generated an excess in the filling generating an emulsion of radiation to all the personnel.
In the case of Chernobyl, it was the experimentation through a series of tests to reduce the power, during which a series of imbalances occurred in the reactor 4 of this nuclear power plant, which led to the uncontrolled overheating of the reactor core nuclear-
In the case of Three Mile Island it was a design error in which the water level was underestimated, believing that the required level was available but in the end it was noted that said water level was not sufficient which caused the melting of a water dipstick.
All this leaves us with reflections on the security protocols followed for the construction or management of these nuclear power plants, and calls into question the human capacity to react to one of these catastrophes. Today, these plants not only put human health at risk but also generate waste that pollutes the planet. Alternatives such as renewable and clean energy already become more popular every day and are about to leave nuclear energy in the past to give rise to a new human stage.
This question relates to the practicality of searching for intelligent life in other solar systems by detecting their radio broadcasts (or aliens find us from ours). The closest stars are 4 light years away from us. How far away must you be from a 781 kHz radio station with power 50.0 kW for there to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Answer:
[tex]2.77287\times 10^{15}\ m[/tex]
Explanation:
P = Power = 50 kW
n = Number of photons per second
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\nu[/tex] = Frequency = 781 kHz
r = Distance at which the photon intensity is i = 1 photon/m²
Power is given by
[tex]P=nh\nu\\\Rightarrow n=\dfrac{P}{h\nu}\\\Rightarrow n=\dfrac{50000}{6.626\times 10^{-34}\times 781000}\\\Rightarrow n=9.66201\times 10^{31}\ photons/s[/tex]
Photon intensity is given by
[tex]i=\dfrac{n}{4\pi r^2}\\\Rightarrow 1=\dfrac{9.66201\times 10^{31}}{4\pi r^2}\\\Rightarrow r=\sqrt{\dfrac{9.66201\times 10^{31}}{4\pi}}\\\Rightarrow r=2.77287\times 10^{15}\ m[/tex]
The distance is [tex]2.77287\times 10^{15}\ m[/tex]
You must stay at a distance of [tex]2.77287*10^1^5m[/tex]
How can we arrive at this result?First, we have to find the number of protons per second. This will be done using the equation: [tex]n= \frac{P}{h*v}[/tex]In this equation, the "h" represents Planck's constant and will take on the value of [tex]6.626*10^-^3^4m^2\frac{Kg}{s}[/tex]
The "r" will be equal to 1 photon/m² and the "P' will be equal to 50 kW.
Therefore, we will solve the equation as follows:
[tex]n= \frac{50000}{(6.626*10^-^3^4*781000)}= 9.66201*10^3^1 \frac{protons}{s}[/tex]
From this value, we can calculate the appropriate distance for you to position yourself. For this, we will use the equation:[tex]r^2=\frac{n}{4*\pi} \\r= \sqrt{\frac{9.6621*10^3^1}{4*\pi } } = 2.77287*10^1^5m[/tex]
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Starting from rest, your friend dives from a high cliff into a deep lake below, yelling in excitement at the thrill of free-fall on her way down. You watch her, as you stand on the lake shore, and at a certain instant your keen hearing recognizes that the usual frequency of her yell, which is 919 Hz, is shifted by 55.9 Hz. How long has your friend been in the air when she emits the yell whose frequency shift you hear? Take 342 m/s for the speed of sound in air and 9.80 m/s2 for the acceleration due to gravity.
Answer:
2 seconds
Explanation:
f = Frequency of yell = 919 Hz
[tex]\Delta f[/tex] = Shifted frequency = 55.9 Hz
v = Speed of sound in air = 342 m/s
[tex]v_r[/tex] = Velocity of friend
a = Acceleration due to gravity = 9.81 m/s²
From the Doppler shift formula we have
[tex]\dfrac{f+\Delta f}{f}=\dfrac{v}{v-v_r}\\\Rightarrow v_r=v-\dfrac{vf}{f+\Delta f}\\\Rightarrow v_r=342-\dfrac{342\times 919}{919+55.9}\\\Rightarrow v_r=19.61\ m/s[/tex]
The velocity of the my friend is 19.61 m/s
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{19.61-0}{9.8}\\\Rightarrow t=2\ s[/tex]
The time my friend is in the air is 2 seconds
A seaside cliff is 30 m above the ocean surface, and Sam is standing at the edge of the cliff. Sam has three identical stones. The first stone he throws off the cliff at 30° above the horizontal. The second stone he throws vertically downward into the ocean. The third stone he drops into the ocean.
1. In terms of magnitude, which stone has the largest change in its velocity over a one second time interval after its release? (Sam’s throwing speed is 10 m/s.)
Answer:
In terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
Explanation:
Stone 1:
vi = 10 m/s
vix = vi*Cos ∅ = (10 m/s)*Cos 30° = 8.66 m/s = vx
viy = vi*Sin ∅ = (10 m/s)*Sin 30° = 5 m/s
vy = viy - g*t = (5 m/s) - (9.8m/s²)*(1 s) = -4.8
then
v = √(vx²+vy²) = √((8.66)²+(-4.8)²) = 9.90 m/s
Δv = v - vi = 9.902 m/s - 10 m/s
⇒ Δv = -0.098 m/s
Stone 2:
vi = 10 m/s
v = vi + g*t = (10 m/s) + (9.8m/s²)*(1 s) = 19.8 m/s
Δv = v - vi = (19.8 m/s) - (10 m/s)
⇒ Δv = 9.8 m/s
Stone 3:
vi = 0 m/s
v = g*t = (9.8m/s²)*(1 s) = 9.8 m/s
Δv = v - vi = (9.8 m/s) - (0 m/s)
⇒ Δv = 9.8 m/s
Finally, in terms of magnitude, the stones 2 and 3 have the largest change in its velocity over a one second time interval after their release.
1. A 70-kg swimmer dives horizontally off a 500-kg raft. The diver's speed immediately after leaving the raft is 6.0 m/s. A micro-sensor system attached to the edge of the raft measures the time interval during which the diver applies an impulse to the raft just prior to leaving the raft surface. If the time interval is read as 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?
To solve this problem it is necessary to apply the concepts related to momentum theorem.
The equation for impulse is given as
[tex]I = Ft[/tex]
Where
I = Force
t = Time
At the same time we have the equation for momentum is given as
[tex]p = mv[/tex]
The impulse momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Therefore
I = p
Ft = mv
Solving to find the force
[tex]F = \frac{mv}{t}[/tex]
[tex]F = \frac{(70)(6)}{0.25}[/tex]
[tex]F = 1680N[/tex]
Therefore the magnitude of the average horizontal force by diver on the raft is 1680N
The magnitude of the average horizontal force exerted by the diver on the raft is 1680 N.
Explanation:To find the magnitude of the average horizontal force exerted by the diver on the raft, we need to start by calculating the change in momentum of the diver. The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. The change in momentum is equal to the impulse, which is given by the equation J = Δp = mΔv.
Since the swimmer dives horizontally, the change in velocity is equal to the initial velocity of the swimmer. Therefore, Δv = 6.0 m/s. Substituting the values, we get J = (70 kg)(6.0 m/s) = 420 kg·m/s.
The impulse is equal to the average force multiplied by the time interval, so we can rearrange the equation to solve for the average force. F = J / Δt = 420 kg·m/s / 0.25 s = 1680 N.
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wo parallel-plate capacitors have the same dimensions, but the space between the plates is filled with air in capacitor 1 and with plastic in capacitor 2. Each capacitor is connected to an identical battery, such that the potential difference between the plates is the same in both capacitors. Compare the magnitudes of the electric fields between the plates, ????1 and ????2, and the magnitudes of the free charges on the plates, ????1 and ????2.
Final answer:
The electric field between the plates will be the same in both capacitors, but the charge on the plates of capacitor 2 with plastic between the plates will be greater than the charge on the plates of capacitor 1 with air between the plates.
Explanation:
When comparing the magnitudes of the electric fields between the plates, we can use the formula E = Q / (ε0 * A), where E is the electric field, Q is the charge on the plates, ε0 is the permittivity of free space, and A is the area of the plates. In this case, since the capacitors have the same dimensions and the same charge, the electric field between the plates in both capacitors will be the same.
Regarding the magnitudes of the free charges on the plates, we know that Q = C * V, where Q is the charge, C is the capacitance, and V is the potential difference. Since the capacitance is directly proportional to the dielectric constant of the material between the plates, and the charge is the product of the capacitance and potential difference, the charge on the plates of capacitor 2 with plastic between the plates will be greater than the charge on the plates of capacitor 1 with air between the plates, because the dielectric constant of plastic is greater than 1.
Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis oriented at 0°. A second polarizer has its transmission axis oriented at 45° and a third polarizer oriented with its axis at 90°. Determine the fraction of light intensity exiting the third sheet with and without the second sheet present.
Answer:
Explanation:
Given
Initial Intensity of light is S
when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.
When it is passed through a second Polarizer with its transmission axis [tex]\theat =45^{\circ}[/tex]
[tex]S_1=S_0\cos ^2\theta [/tex]
here [tex]S_0=\frac{S}{2}[/tex]
[tex]S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}[/tex]
[tex]S_1=\frac{S}{4}[/tex]
When it is passed through third Polarizer with its axis [tex]90^{\circ}[/tex] to first but [tex]\theta =45^{\circ}[/tex] to second thus [tex]S_2[/tex]
[tex]S_2=S_0\cos ^2\theta [/tex]
[tex]S_2=\frac{S}{4}\times \frac{1}{2}[/tex]
[tex]S_2=\frac{S}{8}[/tex]
When middle sheet is absent then Final Intensity will be zero
Final answer:
The first polarizing sheet reduces the intensity of unpolarized light to 50%. The second polarizing sheet further reduces the intensity to 25%. The third polarizing sheet, with the second sheet present, does not allow any light to pass through.
Explanation:
When unpolarized light passes through a polarizing sheet, the intensity of the light reduces by half. The first polarizing sheet reduces the intensity to 50% of the original intensity. The second polarizing sheet, oriented at an angle of 45° to the first sheet, further reduces the intensity by 50%. So, the intensity exiting the second sheet is 25% of the original intensity (50% x 50% = 25%).
However, the third polarizing sheet, oriented at an angle of 90° to the first, does not allow any light to pass through because the transmission axis of the third sheet is perpendicular to the polarization direction of the light. Therefore, the fraction of light intensity exiting the third sheet, with the second sheet present, is 0%.
An 80 kg astronaut has gone outside his space capsule to do some repair worlc Unfortunately, he forgot to lock his safetytether in place, and he has drifted 5.0 m away from the capsule. Fortunately, he has a 1000 w portable laser with freshbatteries that will operate it for 1.0 hr His only chance is to accelerate himself toward the space capsule by firing the laser inthe opposite direction. He has a 10-h supply of oxygen. How long will it take him to reach safety?
Final answer:
The question involves using the principles of conservation of momentum and kinematics in space to rescue an astronaut adrift from their space capsule using a laser for propulsion. However, the time required for the astronaut to reach safety cannot be calculated without the thrust or force specifics of the laser.
Explanation:
The scenario described involves the principles of conservation of momentum and the astronaut's ability to utilize a laser as a propulsion device. The astronaut's mass is 80 kg, and they need to use the laser to accelerate towards the space capsule. Without any external forces, momentum is conserved, and hence, the astronaut can generate thrust in space by expelling photons in the opposite direction of the desired movement, albeit very weak thrust. To determine how long it will take him to reach safety, we would need to calculate the actual propulsion force of the laser and the resulting acceleration. This can be derived from the conservation of momentum and Newton's second law of motion. However, with the current information provided, it is impossible to provide an accurate estimate without knowing the momentum or thrust provided by the laser.
Without such crucial information, assuming an ideal scenario and if the astronaut could somehow generate a constant force to obtain a tangible acceleration, we can use kinematics equations to estimate travel time once the acceleration is known. Yet in this specific situation, it's not feasible to calculate the time required for the astronaut to reach the space capsule without additional details about the actual force the laser can exert on the astronaut.
There is a skier at the top of a ski slope. The skier has potential energy. What gives the skier his potential energy? A. his speed B. friction C. gravity D. his efficiency
Answer:C
Explanation:
Skier at the top of a ski has Potential Energy due to gravity.
Potential Energy is the Energy Possessed by an object when it attains a height concerning some zero level Position.
During the process of attaining the height, some work has to be done against gravity and this energy stored within the object after attaining some height w.r.t relative zero position.
The answer is C: gravity
An inventor develops a stationary cycling device by which an individual, while pedaling, can convert all of the energy expended into heat for warming water.
How much mechanical energy is required to increase the temperature of 300 g of water (enough for 1 cup of coffee) from 20°C to 95°C?
(1 cal = 4.186 J, the specific heat of water is 4 186 J/kg⋅°C)
a. 94 000 J
b. 22 000 J
c. 5 400 J
d. 14 J
Answer:
[tex]Q=94185\ J[/tex]
Explanation:
Given:
mass of water, [tex]m=0.3\ kg[/tex]initial temperature of water, [tex]T_i=20^{\circ}C[/tex]final temperature of water, [tex]T_f=95^{\circ}C[/tex]specific heat of water, [tex]c=4186\ J.kg^{-1}.K^{-1}[/tex]Now the amount of heat energy required:
[tex]Q=m.c.\Delta T[/tex]
[tex]Q=0.3\times 4186\times (95-20)[/tex]
[tex]Q=94185\ J[/tex]
Since all of the mechanical energy is being converted into heat, therefore the same amount of mechanical energy is required.
In the experiment to measure specific heat of a metal, you get the following data. What is the specific heat of the metal in J/kg.C?
(answer within +-3% will considered correct) MASSES: water= 259 g, metal=159 g, calorimeter=97 g INITIAL TEMPERATURES: water+calorimeter=21ºC, metal=98ºC Final equilibrium temp=31ºC Calorimeter is made of aluminum.
Answer:
c = 1,100 J/kgºC
Explanation:
Assuming that all materials involved, finally arrive to a final state of thermal equilibrium, and neglecting any heat exchange through the walls of the calorimeter, the heat gained by the system "water+calorimeter" must be equal to the one lost by the unknown metal.
The equation that states how much heat is needed to change the temperature of a body in contact with another one, is as follows:
Q = c * m* Δt
where m is the mass of the body, Δt is the change in temperature due to the external heat, and c is a proportionality constant, different for each material, called specific heat.
In our case, we can write the following equality:
(cAl * mal * Δtal) + (cH₂₀*mw* Δtw) = (cₓ*mₓ*Δtₓ)
Replacing by the givens , and taking cAl = 0.9 J/gºC, we have:
Qg= 0.9 J/gºC*97g*10ºC + 4.186 J/gºC*259g*10ºC = 11,715 J(1)
Ql = cₓ*159g*67ºC (2)
Based on all the previous assumptions, we have:
Qg = Ql
So, we can solve for cx, as follows:
cx = 11,715 J / 159g*67ºC = 1.1 J/gºC (3)
Expressing (3) in J/kgºC:
1.1 J/gºC * (1,000g/1 kg) = 1,100 J/kgºC
Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2, and oxygen gas, O2 to produce the following combustion reaction: 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) Imagine that you have a 6.50 L gas tank and a 4.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
Answer:
Pressure in acetylene gas tank will be 74.8atm
Explanation:
Step 1: Using the ideal gas equation, determine the number of moles of oxygen
[tex]n=\frac{PV}{RT}[/tex]
[tex]P_O=115atm[/tex]
[tex]V_O=6.50L[/tex]
[tex]n_O=\frac{P_OV_O}{RT}[/tex]
[tex]n_O=\frac{115*6.50}{RT}[/tex]
[tex]n_O=\frac{747.5}{RT}[/tex]
As temperature is unknown and assumed to be the same for both gases, and the ideal gas constant will be the same for both cases, these values are left as constants 'T' and 'R'.
Step 2: Determine the proportionate number of moles of acetylene required based on the chemical equation
2 moles of acetylene require 5 moles of oxygen for complete combustion
Thus, 0.4 moles of acetylene are required per mole of oxygen
[tex]n_A=0.4n_O[/tex]
[tex]n_A=0.4*\frac{747.5}{RT}[/tex]
[tex]n_A=\frac{299}{RT}[/tex]
Step 3: Determine the pressure of acetylene tank required in 4.00L tank
[tex]n_A=\frac{P_AV_A}{RT}[/tex]
[tex]\frac{299}{RT}=\frac{P_A*4.00}{RT}[/tex]
[tex]P_A=\frac{299}{4.00}[/tex]
[tex]P_A=74.75 atm[/tex]
Assumptions:
Temperature is the same for both gases and constant
The ideal gas constant is the same for both gases
The combustion reaction is complete and there are no limiting factors
A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is applied a damping force (in Newtons) of the form F = –0.1v, with v the velocity in m/s. The spring is stretched a distance xm and released. After four complete oscillations, what fraction of the mechanical energy is retained by the system?
Answer:
0.5% per oscillation
Explanation:
The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.
Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).
Where m= Mass, v= velocity.
Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).
Where k= force constant, X= displacement.
Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).
Therefore, we use equation (3) to get the resonance frequency,
W^2= k/m--------------------------------------------------------------------------------------(3)
Slotting values into equation (3).
= 10/2.5.
= ✓4.
= 2 s^-1.
Recall that, F= -kX
F^2= (-0.1)^2
Potential energy,PE= 1/2 ×0.01
Potential energy= 0.05 ×100
= 0.5% per oscillation.
In a damped harmonic oscillator, the fraction of mechanical energy retained by the system after multiple oscillations can be calculated by comparing the initial and final potential energies. The fraction of mechanical energy retained is equal to the ratio of the final amplitude squared to the initial amplitude squared.
Explanation:In a damped harmonic oscillator, the mechanical energy is gradually lost due to the damping force. The fraction of mechanical energy retained by the system after four complete oscillations can be determined by comparing the initial mechanical energy to the final mechanical energy. The initial mechanical energy is the sum of the potential energy and kinetic energy, while the final mechanical energy is only the potential energy.
Using the equation for the potential energy of a spring, U = ½kx², we can calculate the initial and final potential energies. The initial potential energy can be calculated using the initial amplitude, A, and the spring constant, k. The final potential energy can be calculated using the final amplitude, A', and the same spring constant, k.
The fraction of mechanical energy retained by the system is equal to the ratio of the final potential energy to the initial potential energy. This can be calculated using the equation: fraction of mechanical energy retained = (A'/A)².
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After coming down a steep hill at a constant speed of 43 m/s, a car travels along the circumference of a vertical circle of radius 618 m until it begins to climb another hill. r x What is the magnitude of the net force on the 34 kg driver of the car at the lowest point on this circular path? Answer in units of kN.
Answer:
F=0.101 kN
Explanation:
Newton's 2nd law, F = ma, but this is circular path, the acceleration (a) is the centripetal acceleration.
a = (v²) / r
F = (m×v²) /r
F=(34 kg)×(43 m/s)² / 618 m
F=101.72 N
To convert Newtons into kilo-Newtons divide it 1000.
F=0.101 kN
At high noon, the sun delivers 1 000 W to each square meter of a blacktop road.
What is the equilibrium temperature of the hot asphalt, assuming its emissivity e = 1? (σ = 5.67 × 10−8W/m2⋅K4) .
a. 75°Cb. 84°Cc. 91°Cd. 99°C
Answer:
The correct answer is c, T = 91.3°C
Explanation:
For this exercise let's use Stefan's equation on the emission of a black body
P = σ A e T⁴
Where σ is the Stefan-Boltzmann constant, A the area, and 'e' emissivity and T the absolute temperature
In this case give the absorbed power is 1000W per square meter, let's clear the temperature equation
T⁴ = (P / A) 1/σ e
Let's calculate
T⁴ = 1000 1 / (5.67 10⁻⁸ 1)
T⁴ = 176.37 10⁸
T =[tex]\sqrt[4]{176.37 10^8}[/tex]
T = 3.6442 10² K
Let's reduce to degrees Celsius
T = 364.42 -273.15
T = 91.3 ° C
The correct answer is c
The equilibrium temperature of the hot asphalt, assuming it behaves as a perfect blackbody with emissivity of 1, can be calculated using the Stefan-Boltzmann law, resulting in approximately 91°C, making choice (c) the correct answer.
To determine the equilibrium temperature of the hot asphalt, we can use the concept of blackbody radiation.
Since the emissivity (e) is 1, the asphalt behaves as a perfect blackbody, which means it absorbs and emits radiation efficiently.
The power per unit area absorbed by the asphalt is:
P = 1000 W/m²According to the Stefan-Boltzmann law, the power radiated per unit area by a blackbody is given by:
P = σeT⁴where
σ is the Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²·K⁴), and T is the equilibrium temperature in KelvinGiven that e = 1 for a perfect blackbody, we can set the absorbed power equal to the emitted power:
1000 = 5.67 × 10⁻⁸ T⁴Solving for T:
T⁴ = 1000 / 5.67 × 10⁻⁸T⁴ ≈ 1.76 × 10¹⁰T ≈ (1.76 x 10¹⁰)∧1/4T ≈ 278.8 KConverting to Celsius:
T ≈ 278.8 K - 273.15 ≈ 5.65°CThis result does not match any answer choices, which suggests a potential issue. The correct calculations should yield a higher temperature due to an error in an earlier assumption or value misunderstanding. Revisiting the calculations correctly:
Solving again for higher accuracy:
T ≈ (1000 / 5.67 x 10⁻⁸)∧1/4 ≈ 364 K ≈ 91°CTherefore, the correct equilibrium temperature is 91°C, making the correct choice: (c) 91°C
A refrigerator is being pulled up a ramp with a horizontal force P, which acts at the top corner. The refrigerator has a mass of 75 kg, acting through point G. The ramp is inclined at 20º, and the coefficient of static friction is 0.3 between the refrigerator and the ramp.
(a) Find the force P required to move the refrigerator.
(b) Does the refrigerator tip or slide?
Answer:
(a) P = 459.055 N.
(b) the refrigerator tips.
Explanation:
Given, the angle of ramp is 20°.
When the weight of refrigerator is resolved in directions parallel and perpendicular to ramp, 75×g×sin(20°) and 75×g×cos(20°).
⇒ normal contact force is 75×g×cos(20°).
⇒ frictional force is 0.3×75×g×cos(20°) = 207.414 N
so, total opposite force is 207.414 + 75×g×sin(20°) = 459.055 N.
so, the force needed is P = 459.055 N
And as the moment due to both opposite force and P force are in same direction the refrigerator tips rather than just sliding.
Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What percent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 ∘ C and 721 mmHg ?
Answer:
The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%
Explanation:
From General gas equation.
PV = nRT............................... Equation 1
Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.
n = mass/molar mass .................. Equation 2
substituting equation 2 into equation 1.
PV = (mass/molar mass)RT
⇒ Mass/molar mass = PV/RT..................... Equation 3
But mass = Density × Volume
⇒ M = D × V.................... Equation 4
Where D = density, M = mass
Substituting equation 4 into equation 3
DV/molar mass = PV/RT............ Equation 5
Dividing both side of the equation by Volume (V) in Equation 5
D/molar mass = P/RT .............. Equation 6
Cross multiplying equation 6
D × RT = P × molar mass
∴ Molar mass = (D × RT)/P.................. Equation 7
Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,
T = 25°C = 25 + 273 = 298 K,
P =721 mmHg = (721/760) atm= 0.949 atm
Substituting these values into equation 7
Molar mass = (0.518 × 0.0821 × 298)/0.949
Molar mass = 13.35 g/mole
The molar mass of the mixture is =13.35 g/mole
Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.
∴ 13.35 = 4(y) + 32(1-y)
13.35 = 4y + 32 - 32y
Collecting like terms in the equation,
32y - 4y = 32 - 13.35
28y = 18.65
y = 18.65/28
y =0.666
y = 0.666 × 100 = 66.6%
∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%
The percent (by moles) of He in the helium-oxygen mixture with a density of 0.518 g/L at 25 ∘C and 721 mmHg is 13.6%.
Explanation:To determine the percent (by moles) of He in the helium-oxygen mixture, we need to use the Ideal Gas Law equation:
PV = nRT
Where:
P is the pressure (721 mmHg)V is the volume (0.518 L)n is the number of moles of the gas we're interested in (He in this case)R is the ideal gas constant (0.0821 L·atm/mol·K)T is the temperature in Kelvin (25 + 273 = 298 K)Rearranging the equation to solve for n:
n = PV / RT
Substituting the values:
n = (721 mmHg * 0.518 L) / (0.0821 L·atm/mol·K * 298 K)
n = 0.136 mol
Since the total number of moles in the mixture is 1 (as it is a binary mixture), the percent (by moles) of He can be calculated as:
Percent of He = (0.136 mol of He / 1) * 100%
Percent of He = 13.6%
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When the daughter nucleus produced in a radioactive decay is itself unstable, it will eventually decay and form its own daughter nucleus. If the newly formed daughter nucleus is also unstable, another decay will occur, and the process will continue until a nonradioactive nucleus is formed. Such a series of radioactive decays is called a decay chain.
A good example of a decay chain is provided by 232 90Th, a naturally occurring isotope of thorium.
What is the energy Q released in the first step of the thorium-232 decay chain? The atomic mass of 232 90Th is 232.038054 u and the atomic mass of 228 88Ra is 228.0301069 u.
Answer in (MeV) and show your work
Answer:
4.981 MeV
Explanation:
The quantity of energy Q can be calculated using the formula
Q = (mass before - mass after) × c²
Atomic Mass of thorium = 232.038054 u, atomic of Radium = 228.0301069 u and mass of Helium = 4.00260. The difference of atomic number and atomic mass between the thorium and radium ( 232 - 228) and ( 90 - 88) show α particle was emitted.
1 u = 931.494 Mev/c²
Q = (mass before - mass after) × c²
Q = ( mass of thorium - ( mass of Radium + mass of Helium ) )× c²
Q = 232.038054 u - ( 228.0301069 + 4.00260) × c²
Q = 0.0053471 u × c²
replace 1 u = 931.494 MeV/ c²
Q = 0.0053471 × c² × (931.494 MeV / c²)
cancel c² from the equation
Q = 0.0053471 × 931.494 MeV = 4.981 MeV
A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg. (a) What is the angular velocity of the arm immediately after catching the ball as measured from the arm socket? (b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?
To solve the problem, we require an understanding of physics concepts like angular velocity, moment of inertia, and torque. The catcher catching the ball changes its angular momentum, resulting in an angular velocity. The torque experienced when the arm stops the rotation can be computed using known equations.
Explanation:This question involves concepts of physics like angular velocity, moment of inertia, and torque. Initially, with the catcher's arm at the ready position, the system (arm and ball) has zero angular velocity. Then when the catcher catches the ball, he applies an impulse to it and changes not just the linear momentum but the angular momentum about the shoulder as well.
The change in angular momentum (angular impulse) will be equal to the product of the mass of the baseball, its velocity, and the arm's length, i.e., 0.145kg × 40m/s × 0.5m= 2.9 kg m²/s. This change in angular momentum over time will induce an angular velocity, which can be calculated by dividing the change in angular momentum by the moment of inertia of the system (arm and ball).
For part (b), the torque experienced by the arm when it stops the rotation can be computed from the known equation Torque = (Moment of Inertia × Angular Acceleration). The angular acceleration is determined by the change in angular velocity divided by the time taken which in this case is 0.3 seconds. Taking all these physics concepts into account will yield the correct numerical solutions for parts (a) and (b).
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To find the angular velocity of the arm after catching the ball, conservation of angular momentum was used, resulting in an angular velocity of 7.85 rad/s. The torque needed to stop the arm's rotation in 0.3 seconds is -9.66 N - m, calculated using the angular deceleration. The answer involves concepts of angular momentum and torque.
A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg.
(a) What is the angular velocity of the arm immediately after catching the ball?
To find the angular velocity of the arm immediately after catching the ball, we need to use the principle of conservation of angular momentum. The initial angular momentum of the ball can be calculated using:
L_initial = m_ball * v_ball * r_arm
where m_ball = 0.145 kg, v_ball = 40 m/s, and r_arm = 0.5 m.
L_initial = 0.145 kg * 40 m/s * 0.5 m = 2.9 kg·m²/s
The moment of inertia of the arm plus the ball (approximated as point mass at the end) is:
I_total = I_arm + m_ball * r_arm²
Using the formula for the moment of inertia of a rod about one end: I_arm = (1/3) * m_arm * (r_arm)², where m_arm = 4.0 kg and r_arm = 0.5 m:
I_arm = (1/3) * 4.0 kg * (0.5 m)² = 0.333 kg·m²
Adding the moment of inertia of the ball:
I_total = 0.333 kg·m² + 0.145 kg * (0.5 m)² = 0.333 kg·m² + 0.03625 kg·m² = 0.36925 kg·m²
Since angular momentum is conserved, L_initial = I_total * ω, where ω is the angular velocity:
ω = L_initial / I_total = 2.9 kg·m²/s / 0.36925 kg·m² = 7.85 rad/s
(b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?
Torque (τ) can be calculated using the relationship between torque, angular deceleration (α), and moment of inertia (I):
τ = I_total * α
First, we find the angular deceleration. The arm stops, meaning final angular velocity is 0.
Using the angular kinematic equation: ω_final = ω_initial + α * t, where ω_final = 0 and t = 0.3 s:
0 = 7.85 rad/s + α * 0.3 s
α = -7.85 rad/s / 0.3 s = -26.17 rad/s²
Now, calculate torque:
τ = 0.36925 kg·m² * (-26.17 rad/s²) = -9.66 N·m
The negative sign indicates that the torque is in the direction opposite to the rotation.
A slice of bread contains about 100 kcal. If specific heat of a person were 1.00 kcal/kg·°C, by how many °C would the temperature of a 70.0-kg person increase if all the energy in the bread were converted to heat?a. 2.25°Cb. 1.86°Cc. 1.43°Cd. 1.00°C
Answer:
(c) 1.43°C
Explanation:
If the energy in the bread are converted to heat.
Then, The heat transferred from the bread to person = 100 kcal.
From specific heat capacity,
Q = cmΔT............................ equation 1
Where Q = quantity of heat, m = mass of the person, c = specific heat capacity of the person, Δ = increase in temperature.
Making ΔT the subject the equation 1,
ΔT = Q/cm........................ equation 2
Where Q = 100 kcal, c= 1.00 kcal/kg.°C, m = 70.0 kg
Substituting these values into equation 2,
ΔT = 100/(1×70)
ΔT = 100/70
ΔT = 1.428
ΔT ≈ 1.43°C
The increase in temperature of the body is = 1.43°C
The right option is (c) 1.43°C
The deflection of air masses to the right or left (depending on latitude) as they move from one latitude to another is called the:
a. Coriolis effect.
b. Hadley cell.
c. Saffir-Simpson scale.
d. Cyclonic effect.
e. Ekman spiral.
The coriolis effect is the force produced by the rotation of the Earth in space, which tends to deflect the trajectory of objects that move on the surface of the earth; to the right in the northern hemisphere and to the left, in the south. Said 'object' for this particular case is the mass of air. Therefore the correct answer is A: Coriolis effect.
Explain in detail why it may take longer for you to fight off the flu virus the first time this season than it would if you caught the same strain a second time.
Answer:
"when an infection has cleared, a small number of B and T cells are present in the blood with memory of the virus, allowing them to activate and destroy viruses more quick and fast next time when they enter the body."which is known as immune system."
Explanation:
Our immune system is not only designed to destroy disease-causing codes but remembers codes it's encountered so they are able to fight against them with when they return. when we all come across these types of invaders our layer of protection stops it from entering our body. When body senses a virus or other infections, like bacteria, the body tends to try and destroy the foreign invaders. The first step is the macrophages is activated. Macrophages destroy the foreign invaders. If the viruses penetrates deep into the body it result into infection. Where the T cells and B cells are activated to fight against these viruses.T & B cells contains antibodies that attach to the virus and label it as foreign for other cells to destroy.
Near the top of the Citigroup Bank building in New York City, there is a 4.00 105 kg mass on springs having adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven — the driving force is transferred to the mass, which oscillates instead of the building.
(a) What effective force constant should the springs have to make the mass oscillate with a period of 3.00 s? N/m
(b) What energy is stored in the springs for a 2.00 m displacement from equilibrium?
Answer:
(a) k = 1.76× 10⁶ N/m
(b) E = 3.52 × 10⁶ J
Explanation:
(a)
The period (T) of a spring = 2π√(m/k)
where m = mass of the spring in kg, k = spring constant.
T = 2π√(m/k)..................... equation 1
making k the subject of the equation,
k = 4π²(m)/T².............................. equation 2
Where m = 4.00 × 10⁵ kg, T = 3.00 s, π = 3.143
Substituting these values into equation 2
k = 4(3.143)²(4.0×10⁵)/3²
k = (1.58 × 10⁷)/9
k = 1.76× 10⁶ N/m
(b)
The energy stored(E) in a spring = 1/2ke²
Where k = spring constant, e = extension.
E = 1/2ke²
k = 1.76× 10⁶ N/m, e= 2.00 m
∴E = 1/2(1.76× 10⁶)(2)²
E = 2 × 1.76 × 10⁶
E = 3.52 × 10⁶ J
You are comparing two diffraction gratings using two different lasers: a green laser and a red laser. You do these two experiments
1. Shining the green laser through grating A you see the first maximum 1 meter away from the center
2. Shining the red laser through grating B, you see the first maximum 1 meter away from the center
In both cases, the gratings are the same distance from the screen.
(a) What can you deduce about the gratings?
(b) What would you observe if you shone the green laser through grating B:?
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
b. (a) grating B has more lines/mm; (b) the first maximum less than 1 meter away from the center
c.(a) grating B has more lines/mm; (b) the first maximum more than 1 meter away from the center
d. (a) grating B has more lines/mm: (b) the first maximum 1 meter away from the center
e. (a) grating A has more lines/mm; (b) the first maximum more than 1 meter away from the center
f. (a) grating A has more lines/mm; (b) the first maximum 1 meter away from the center
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Final answer:
Comparison of the diffraction patterns for two lasers and two gratings reveals that Grating B must have more lines per millimeter, and if the green laser were shone through Grating B, the first maximum would be more than 1 meter away from the center.
Explanation:
To analyze this problem, we need to apply our knowledge of diffraction gratings and wavelength of light. The position of the maxima on the screen depends on the grating spacing (number of lines per millimeter) and the wavelength of the light. The formula for the angle of the maxima for a diffraction grating is:
nλ = d sin θ,
where:
When comparing two diffraction gratings with different lasers:
A green laser (shorter wavelength) producing a first maximum at 1 meter suggests that the spacing between lines (d) in grating A supports this particular maximum for that wavelength.A red laser (longer wavelength) producing a first maximum at the same distance suggests that grating B must have a smaller d (more lines per mm) to compensate for its longer wavelength to produce a maximum at the same distance.Hence, the answer is (c):
a) Grating B has more lines/mm; because it compensates for the longer wavelength of the red light to still create a maximum at the same position as the green light with grating A.
b) If the green laser (shorter wavelength) were shone through grating B (more lines/mm), the first maximum would be more than 1 meter away from the center, since a grating with more lines per millimeter spreads the maxima further apart for the same wavelength, compared to a grating with fewer lines per millimeter.
Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for problems like this one. A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.130 m, an angular speed of 78.0 rad/s, and a moment of inertia of 1.25 kg·m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of 6 during a time of 3.00 s. (a) Find the magnitude of the angular deceleration of the cylinder. (b) Find the magnitude of the force of friction applied by the brake shoe.
Answer:
21.67 rad/s²
208.36538 N
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = [tex]\dfrac{1}{6}78=13\ rad/s[/tex]
[tex]\omega_i[/tex] = Initial angular velocity = 78 rad/s
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation
t = Time taken
r = Radius = 0.13
I = Moment of inertia = 1.25 kgm²
From equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2[/tex]
The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875[/tex]
Frictional force is given by
[tex]F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N[/tex]
The magnitude of the force of friction applied by the brake shoe is 208.36538 N
The angular deceleration of the cylinder is 46.0 rad/s².
The force of friction applied by the brake shoe is 1180 N.
Here's how we can approach it:
(a) Angular Deceleration:
Initial Angular Speed (ω₀): 92.0 rad/s
Final Angular Speed (ωf): ω₀/2 = 92.0 rad/s / 2 = 46.0 rad/s
Time (Δt): 4.00 s
We can use the following equation to find the angular deceleration (α):
α = (ωf - ω₀) / Δt
Substituting the values:
α = (46.0 rad/s - 92.0 rad/s) / 4.00 s
α = -46.0 rad/s² (negative sign indicates deceleration)
Therefore, the magnitude of the angular deceleration of the cylinder is 46.0 rad/s².
(b) Force of Friction:
Moment of Inertia (I): 1.36 kg·m²
Angular Deceleration (α): 46.0 rad/s²
The net torque (τ) acting on the cylinder is equal to the product of its moment of inertia and angular deceleration:
τ = I * α
The frictional force (F) applied by the brake shoe creates a torque that opposes the cylinder's rotation. This torque is equal to the force multiplied by the radius of the cylinder (r):
τ = F * r
Since the net torque is caused solely by the frictional force, we can equate the two torque equations:
I * α = F * r
Solving for the force of friction:
F = I * α / r
Substituting the values:
F = 1.36 kg·m² * 46.0 rad/s² / 0.0530 m
F = 1180 N
Therefore, the magnitude of the force of friction applied by the brake shoe is 1180 N.
What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?
Answer:
Absolute zero = 0 K or - 273°C
Explanation:
Absolute zero :
When the entropy and enthalpy of the ideal system reach at the minimum value then the temperature at that condition is known as absolute zero condition.
Absolute temperature is the minimum temperature in the temperature scale.The value of absolute zero is 0 K.
We know that
[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}[/tex]
F=Temperature in Fahrenheit scale
K=Temperature in Kelvin scale
C=Temperature in degree Celsius scale
When K = 0
[tex]\dfrac{C-0}{100}=\dfrac{K-273}{100}[/tex]
[tex]\dfrac{C-0}{100}=\dfrac{0-273}{100}[/tex]
C= - 273°C
Absolute zero = 0 K or - 273°C
Absolute zero is the lowest possible temperature, defined as 0 K on the Kelvin scale and -273.15°C on the Celsius scale. At this temperature, particles have minimal vibrational motion.
Explanation:Absolute zero is the lowest possible temperature where nothing could be colder and no heat energy remains in a substance. It is the point at which the fundamental particles of nature have minimal vibrational motion, retaining only quantum mechanical, zero-point energy-induced particle motion.
On the Kelvin scale, absolute zero is defined as 0 K. This is not stated in degrees, as it is an absolute measure. On the Celsius scale, absolute zero is equivalent to -273.15°C.
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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black and are maintained at a temperature of 450 K. The back sides of the disks are insulated, and the environment that the disks are in can be considered to be a blackbody at 300 K. Determine the net rate of radiation heat transfer from the disks to the environment.
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:
[tex]Q = FA\sigma\Delta T^4[/tex]
Where,
F =View Factor
A = Cross sectional Area
[tex]\sigma =[/tex] Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m
[tex]L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K[/tex]
The view factor between two coaxial parallel disks would be
[tex]\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33[/tex]
[tex]\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75[/tex]
Then the view factor between base to top surface of the cylinder becomes [tex]F_{12} = 0.26[/tex]. From the summation rule
[tex]F_{13} = 1-0.26[/tex]
[tex]F_{13} = 0.74[/tex]
Then the net rate of radiation heat transfer from the disks to the environment is calculated as
[tex]\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}[/tex]
[tex]\dot{Q_3} = 2\dot{Q_{13}}[/tex]
[tex]\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)[/tex]
[tex]\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)[/tex]
[tex]\dot{Q_3} = 780.76W[/tex]
Therefore the rate heat radiation is 780.76W
The net radiation heat transfer from the disks to the environment is computed by applying the Stefan-Boltzmann law for the radiation heat transfer of black bodies. The area of one side of the disk and the given temperatures are substituted into the law's formula to obtain the desired value.
Explanation:The physical concept relevant to the question is the Stefan-Boltzmann law related to radiation heat transfer. Since both disks are black, they are considered perfect black bodies with an emissivity (e) of 1.
Firstly, we calculate the area (A) of one disk as A = π(D/2)² (because the back sides are insulated on both disks, we only need to consider the radiation from one side of each disk). Then, using Stefan-Boltzmann law formula: Qnet = 2σeA(T₁⁴ - T₂⁴) (the factor 2 is due to having two disks), where T₁ = 450K (temperature of the disks) and T₂ = 300K (temperature of the surrounding environment) is used to find the desired rate of heat transfer (Qnet). The Stefan-Boltzmann constant (σ) is known to be 5.67 × 10⁻⁸ J/s.m².K⁴.
With the value of A calculated from the given diameter and the above values substituted, we can calculate Qnet.
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While skiing in Jackson, Wyoming, your friend Ben (of mass 69.8 kg) started his descent down the bunny run, 16.8 m above the bottom of the run. If he started at rest and converted all of his gravitational potential energy into kinetic energy, what is Ben’s kinetic energy at the bottom of the bunny run? Use g = 9.8 m/s 2 Answer in units of J.
Answer:
11484.18 J
Explanation:
Applying conservation of energy we get
potential energy = kinetic energy
mgh = 0.5mv^2
gh= 0.5v^2
[tex]v= \sqrt{2gh}[/tex]
h= 16.8 m
g= 9.8 m/s^2
[tex]v= \sqrt{2\times16.8\times9.8}[/tex]
v=18.14 m/s at the bottom of the bunny run
therefore, Kinetic energy at the bottom of the bunny run
K= [tex]\frac{1}{2} mv^2=\frac{1}{2}69.8\times18.14^2[/tex]
therefore K= 11484.18 J
= 11.48 KJ
Answer: What is his final velocity?
Answer in units of m/s.
Explanation: i need help
A horizontal tube consists of a 7.0-cm-diameter pipe that narrows to a 2.0-cm-diameter throat. In the pipe, the water pressure is twice atmospheric pressure and the water flows with a speed of 0.40 m/s. What is the pressure in the throat, assuming that the water behaves like an ideal fluid? The density of water is 1000 kg/m3 , and atmospheric pressure is 1.01 × 105 Pa.
Answer:
[tex]1.9\times10^{5} Pa[/tex]
Explanation:
[tex]d_{p}[/tex] = diameter of the pipe = 7 cm
[tex]v_{p}[/tex] = speed of water in the pipe = 0.40 m/s
[tex]A_{p}[/tex] = Area of cross-section of pipe = [tex](0.25)\pi d_{p}^{2}[/tex]
[tex]d_{t}[/tex] = diameter of the throat = 2 cm
[tex]v_{t}[/tex] = speed of water in the throat
[tex]A_{t}[/tex] = Area of cross-section of throat = [tex](0.25)\pi d_{t}^{2}[/tex]
Using equation of continuity
[tex]A_{p} v_{p} = A_{t} v_{t} \\(0.25)\pi d_{p}^{2} v_{p} = (0.25)\pi d_{t}^{2} v_{t} \\(7)^{2} (0.40) = (2)^{2} v_{t}\\v_{t} = 4.9 ms^{-1}[/tex]
[tex]P_{o}[/tex] = atmospheric pressure = 1.01 x 10⁵ Pa
[tex]P_{p}[/tex] = Pressure in the pipe = [tex]2 P_{o}[/tex] = 2.02 x 10⁵ Pa
[tex]P_{t}[/tex] = Pressure in the throat
Using Bernoulli's theorem
[tex]P_{t} + (0.5)\rho v_{t}^{2} = P_{p} + (0.5)\rho v_{p}^{2}\\P_{t} + (0.5)(1000) (4.9)^{2} = 2.02\times10^{5} + (0.5)(1000) (0.40)^{2}\\P_{t} + 12005 = 202080\\P_{t} = 190075 Pa\\P_{t} = 1.9\times10^{5} Pa[/tex]
The pressure in the throat of the pipe, assuming that the water behaves like an ideal fluid, can be obtained using the continuity equation and Bernoulli's equation. The velocity at the throat is first calculated using the continuity equation and then Bernoulli's equation is used to find the pressure. The pressure comes out to be 1.94 x 105 Pa.
Explanation:This problem can be resolved using the continuity equation and Bernoulli's equation, both of which refer to the conservation laws of mass and energy respectively in fluid dynamics. Bernoulli's equation states that, along a streamline, the sum of all forms of fluid energy is constant with respect to any increase in height. The continuity equation states that the mass flow rate must remain constant in the pipe i.e., the velocity of fluid at the wide end of the pipe multiple by its area will be equal to the velocity at the narrow end multiplied by its area.
The velocity at the throat can be found using the continuity equation: (7)^2 x (0.40) = (2)^2 x V2. Solving this gives V2 = 2.45 m/s.
Next, we'll use Bernoulli's equation, P1 + 0.5 x rho x (V1)^2 = P2 + 0.5x rho(V2)^2, where P1 is the pressure at the wide end (which is given as twice of atmospheric pressure), rho is density of water, V1 is velocity at the wide end, P2 is the pressure at the narrow end (which we are to find), and V2 is the velocity at the narrow end.
Substituting the given values, we have (2 x 1.01e5) + 0.5 x 1000 x (0.4)^2 = P2 + 0.5*1000*(2.45)^2. Solving this equation gives P2 = 1.94 x 105 Pa.
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