Answer:
#_photon = 5 10²⁰ photons / s
Explanation:
For this exercise let's calculate the energy of a single quantum of energy, use Planck's law
E = h f
c= λ f
E = h c / λ
λ= 1000 nm (1 m / 109 nm) = 1000 10⁻⁹ m
Let's calculate
E₀ = 6.6310⁻³⁴ 3 10⁸/1000 10⁻⁹
E₀ = 19.89 10⁻²⁰ J
This is the energy emitted by a photon let's use a proportions rule to find the number emitted in P = 100 w
#_photon = P / E₀
#_photon = 100 / 19.89 10⁻²⁰
#_photon = 5 10²⁰ photons / s
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 17 cm away. The force of attraction is then 68nM .
How many electrons were transferred from one ball to the other?
Answer:
2.92 x 10¹² electrons
Explanation:
given,
distance between two plastic ball, r = 17 cm
r = 0.17 m
Force of attraction = F = 68 mN
F = 0.068 N
number of electron transferred from one ball to another.
using Coulomb Force equation
[tex]F = \dfrac{kq^2}{r^2}[/tex]
[tex]0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}[/tex]
q² = 2.1835 x 10⁻¹³
q = 4.67 x 10⁻⁷ C
now, number of electron
[tex]N = \dfrac{q}{e}[/tex]
e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C
[tex]N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}[/tex]
N = 2.92 x 10¹² electrons
electrons were transferred from one ball to the other is 2.92 x 10¹²
Field strength is directly proportional to the square of charge. The charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex]
From Coulomb law:[tex]F = \dfrac {kQ^2}{r^2}[/tex]
Where,
[tex]F[/tex]- field strength = 68 mN = 0.06 N
[tex]Q[/tex]- charge
[tex]r[/tex]- distance = 17 cm =0.17 m
[tex]k[/tex]- constant = 9x10⁹
Put the values in the formula,
[tex]0.068 = \dfrac{9\times 10^9 \rimes Q^2}{0.017^2}\\\\q^2 = 2.1835 \times 10^{-13}\\\\q = 4.67 \times 10^{-7} \rm \ C[/tex]
Therefore, the charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex].
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A classical estimate of the vibrational frequency is ff = 7.0×10137.0×1013 HzHz. The mass of a hydrogen atom differs little from the mass of a proton. If the HIHI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring?
Answer:
The force constant of the spring is 317.8 N/m.
Explanation:
Given that,
Frequency [tex]f=7.0\times10^{13}\ Hz[/tex]
We need to calculate the reduced mass
Using formula of reduced mass
[tex]\mu=\dfrac{m_{H}m_{I}}{m_{H}+m_{I}}[/tex]
Where, [tex]m_{H}[/tex]= atomic mass of H
[tex]m_{I}[/tex]= atomic mass of I
Put the value into the formula
[tex]\mu=\dfrac{1\times126.9}{1+126.9}[/tex]
[tex]\mu=0.99\ u[/tex]
[tex]\mu=0.99\times1.66\times10^{-27}\ Kg[/tex]
[tex]\mu=1.643\times10^{-27}\ kg[/tex]
We need to calculate the force constant of the spring
Using formula of frequency
[tex]f=\dfrac{1}{2\pi}\times\sqrt{\dfrac{k}{\mu}}[/tex]
[tex]k=f^2\times 4\pi^2\times\mu[/tex]
Put the value into the formula
[tex]k=(7.0\times10^{13})^2\times4\pi^2\times1.643\times10^{-27}[/tex]
[tex]k=317.8\ N/m[/tex]
Hence, The force constant of the spring is 317.8 N/m.
Answer:
The force constant of the spring is 317.8 N/m.
Explanation:
hope it helped <3
A 747 jumbo jet of mass 300,000 kg accelerates down the runway at 4 m/s2 . What must be the force generated by each of its four engines
Answer:
300000 N
Explanation:
We are given that
Mass of jumbo jet=300,000kg
Acceleration of jet=[tex]4m/s^2[/tex]
We have to find the force generated by each of its four engines.
We know that
Force,F =ma
Total force exert by four engines=4F
[tex]4F=300000\times 4[/tex]
[tex]F=\frac{300000\times 4}{4}=300000N[/tex]
Hence, the force generated by each of its four engines=300000 N
A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnitude, the balls will be attracted to each other.
a. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.
b. The negative charges on the sphere A move to the side closer to sphere B, while the positive charges move to the side further away from sphere B. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.
c. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
d. The negative charges on the sphere B move to the side closer to sphere A, while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Answer:
D
Explanation:
Ball A is a non positively charged non metal while ball B is metal ball.
Given: The ball B positive charge of small magnitude
To prove: Balls will attract each other
IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Hence the correct answer will be D .
If the speed of sound in air is 340 m/s, what is approximately the length of the shortest air column closed at one end that will respond to a tuning fork of frequency 198 Hz?
Answer:
42.9 cm
Explanation:
given,
speed of sound, v = 340 m/s
Frequency of tuning fork, f = 198 Hz
The length of the shortest air column closed at one end.
this is the case of the standing wave.
L = N λ
N = the number of complete sine waves in the standing wave.
The standing wave in the problem is the first harmonic of a pipe open at one end.
N = 1/4
we know,
v = f λ
340 = 198 x λ
λ = 1.717 m
now,
L = N λ
[tex]L = \dfrac{1}{4}\times 1.717[/tex]
[tex]L = 0.429\ m[/tex]
L = 42.9 cm
minimum length of the required pipe is 42.9 cm.
did you find the answer?
Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 1.0 meters above the ground and is trying to spray Ferdinand,who is standing 10.0 meters away. To increase the range of the water, Isabella places her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the cross-sectional area of the hose hole does she have to cover to be able to spray her friend? Assume that the cross section of the hose opening is circular with a radius of 1.5 centimeters. Express your answer as a percentage to the nearest integer.
Answer:
84%
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]y_0=1\ m[/tex]
x = 10 m
t = Time taken
[tex]v_0[/tex] = 3.5 m/s (assumed, as it is not given)
[tex]A_0=\pi 1.5^2[/tex]
We have the equation
[tex]y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s[/tex]
[tex]x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s[/tex]
From continuity equation we have
[tex]Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2[/tex]
Fraction is given by
[tex]f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%[/tex]
The fraction is 84%
A fraction of the cross-sectional area of the hose hole that Isabella must cover is 84%.
Given the following data:
Vertical distance = 1.0 meters.
Horizontal distance = 10.0 meters.
Radius = 1.5 cm.
Vertical speed = 3.5 m/s.
How to calculate the time.The time taken to reach a maximum height is given by this formula:
[tex]H=\frac{1}{2} gt^2\\\\t=\sqrt{\frac{2H}{g} } \\\\t=\sqrt{\frac{2\times 1.0}{9.8} }[/tex]
t = 0.452 seconds.
For the velocity required, we have:
[tex]x=x_o+Vt\\\\10=0+V0.452\\\\V=\frac{10}{0.452}[/tex]
V = 22.12 m/s.
From continuity equation, we have:
[tex]A=\frac{A_0V_0}{V} \\\\A=\frac{\pi r^2V_0}{V}\\\\A=\frac{\pi1.5^2 \times 3.5}{22.12} \\\\A=\frac{24.7433}{22.12}[/tex]
A = 1.118 cm^2.
Now, an expression for the fraction of the cross-sectional area is given by:
[tex]F=\frac{A_0-A}{A_0} \times 100\\\\F=\frac{7.0695-1.118}{7.0695} \times 100\\\\F=\frac{5.9515}{7.0695} \times 100\\\\F=0.8419 \times 100[/tex]
F = 84.19 ≈ 84%.
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A steam power plant operates on an ideal reheat Rankine cycle between the pressure limits of 15 MPa and 10 kPa. The mass flow rate of steam through the cycle is 12 kg/s. Steam enters both stages of the turbine at 500C.
If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10%, determine
(a) The pressure, in kPa ,at which reheating takes place,
(b) The total rate of heat input, in kW, in the boiler,
(c) The thermal efficiency of the cycle. Also, show the cycle on a T-s diagram with respect to the saturation lines.
This is a complex thermodynamics problem about a steam power plant's ideal reheat Rankine cycle. It requires calculation of pressure at reheat stage, total heat input rate and thermal efficiency by using correct thermodynamic practices. Due to the complexity, expert understanding and use of steam tables is necessary.
Explanation:This is a thermodynamics problem associated with the ideal reheat Rankine cycle in a steam power plant. Due to the complex nature of the problem, the aid of thermodynamics and steam table reference books is required to solve it.
Under these ideal cycle conditions, calculating assumed pressure drops and using the steam tables, we can estimate the reheat pressure during the cycle. The exact figures would be subject to critical examination and calculation based on precise cycle parameters following real-life engineering practices.
For part (b), the total rate of heat input can be calculated using the mass flow rate and specific enthalpies at different stages of the cycle. The thermal efficiency for part (c) can be obtained by dividing the net work output by the total heat input.
An actual T-s diagram for the Rankine cycle would aid in visualizing the process but cannot be drawn here. Basically, the cycle is plotted with entropy(s) on the x-axis and temperature(T) on the y-axis, showing the expansion, reheat, and condensation stages of the steam.
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The Ideal Reheat Rankine Cycle is a model for determining efficiency in steam power plants. The pressure at which reheating takes place is typically assumed to be at the midpoint of the high and low-pressure limits. The total heat rate input in the boiler and thermal efficiency cannot be determined without more detailed specification of the conditions.
Explanation:This question is related to the concepts of thermodynamics in the field of Mechanical Engineering. The Ideal Reheat Rankine Cycle is a model used to predict the performance of steam turbine systems, and we will use principles from this model to answer your questions.
(a) The pressure at which reheating takes place cannot be definitively stated without more specific information like the temperature at the exit of the high-pressure turbine or the entropy at key points in the cycle. However, it’s commonly assumed in practice to occur at the mid-point of the high and low-pressure limits, which would be 7.5 MPa or 7500 kPa.
(b) The total rate of heat input in the boiler cannot be definitively stated without more specific information such as the specific enthalpies at key points in the cycle.
(c) Thermal efficiency of the cycle also requires further specific information such as the enthalpy or temperature at key points in the cycle to compute.
Moreover, drawing a T-s (Temperature vs Entropy) diagram with respect to the saturation lines would need a graphical interface, so it's not possible to illustrate it here.
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Pedas hot water spring is located 15 km from Seremban town. The government has decided to build geothermal power plant using hot water from Pedas hot water spring. A geothermal power plant uses geothermal water at 150°C at a rate 210 kg/s as the heat source and produces 8000kW of net power. The geothermal water leaves the plant at 90°C. If the environment temperature is around 25°C. Determine:
a. The actual rate of heat input to this power plant.
b. The actual thermal efficiency and the maximum possible thermal efficiency of this power plant.
c. The actual rate of heat rejection from this power plant.
d The power output if the geothermal water leaves the plant at 40°C, and maintain it thermal efficiency.
Final answer:
To determine the actual rate of heat input to the geothermal power plant, calculate the heat input using the provided formula. The actual thermal efficiency can be found by dividing the net power output by the heat input. The actual rate of heat rejection can then be calculated as the difference between the heat input and the net power output. To determine the power output if the geothermal water leaves the plant at a different temperature while maintaining its thermal efficiency, use the provided formula with the appropriate values.
Explanation:
The actual rate of heat input to the geothermal power plant can be calculated using the formula:
Heat input = mass flow rate * specific heat capacity * (hot water temperature - environment temperature)
Given that the mass flow rate is 210 kg/s, the specific heat capacity is 4.186 J/g°C, the hot water temperature is 150°C, and the environment temperature is 25°C, we can substitute these values into the formula to find the heat input.
The actual thermal efficiency of the power plant can be calculated using the formula:
Thermal efficiency = (Net power output / Heat input) * 100%
Given that the net power output is 8000 kW and we have already calculated the heat input, we can substitute these values into the formula to find the actual thermal efficiency. The maximum possible thermal efficiency can be calculated as the Carnot efficiency, which is the maximum possible efficiency for a heat engine operating between two temperatures.
The actual rate of heat rejection from the power plant can be calculated as the difference between the heat input and the net power output.
To find the power output if the geothermal water leaves the plant at 40°C and maintain its thermal efficiency, we can use the formula:
Power output = (Heat input - Heat rejection) * (Net power output / Heat input)
Substituting the appropriate values into the formula will give us the desired power output.
A body weighs 1000 lbf when exposed to a standard earth gravity of g=32.174 ft/s^2. a. What is its mass in kg? b. What is its weight in N if on moon (gmoon = 1.62 m/s^2)? c. How fast will it accelerate if under a force of 400 lbf on moon and on the earth in m/s.
Answer
given,
weight = 1000 lbf
g = 32.174 ft/s²
mass =[tex] \dfrac{1000}{32.174}[/tex]
m = 31 slug
1 slug = 14.593 kg
m = 31 x 14.593
m = 452.383 Kg
b) weight on moon
W = m g
W = 452.383 x 1.62
W= 732.86 N
c) we know,
F = m a
400 lbf = 31 slugs x a
a = 12.90 ft/s²
1 ft = 0.304 m
a = 3.92 m/s²
Final answer:
This answer covers calculating mass in kg, weight on the moon, and acceleration with a force on both Earth and the moon.
Explanation:
Mass Calculation:
Given weight = 1000 lbf and gravitational acceleration on Earth, g = [tex]32.174 ft/s^2.[/tex]Convert weight to Newtons: 1000 lbf × 4.448 N/lbf = 4448 N.Calculate mass using weight formula: Weight = mass × acceleration due to gravity.Mass = Weight / acceleration = [tex]4448 N / 9.8 m/s^2[/tex] = 454.7 kg.Weight on Moon:
Weight on the moon = mass × moon gravity acceleration = 454.7 kg × [tex]1.62 m/s^2[/tex] = 736.63 N.Acceleration with Force:
On the Moon: acceleration = force / mass = 400 lbf × 4.448 N/lbf / 454.7 kg = 3.91 m/s.On Earth: acceleration = 400 lbf × 4.448 N/lbf / 1000 kg = 1.79 m/s.Calculate the average density of the earth in g/cm3, assuming our planet to be a perfect sphere.
To develop this problem we will apply the relationship of density, such as the unit of mass per unit volume of a body. For this relationship we will use the known constants of the value of the land's mass and its respective radius. These values will be converted from kilograms to grams and meters to centimeters respectively. We will find the volume through the geometric relationship of the sphere using the radius of the earth.
The mass of the Earth is given as
[tex]m_E = 5.9722*10^{24}kg (\frac{1000g}{1kg}) = 5.9722*10^{27}g[/tex]
The radius of the Earth is
[tex]r_E = 6.3781*10^6m (\frac{100cm}{1m}) = 6.3781*10^8cm[/tex]
Using the geometric value of volume for a sphere, and using the radius of the earth, as the radius of that sphere we have to
[tex]V_e = \frac{4}{3} \pi r^3[/tex]
Replacing,
[tex]V_E = \frac{4}{3} \pi (6.3781*10^8)^3[/tex]
[tex]V_E = 1.09*10^{27}cm^3[/tex]
The expression for the density of the Earth is
[tex]\rho = \frac{m_E}{V_E}[/tex]
Replacing,
[tex]\rho = \frac{5.9722*10^{27}}{1.09*10^{27}}[/tex]
[tex]\rho = 5.48g/cm^3[/tex]
Therefore the average density of the earth is [tex] 5.48g/cm^3[/tex]
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 8.5 m/s, the marine biologist measures the number of clicks occuring per second to be at a frequency of 2770 Hz.
1. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
From Doppler effect we have that the frequency observed for the relation between the velocities is equivalent to the frequency observed. That is mathematically,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
Here,
Speed of sound in water [tex]v = 1522m/s[/tex]
The Dolphin swims directly away from the observer with a velocity [tex]v_s = 8.5m/s[/tex]
Observed frequency of the clicks produced by the dolphin [tex]F_r = 2770Hz[/tex]
Replacing we have,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
[tex]F_s = \frac{v+v_s}{v}[/tex]
[tex]F_s = 2770 (\frac{1522}{1522+8.5})[/tex]
[tex]F_s = 2754.61Hz[/tex]
Therefore the frequency emitted by the dolphin is 2754.61Hz
A 50-kg sprinter accelerates from 0 to 11 m/s in 3.0 s. What is the power output for this rapid start?
Answer:
1008.33 W.
Explanation:
Power: This can be defined as the rate at which energy is consumed or dissipated. The S.I unit of power is Watt (W).
P = E/t.......................... Equation 1
Where P = Energy, E = Energy, t = time.
But from the question,
E = 1/2mΔv²................... Equation 2
where m = mass of the sprinter, Δv = change in velocity of the sprinter = final velocity - initial velocity.
Substitute equation 2 into equation 1
P = 1/2mΔv/t....................... Equation 3
Given: m = 50 kg, Δv = 11-0 = 11 m/s, t = 3.0 s.
Substitute into equation 3
P = 1/2(50)(11)²/3
P = 25(121)/3
P = 1008.33 W.
Thus the power output = 1008.33 W.
Final answer:
The power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s is 1008.33 Watts or approximately 1.35 horsepower.
Explanation:
To calculate the power output of a 50-kg sprinter who accelerates from 0 to 11 m/s in 3.0 s, we use the work-energy principle and the definition of power. The work done on the sprinter is equal to the change in kinetic energy, and power is the work done per unit time.
The kinetic energy (KE) at the start is 0 since the sprinter starts from rest. The kinetic energy at the end is KE = 1/2 x mass x velocity² = 1/2 x 50 kg x (11 m/s)². After calculating the kinetic energy, we get KE = 1/2 x 50 x 121 = 3025 Joules.
Now, we find the power by dividing the work by the time interval: Power = Work / Time = 3025 J / 3.0 s = 1008.33 Watts.
To convert watts to horsepower, use the conversion 1 horsepower = 746 watts. Therefore, Power in horsepower = 1008.33 W / 746 W/hp = approximately 1.35 hp.
Which of the following statements are true? Check all that apply. Check all that apply. A satellite's motion is independent of its mass. A geosynchronous satellite has a period of approximately 28 days. A satellite's velocity and orbital radius are independent of each other. The launch speed of a satellite determines the shape of its orbit around Earth.
Answer:
The following are true:
A satellite's motion is independent of its mass.
The launch speed of a satellite determines the shape of its orbit around Earth.
Explanation:
A satellite's motion is independent of its mass.
The speed of a satellite can be determined by means of the Universal law of gravity:
[tex]F = G\frac{M\cdot m}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are masses of the two objects and r is the distance between them.
In equation 1, Newton's second law can be replaced:
[tex]F = m\cdot a[/tex]
[tex]m. a = G\frac{M\cdot m}{r^{2}}[/tex] (2)
Since it is a circular motion, the centripetal acceleration can be used:
[tex]a = \frac{v^{2}}{r}[/tex] (3)
Then, equation 3 can be replaced in equation 2:
[tex]m\frac{v^{2}}{r} = G\frac{M\cdot m}{r^{2}}[/tex] (4)
In this case, m is the satellite's mass and M is the Earth mass. Therefore, v can be isolated from equation 4:
[tex]v^{2} = G\frac{M\cdot m r}{mr^{2}}[/tex]
[tex]v^{2} =\frac{G M}{r}[/tex]
[tex]v = \sqrt{\frac{G M}{r}}[/tex]
Where G is the gravitational constant, M is the mass of the Earth and r is the orbital radius of the satellite.
Notice, how the speed of the satellite does not depend in its mass, only in the Earth mass and its orbital radius.
The launch speed of a satellite determines the shape of its orbit around Earth.
Depending on the launch speed the satellite can reach a lower or higher height above the surface of the Earth and, therefore, its orbital radius will be bigger or smaller according with that height.
Remember that the orbital radius for the satellite will be the sum of the Earth radius and the height above the surface.
Key term:
Geosynchronous satellite: It is a satellite with the same orbital period as Earth rotation period (24 hours).
Tectonic plates are large segments of the Earth's crust that move slowly. Suppose one such plate has an average speed of 2.0 cm per year.
What distance does it move in 80 seconds at this speed?
Answer:
[tex]d=2.5367\times 10^{-6}\ cm[/tex]
Explanation:
Given:
speed of tectonic plate, [tex]v=2\ cm.yr^{-1}=\frac{2}{365\times 24\times 60 \times 60}\ cm.s^{-1}[/tex]time of recording the plate movement, [tex]t=80\ s[/tex]Now the distance moved by the tectonic plate in the stipulated times:
[tex]d=v.t[/tex]
[tex]d=\frac{1}{365\times 12\times 3600} \times 80[/tex]
[tex]d=2.5367\times 10^{-6}\ cm[/tex]
Tectonic plates are the mass rock from of the earth's uppermost mantle and the crust which show similar behavior of movement with respect to the other such groups of lithosphere.
Calculate the linear momentum of photons of wavelength 350 nm. What speed does a hydrogen molecule need to travel to have the same linear momentum?
Answer:
a)[tex]p=1.89x10^{-27}kg.m.s^{-1}[/tex]
b)[tex]v=0.565\frac{m}{s}[/tex]
Explanation:
First, we need to obtain the linear momentum of the photons of wavelength 350nm.
We are going to use the following formula:
[tex]\lambda=\frac{h}{p}\\Where:\\\lambda=wavelength\\h=placnk's\_constant\\p=Linear\_momentum[/tex]
So the linear momentum is given by:
[tex]p=\frac{h}{\lambda}\\\\p=\frac{6.626x10^{-34}J.s}{350x10^{-9}m}\\\\p=1.89x10^{-27}kg.m.s^{-1}[/tex]
Having the linear momentum of the photon, we can calculate the speed of the hydrogen molecule to have the same momentum, we can use the classic formula for that:
[tex]p=m.v[/tex]
[tex]where:\\m=mass\\v=speed\\p=linear\_momentum[/tex]
The mass of the hydrogen molecule is given by:
[tex]m=2*(1.0078x10^{-3}\frac{kg}{mol}})x\frac{1}{6.022x10^{23}mol}[/tex]
[tex]3.35x10^{-27}kg[/tex]
What we've done here is to use the molecular weight of the hydrogen, and covert it kilograms, we had to multiply by two because the hydrogen molecule is found in pairs.
so:
[tex]v=\frac{p}{m}\\\\v=\frac{1.89x10^{-27}kg.m.s^{-1}}{3.35x10^{-27}kg}\\\\v=0.565\frac{m}{s}[/tex]
(a) The linear momentum of the photon is 1.89 x 10⁻²⁷ kgm/s.
(b) The speed of hydrogen atom with same momentum is 2,079.35 m/s.
Linear Momentum of the photon
The linear momentum of the photon is calculated as follows;
P = h/λ
P = (6.63 x 10⁻³⁴) / (350 x 10⁻⁹)
P = 1.89 x 10⁻²⁷ kgm/s
Speed of hydrogen atom with same momentump = mv
v = p/m
v = (1.89 x 10⁻²⁷ ) / (9.11 x 10⁻³¹)
v = 2,079.35 m/s
Thus, the speed of hydrogen atom with same momentum is 2,079.35 m/s.
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A 0.l ‑kilogram block is attached to an initially unstretched spring of force constant k = 40 N/m as shown right. The block is displaced 0.1 m from the equilibrium point and released from rest at time t = 0.
What is the maximum potential energy of the oscillating system?
40 J
0.1 J
0.1 J
0.2 J
Answer:
The maximum potential energy of the system is 0.2 J
Explanation:
Hi there!
When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.
The equation of elastic potential energy (EPE) is the following:
EPE = 1/2 · k · x²
Where:
k = spring constant.
x = stretching distance.
The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.
Then:
EPE = 1/2 · 40 N/m · (0.1 m)²
EPE = 0.2 J
The maximum potential energy of the system is 0.2 J
The maximum potential energy of the spring system, given a spring constant of 40 N/m and a displacement of 0.1 m, is calculated using the formula PEmax = (1/2)kx2, and the result is 0.2 J.
Explanation:When dealing with a spring system in physics, we use Hooke's Law, which states the force exerted by a spring is directly proportional to the amount it is stretched or compressed and is given by F = -kx, where k is the spring constant and x is the displacement from equilibrium. The potential energy stored in such a system can be calculated using the formula PEmax = (1/2)kx2.
In this case, given a spring constant k = 40 N/m and a maximum displacement x = 0.1 m, the potential energy at maximum compression or stretch is:
PEmax = (1/2)kx2 = (1/2)(40 N/m)(0.1 m)2 = (1/2)(40)(0.01) = 0.2 J.
Therefore, the maximum potential energy of the oscillating system is 0.2 J.
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?
To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as
[tex]F = \frac{kq_1q_2}{d^2}[/tex]
Here
k = Coulomb's Constant
[tex]q_{1,2} =[/tex] Charge of each object
d = Distance
Our values are given as,
[tex]q_1 = 1 \mu C[/tex]
[tex]q_2 = 6 \mu C[/tex]
d = 1 m
[tex]k = 9*10^9 Nm^2/C^2[/tex]
a) The electric force on charge [tex]q_2[/tex] is
[tex]F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}[/tex]
[tex]F_{12} = 54 mN[/tex]
Force is positive i.e. repulsive
b) As the force exerted on [tex]q_2[/tex] will be equal to that act on [tex]q_1[/tex],
[tex]F_{21} = F_{12}[/tex]
[tex]F_{21} = 54 mN[/tex]
Force is positive i.e. repulsive
c) If [tex]q_2 = -6 \mu C[/tex], a negative sign will be introduced into the expression above i.e.
[tex]F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}[/tex]
[tex]F_{12} = F_{21} = -54 mN[/tex]
Force is negative i.e. attractive
You throw a glob of putty straight up toward the ceiling, which is 3.00m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 8.40m/s . What is the speed of the putty just before it strikes the ceiling? How much time from when it leaves your hand does it take the putty to reach the ceiling?
The answer provides the speed of the putty just before striking the ceiling and the time it takes for the putty to reach the ceiling. The speed is found to be 5.79 m/s and the time taken is calculated to be 1.5 seconds.
Explanation:The speed of the putty just before it strikes the ceiling:
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is acceleration, and s is displacement, the speed is found to be 5.79 m/s.
The time it takes for the putty to reach the ceiling:
By using the equation v = u + at, where t is time, a is acceleration, u is initial velocity, and v is final velocity, the time taken is calculated to be 1.5 seconds.
A rotating wheel requires 10.0 s to rotate 39.0 revolutions. Its angular velocity at the end of the 10.0-s interval is 92.0 rad/s. What is the constant angular acceleration (in rad/s) of the wheel
Answer:
[tex]13.5 rad/s^2[/tex]
Explanation:
39 revolutions = 39 * 2π = 245 rad
Let [tex]\omega_0[/tex] be the initial angular velocity of the wheel before the 10 s interval. We have the following equation of motion for angular velocity
[tex]\omega = \omega_0 + \alpha t[/tex]
where [tex]\omega = 92 rad/s[/tex] is the final angular velocity of the rotating wheel. [tex]\alpha[/tex] is the constant angular acceleration of the wheel, which we are looking for, and t = 10s is the duration time that the wheel rotates
[tex]92 = \omega_0 + 10\alpha[/tex]
[tex]\omega_0 = 92 - 10\alpha[/tex]
We also have the following equation of motion for angles
[tex]\theta = \omega_0t + \alpha t^2/2[/tex]
where [tex]\theta = 245 rad[/tex] is the total angles rotated
we can also substitute [tex]\omega_0 = 92 - 10\alpha[/tex]
[tex]245 = 10(92 - 10\alpha) + \alpha 10^2/2[/tex]
[tex]245 = 920 - 100\alpha + 50\alpha[/tex]
[tex]50\alpha = 675 [/tex]
[tex]\alpha = 675 / 50 = 13.5 rad/s^2[/tex]
Find the magnitude of the electric force on a 2.0 uC charge in a 100n/C electric field.
Answer:
Electric force will be equal to [tex]2\times 10^{-4}N[/tex]
Explanation:
We have given charge [tex]q=2\mu C=2\times 10^{-6}C[/tex]
Electric field will be [tex]E=100N/C[/tex]
We have to find the magnitude of force
Electric force is the multiplication of electric field and charge
So electric force [tex]F=qE=2\times 10^{-6}\times 100=2\times 10^{-4}N[/tex]
Electric force will be equal to [tex]2\times 10^{-4}N[/tex]
Jessie creates a model of two waves that have the same amplitude as shown.
Which description BEST compares the two waves?
Wave B has less energy because it has a lower frequency and a longer wavelength.
Wave A has less energy because it has a higher frequency and a shorter wavelength.
Wave B has greater energy because it has a higher frequency and a shorter wavelength.
Wave A has greater energy because it has a lower frequency and a longer wavelength.
Answer:
b
Explanation:
Answer:
B
Explanation:
egde test 2023
Gauss's law is usualy written as :
∫Φ_E = E * dA= qencl / ε0
what does qencl refers in Guass's law?
O the net charge inside the closed surface
O the charge residing on insulators inside the closed surface
O all the charge in the physical system
O any charge inside the closed surface that is arranged symmetrically
Answer:
(a) the net charge inside the closed surface.
Explanation:
In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.
A butterfly flies from the top of a tree in the center of a garden to rest on top of a red flower at the garden's edge. The tree is 9.0 m taller than the flower, and the garden is 17 m wide.
Determine the magnitude of the butterfly's displacement.
Express your answer using two significant figures.
Answer:
R=12 m
Explanation:
I have drawn a vector form which I attached here.Please first go through that
Given Data
y (Tree Taller) = 9.0 m
x(Distance from center to edge is)= 8.5 m
To find
R(magnitude of the butterfly's displacement
)
Solution
[tex]R=\sqrt{x^{2} +y^{2} }\\R=\sqrt{9^{2}+8.5^{2} }\\ R=12.379m[/tex]
Convert to two significant figures
R=12 m
The butterfly's displacement, when it moves from the tree to the flower, is approximately 19m. This is determined using the Pythagorean theorem, which is applied to the '9m' height difference between the tree and flower, and '17m' width of the garden.
Explanation:To calculate the magnitude of the butterfly's displacement, we need to use the Pythagorean theorem, which establishes a relationship in any right triangle. The vertical height of the tree is 9m, which forms one side of the right triangle. The horizontal distance across the garden (or width of the garden) is 17m, forming the other side of the right triangle.
Applying the Pythagorean theorem (a^2 + b^2 = c^2), where 'a' and 'b' are the sides and 'c' is the hypotenuse (in this case, the butterfly's displacement), we get:
Displacement = √[(9m)^2 + (17m)^2]
So, the displacement of the butterfly is approximately 19 m, when rounded to two significant figures.
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The temperature of a system rises by 45 °C during a heating process. Express this rise in temperature in Kelvin and Fahrenheit.
Answer:
318K
113°F
Explanation:
45°C = 45 + 273K = 318K
45°C = (45 × 9/5) + 32 = 81 + 32 = 113°F
You crouch from a standing position, lowering your center of mass 18 cm in the process. Then you jump vertically. The average force exerted on you by the floor while you jump is three times your weight. What is your upward speed as you pass through your standing position in leaving the floor?
To solve this problem we will apply the concepts related to balance. Since the force applied is 3 times the weight, and the weight is defined as the multiplication between mass and gravity, we will have that the dynamic equilibrium ratio would be given by the relation,
[tex]\sum F = ma[/tex]
[tex]3mg-mg = ma[/tex]
Rearranging to find a,
[tex]a = 2g[/tex]
Using the linear motion kinematic equations, which express that the final velocity of the body, and in the absence of initial velocity, is equivalent to the product between 2 times the acceleration by the distance traveled, that is
[tex]v^2 = 2as[/tex]
[tex]v^2 = 2(2g)(0.18)[/tex]
[tex]v^2 = 2(2*9.8)(0.18)[/tex]
[tex]v = 2.66m/s[/tex]
Therefore the upward speed is 2.66m/s
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity (in m/s) of the blast wave? (Enter the magnitude.) x m/s (b) Compare this with the speed of sound, which is 343 m/s at sea level. Volast wave = sound
Answer:
156.67 m/s
0.45676 times the speed of sound
Explanation:
Distance from the ground = 23.5 km = 23500 m
Time taken by the blast waves to reach the ground = [tex]2\ minutes\ 30\ seconds=2\times 60+30=150\ s[/tex]
Spedd of the wave would be
[tex]Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s[/tex]
The velocity of the blast wave is 156.67 m/s
v = Velocity of sound = 343 m/s
[tex]\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v[/tex]
The blast wave is 0.45676 times the speed of sound
The average velocity of the blast wave can be calculated using the distance traveled and time taken, resulting in 156.7 m/s. This velocity is less than half the speed of sound at sea level.
The average velocity of the blast wave can be calculated by dividing the distance traveled by the time taken. The distance is the altitude where the explosion occurred, 23.5 km. Converting 2 minutes 30 seconds to seconds gives a time of 150 seconds.
So, velocity = distance / time = 23.5 km / 150 s = 0.1567 km/s = 156.7 m/s.Comparing this to the speed of sound at sea level, which is 343 m/s, the blast wave's velocity of 156.7 m/s is less than half the speed of sound.
a raft of dimensions 2m wide by 3m long by .5m thick is floating in a pond if one fourth of the raf is submerged find the density of the raft material
Answer:
250 kg/m3
Explanation:
The total volume of the raft is length times width times height
V = lwh = 2 * 3 * 0.5 = 3 m cubed
The volume of the raft that is submerged in water is 1/4 of total volume
3 /4 = 0.75 m cubed
Let water density = 1000 kg/m cubed and g = 10 m/s2
The buoyant force is equal to the weight of water displaced by the raft
F = 0.75 * 1000 * 10 = 7500 N
This force is balanced by the raft weight, so the weight of the raft is also 7500N
Mass of raft is 7500 / g = 7500 / 10 = 750 kg
Raft density is mass divided by volume = 750 / 3 = 250 kg/m3
An electrically neutral balloon is rubbed on cloth and becomes positively charged. What can be said about its mass:_______
Answer:
The charged balloon’s mass will be almost equal to the original mass
Explanation:
When an electrically neutral balloon is rubbed on cloth, it losses some electron and becomes positively charged.
The balloon becomes positively charged by the removal of electrons (electrons from the cloth repels electron in the balloon).
The mass of positively charged balloon, will be equal to the original mass minus the mass of electrons removed.
Since Electrons have negligible mass, the removal of electrons from the neutral balloon will have little effect on the original mass of the balloon.
Therefore, the charged balloon’s mass will be almost equal to the original mass
What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction
Answer:
The magnitude of the magnetic field vector is 1.91T and is directed towards the east.
The steps to the solution can be found in the attachment below.
Explanation:
For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.
Fm = Fg
qvBSin(theta) = mg
Where q = magnitude of charge
v = magnitude of the velocity vector = 4 x10^4 m/s
B = magnitude of the magnetic field vector
theta = the angle between the magnetic field and velocity vectors = 90°
m = mass of the charge = 0.195g
g = acceleration due to gravity =9.8m/s²
On substituting the respective values of all variables in the equation (1) above
B = 1.91T
The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.
Below is the step by step procedure to the solution.
Final answer:
The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. We need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle. By setting up the equation qvBsinθ ≥ mg and solving for B, we can find the minimum magnetic field strength required.
Explanation:
The magnitude of the minimum magnetic field that will keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction can be calculated using the equation for magnetic force. The force on a moving charged particle in a magnetic field is given by the equation F = qvBsinθ, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. In this case, we can assume that the angle between the velocity and the magnetic field is 90 degrees, as the particle is moving horizontally northward. Therefore, to find the minimum magnetic field that keeps the particle moving in this direction, we need to find the minimum value of B that produces a force greater than or equal to the gravitational force acting on the particle.
Let's take the example given in Example 22.1, where a glass rod with a positive charge of 20 nC is thrown with a horizontal velocity of 10 m/s due west in a place where the Earth's magnetic field is due north parallel to the ground. The force on the rod due to the Earth's magnetic field can be calculated using the equation F = qvBsinθ, where q is the charge of the rod (20 nC), v is the velocity of the rod (10 m/s), B is the magnetic field strength (unknown), and θ is the angle between the velocity and the magnetic field (90 degrees). The force on the rod should be equal to or greater than the gravitational force acting on it to keep it moving in the same horizontal, northward direction. Therefore, we can set up the equation F ≥ mg, where F is the force on the rod, m is the mass of the rod, and g is the acceleration due to gravity. Substituting the values, we get qvBsinθ ≥ mg. Solving for B, we get B ≥ mg / qvsinθ. Plugging in the given values, we can find the minimum magnetic field strength required to keep the particle moving in the Earth's gravitational field in the same horizontal, northward direction.
A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
Answer: 4.8 W/m^2
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
Answer: 2.1 W/m^2
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
Answer: 13.8 W/m^2
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
d= ___.
A) Intensity: [tex]4.8 W/m^2[/tex]
B) Intensity: [tex]2.1 W/m^2[/tex]
C) Total intensity: [tex]13.8 W/m^2[/tex]
D) Distance: [tex]1.38\cdot 10^5 m[/tex]
Explanation:
A)
The relationship between intensity of a sound and power emitted is given by
[tex]I=\frac{P}{A}[/tex]
where
I is the intensity
P is the power
A is the surface area at the distance considered
For the speaker in this problem:
P = 60 W
The distance is r = 1.0 m, so the area to consider is the surface of a sphere with this radius:
[tex]A=4\pi r^2 = 4\pi(1.0)^2=12.6 m^2[/tex]
Therefore, the intensity is
[tex]I=\frac{60}{12.6}=4.8 W/m^2[/tex]
B)
For this problem, we can use the same equation we used before:
[tex]I=\frac{P}{A}[/tex]
where in this case, we have:
P = 60 W is the power of the speaker
r = 1.5 m is the distance considered
Therefore, the area of the spherical surface is
[tex]A=4\pi r^2 = 4\pi(1.5)^2=28.3 m^2[/tex]
And so, the intensity here is
[tex]I=\frac{60}{28.3}=2.1 W/m^2[/tex]
C)
In this problem, we have to consider 4 speakers. In order to find the total intensity, we have to add the intensity of each speakers.
For the two front speakers, we have
[tex]P=60 W[/tex]
r = 1.0 m
So we have already calculated their intensity in part A), and it is
[tex]I_1 = I_2 = 4.8 W/m^2[/tex]
For the two rear speakers, we have
[tex]P=60 W[/tex]
r = 1.5 m
So their intensity has been calculated in part B),
[tex]I_3 = I_4 = 2.1 W/m^2[/tex]
Therefore, the total intensity is
[tex]I=I_1 + I_2 + I_3 + I_4 = 4.8+4.8+2.1+2.1=13.8 W/m^2[/tex]
D)
In this case, we want to find the distance r at which the intensity is
[tex]I=10^{-12} W/m^2[/tex]
Using four speakers with power
P = 0.06 W
Since we have 4 speakers, the intensity due to each speaker at the location considered must be
[tex]I'=\frac{I}{4}=\frac{10^{-12}}{4}=2.5\cdot 10^{-13} W/m^2[/tex]
Using the usual equation, we find the area:
[tex]A=\frac{P}{I'}=\frac{0.06}{2.5\cdot 10^{-13}}=2.4\cdot 10^{11} m^2[/tex]
And so, we can find what is the corresponding distance:
[tex]A=4\pi r^2\\r=\sqrt{\frac{A}{4\pi}}=\sqrt{\frac{2.4\cdot 10^{11}}{4\pi}}=1.38\cdot 10^5 m[/tex]
So, the person must be at 138 km.
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The intensity of sound waves can be computed by using the equation I = P / (4πr²). For the minimum discernible intensity, we employ a rearranged version of the same formula to calculate the distance from the car stereo at the threshold of hearing.
Explanation:The question is about the calculation of sound intensity produced by a car stereo featuring four speakers, each rated at 60 Watts, at various distances. To calculate the intensity (I) of the sound waves at different distances, we use the equation I = P / (4πr²) where P represents power and r is the distance.
For part D, to find the distance (d) at which sound from the stereo could still be discerned, we rearrange the intensity equation to solve for distance, d = √(P / (4πI)). We substitute P with the sound produced (0.06 W) and I with the minimum discernible intensity (10-12 W/m²). Disclaimer: Since external factors like absorption or diffusion are not taken into account, and the car stereo is not treated as a perfect point source emitter, these values serve as approximation.
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