The number of photons per square meter per second falling on the earth's surface, given the solar radiation rate and average wavelength is calculated to be 5.36 x 10²¹ photons/m2·s.
Explanation:The energy of the average visible photon is calculated using the formula E= hc/λ, where h is Planck's constant, c is the speed of light and λ is the wavelength. With the given values, we get the energy E by substituting h=6.62607 × 10⁻³⁴ J·s, c=3 × 108 m/s, and λ=560 × 10⁻⁹ m resulting in E= 3.546 × 10⁻¹⁹ J. The number of photons per meter square per second can be calculated by dividing the rate of solar radiation, 1900 W/m2 or 1900 J/s·m2 by the energy per photon, yielding 5.36 x 10²¹ photons/m2·s.
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A 10 µf capacitor is charged to 108 v and is then connected across a 328 ω resistor. what is the initial charge on the capacitor? answer in units of µc.
If a spaceship of proper length 40 m is measured to have a length 30 m, how fast is it moving?
When violet light of wavelength 415 nm falls on a single slit, it creates a central diffraction peak that is 9.90 cm wide on a screen that is 2.53 m away. how wide is the slit?
Final answer:
The width of the slit when violet light of 415 nm wavelength creates a central diffraction peak of 9.90 cm on a screen 2.53 m away is approximately 10.6 μm.
Explanation:
The wavelength of violet light is given as 415 nm, and it produces a diffraction pattern with a central peak width of 9.90 cm on a screen 2.53 meters away. We can find the width of the slit using the formula for single-slit diffraction:
Δy = λL/a
where Δy is the width of the central peak, λ is the wavelength, L is the distance to the screen, and a is the width of the slit. Rearranging the formula to solve for a, we have:
a = λL/Δy
Substituting the provided values:
a = (415 x 10^-9 m)(2.53 m) / (9.90 x 10^-2 m)
After calculating, we find that the width of the slit (a) is approximately:
a ≈ 1.06 x 10^-5 m or 10.6 μm
Using the single-slit diffraction formula and given values, we calculate the width of the slit to be 21.1 μm.
To solve this problem, we'll use the formula for the width of the central peak in a single-slit diffraction pattern:
w = 2Lλ / aWhere:
w = width of the central peak (9.90 cm = 0.099 m)L = distance from the slit to the screen (2.53 m)λ = wavelength of violet light (415 nm = 415 x 10⁻⁹ m)a = width of the slit (what we need to find)Rearranging the formula to solve for a:
a = 2Lλ / wSubstituting the given values:
a = (2 * 2.53 m * 415 x 10⁻⁹ m) / 0.099 mNow, calculate:
a = 2.11 x 10⁻⁵ m or 21.1 μmTherefore, the width of the slit is 21.1 μm.
What is the frequency of a photon that has the same momentum as a neutron moving with a speed of 1.90 × 103 m/s?
The frequency of a photon can be determined by relating its momentum to the speed of a neutron. The resulting frequency is 4.8 × 10¹4 Hz.
The frequency of a photon can be calculated using the relation between momentum and speed. Given that the speed of a neutron is 1.90 × 103 m/s, the frequency of the photon with the same momentum would be 4.8 × 10¹4 Hz.
A diver 40 m deep in 10∘C fresh water exhales a 1.0-cm-diameter bubble.
What is the bubble's diameter just as it reaches the surface of the lake, where the water temperature is 20∘C? Assume that the air bubble is always in thermal equilibrium with the surrounding water.
Express your answer to two significant figures and include the appropriate units.
The diameter of the air bubble when it reaches the surface is about 1.7 cm
[tex]\texttt{ }[/tex]
Further explanationThe basic formula of pressure that needs to be recalled is:
Pressure = Force / Cross-sectional Area
or symbolized:
[tex]\large {\boxed {P = F \div A} }[/tex]
P = Pressure (Pa)
F = Force (N)
A = Cross-sectional Area (m²)
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
In this problem , we will use Ideal Gas Law as follows:
Given:
initial diameter of bubble = d₁ = 1.0 cm
initial depth of the diver = h = 40 m
initial temperature = T₁ = 10 + 273 = 283 K
atmospheric pressure = Po = 1.0 atm = 10⁵ Pa
final temperature = T₂ = 20 + 273 = 293 K
density of water = ρ = 1000 kg/m³
Unknown:
final diameter of bubble = d₂ = ?
Solution:
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
[tex]\frac{P_1 (\frac{1}{6} \pi (d_1)^3)}{T_1} = \frac{P_2(\frac{1}{6} \pi (d_2)^3)}{T_2}[/tex]
[tex]\frac{P_1 (d_1)^3}{T_1} = \frac{P_2 (d_2)^3}{T_2}[/tex]
[tex]\frac{(P_o + \rho g h ) (d_1)^3}{T_1} = \frac{P_o (d_2)^3}{T_2}[/tex]
[tex]\frac{(10^5 + 1000(9.8)(40) ) (1.0)^3}{283} = \frac{10^5(d_2)^3}{293}[/tex]
[tex]\frac{(492000 ) (1.0)^3}{283} = \frac{10^5(d_2)^3}{293}[/tex]
[tex]d_2 \approx 1.7 \texttt{ cm}[/tex]
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Pressure
By applying the ideal gas law, we find the diameter at the surface to be approximately 1.7 cm.
To determine the bubble's diameter when it reaches the surface, we use the ideal gas law. Since temperature and volume are directly proportional under constant pressure, we can express this relationship as:
(V₁/T₁) = (V₂/T₂)
where,
V₁ = initial volume of the bubble T₁ = initial temperature of the water (10°C = 283 K)V₂ = final volume of the bubbleT₂ = final temperature of the water (20°C = 293 K)Given that V₁ is π(0.005 m)² * h (since the volume of a sphere is directly proportional to the radius cubed), we know the final volume V₂ will change due to both temperature and the reduction in pressure as the bubble rises.
Hydrostatic pressure P₁ at 40 meters depth is given by:
P₁ = P₀ + ρgh
P₀ = atmospheric pressure = 1 atmρ = density of water ≈ 1000 kg/m³g = 9.8 m/s²h = 40 mP₁ ≈ 1 + (1000 * 9.8 * 40) / 101325 ≈ 4.93 atm
At the surface, the only pressure is P₀ (1 atm).
Using the combined gas law P₁V₁/T₁ = P₂V₂/T₂:
4.93V₁/283 = 1V₂/293
Solving for V₂ we get:
V₂ = 4.93 * V₁ * 293 / 283
V₂ ≈ 5.09 * V₁
Since the volume ratio (D₂/D₁)³ = 5.09, taking the cube root:
D₂ ≈ 1.71 * D₁
So, if initial diameter D₁ is 1.0 cm:
D₂ = 1.71 * 1.0 cm ≈ 1.7 cm
Therefore, the bubble's diameter at the surface is approximately 1.7 cm.
help asap please!! An aluminum block of mass 12.00 kg is heated from 20 C to 118 C. If the specific heat of aluminum is 913 J-1 kg K-1 then how much energy is required?
A. 10.956 kJ
B. 1073.688 kJ
C. 64.640 kJ
D. 7456.167 kJ
Find the kinetic energy of an electron whose de broglie wavelength is 34.0 nm.
Kinetic energy is a property of a moving item that is affected by both mass and velocity. The electron's kinetic energy with a de-Broglie wavelength of 34nm is 2.06 × [tex]10^-^2^2[/tex]J.
First, we have to convert de Broglie wavelength in m to satisfy the dimensions.
λ = 34.0nm =34×[tex]10^-^9[/tex].
To find kinetic energy we need to find velocity and for that, we need to find momentum by using the formulae:
P= h/λ = 6.6×[tex]10^-^3^4[/tex]/34×[tex]10^-^9[/tex] = 1.94×[tex]10^-^2^6[/tex] kgm/s.
After getting momentum we need to find the velocity
V=p/m = 1.94×[tex]10^-^2^6[/tex] / 9.1×[tex]10^-^3^1[/tex]= 2.13×[tex]10^4[/tex] m/s.
Now we have the value of velocity and by applying it [tex]k=1/2mv^2[/tex], we can find easily find the kinetic energy
[tex]k=1/2mv^2[/tex] = 1/2 (9.1×[tex]10^-^3^1[/tex] kg)×(2.13×[tex]10^4[/tex] m/s[tex])^2[/tex] = 2.06×[tex]10^-^2^2[/tex]J.
Therefore, the kinetic energy of the particle is 2.06×[tex]10^-^2^2[/tex]J.
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Make a rule: how would you find the resistance of a parallel circuit with n identical resistors?
To find the resistance of a parallel circuit with n identical resistors, divide the resistance of one resistor by n.
Explanation:The rule to find the resistance of a parallel circuit with n identical resistors is to divide the value of one resistor by the number of resistors (n). So, the formula is Requiv = R/n, where R is the resistance of one resistor and n is the number of resistors in parallel.
When you have n identical resistors in parallel, you can find the equivalent resistance (Requiv) using the formula:
Requiv = R / n
Where:
Requiv is the equivalent resistance of the parallel combination.
R is the resistance of one individual resistor.
n is the number of identical resistors in parallel.
This formula simplifies the calculation and is useful when you want to determine the overall resistance in a parallel circuit, which is a common scenario in electrical circuits and electronics.
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a brick is suspended above the ground at a height of 6.6 m. it has a mass of 5.3 kg. what is the potential energy of the brick
Conventionally, the field strength around a charged object is the direction of the force acting on a .
Answer:
Field Strength:Any charged body has the capacity to effect any test charge that comes inside its field or region. It can be also defined as,
"The total amount or magnitude of force,F or intensity felt by any unit test charge when it enters an electromagnetic or electric field."
For an electric field its unit will be, volt per meter or simply V/m.Explanation:
A unit test charge inside an electric field:
When a unit test charge enters a given parameters or area set by the charged particle then it will surely experience a force,F equal to the magnitude of that charge body and it will be different as it continues to move closer or far inside the field.
As, we have, "E=F/q",(where "F" is the field strength and "q" is the unit test charge placed inside the field).Which choice shows the size range for nanotechnology?
Answer:
C
Explanation:
10–9 m to 10–7 m
The size range for nanotechnology is 10⁻⁹ m to 10⁻⁷ m.
What is nanotechnology?Nanotechnology improves and revolutionize, The new technology used in industry sectors like information technology, homeland security, medicine, transportation, energy, food safety, and environmental science.
A material with its dimensions less than 100 nanometers or ranging between 1 to 100nm are known as nanomaterials.
Thus, the correct choice is C.
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Benjamina started her walk from the front door of her ground floor apartment. She walked 6 meters to the corner of the building and then turned the corner and walked 10 meters to her friend’s apartment. Identify the difference between the distance she walked and her displacement.
Answer:
The change in her position (from beginning to end), along with the direction, is considered her displacement. This value is a measurement and a direction. The total distance along the path from her starting point to her end point is considered her distance. This value is a measurement only.
Explanation:
A monochromatic laser is exciting hydrogen atoms from the n=2 state to the n=5 state. part a what is the wavelength λ of the laser?
What is the maximum power consumption of a 3.0-v portable cd player that draws a maximum of 330 ma of current?
Final answer:
The maximum power consumption of a 3.0-v portable CD player drawing 330 mA of current is calculated using the formula P = IV, giving a result of 0.99 W, which is not directly listed in the provided options.
Explanation:
The question asks for the maximum power consumption of a 3.0-volt portable CD player that draws a maximum of 330 milliamps of current. To find the power, we use the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts.
In this case, I = 330 mA = 0.33 A (since 1A = 1000mA), and V = 3.0 V. Substituting these values into the formula gives:
P = 0.33 A × 3.0 V = 0.99 W.
The absolute (or total) pressure at the bottom of a cylindrical container with a cross-sectional area of 47.0 cm2 and holding a fluid of density 560 kg/m3 is 115 kpa. (a) determine the depth of the fluid.
The fluid's depth in the cylinder can be determined using the equation for fluid pressure, h = P / (pg), and plugging in given values, resulting in an approximated depth of 20.9 meters.
Explanation:The total absolute pressure at a point in a fluid is the sum of the atmospheric pressure and the pressure due to the fluid above the point of reference. The latter is given by the equation P = pgh, where p is the density of the fluid, g is the acceleration due to gravity and h is the height (or depth) of the fluid column above the point of reference. In this scenario, we know P (absolute pressure, 115 kPa), p (density, 560 kg/m3), and g (standard gravity, roughly 9.81 m/s2), and we want to find h.
To find h, we can rearrange our equation: h = P / (pxg) . Plugging the given values we have h=115000Pa/(560kg/m3*9.8m/s2), which gives h = 20.9 m.
This means the depth of the fluid in the cylinder is approximately 20.9 meters.
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If 3.0 × 10^15 electrons flow through a section of a wire of diameter 2.0 mm in 4.0 s, what is the current in the wire? (e = 1.60 × 10^-19 c)
The current through the wire is of 0.00012 amperes.
How to find the current?By definition, current will be equal to the quotient between the total charge that flows and the time in which it flows.
Here we have 3.0*10^15 electrons, each one with charge:
e = 1.6*10^(-19) C
Then the total charge that we have is:
Q = (1.6*10^(-19) C)*(3.0*10^15) = 0.00048 C
Finally, the current is:
I = Q/T
where:
Q = 0.00048 CT = 4.0 sI = (0.00048 C)/(4s) = 0.00012 A
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Assume that you have two objects, one with a mass of 6 kg and the other with a mass of 17 kg, each with a charge of −0.027 c and separated by a distance of 3 m. what is the electric force that these objects exert on one another? answer in units of n. what is the gravitational force between them? answer in units of n.
Electric force is the attractive or reflective force of interception between two charged object.
A) The electric force that these objects exert on one another is [tex]7.28\times 10^5[/tex] N.B) The gravitational force between them is [tex]7.56\times10^{-10}[/tex] N.Given information-
The mass of first object is 6 kg.
The mass of second object is 17 kg.
The charge on both the object is 0.027 C.
The distance between the two object is 3 m.
What is electric force?Electric force is the attractive or reflective force of interception between two charged object. It can be calculated using the Coulomb's law as,
[tex]F=k_e\dfrac{q_1\q_2}{r^2}[/tex]
Here, [tex]q[/tex] is the charge on the object and [tex]r[/tex] is the distance between the objects. [tex]k_e[/tex] is coulombs constant [tex](8.987\times 10^9)[/tex] N-m squared per C squared.
A) Electric force that these objects exert on one another-Put the values in above formula to find out the electric force between given objects-
[tex]F=8.987\times10^9\dfrac{(-0.027)\times(-0.027)}{3^2}\\F=7.28\times 10^5[/tex]
Hence the electric force that these objects exert on one another is [tex]7.28\times 10^5[/tex] N.
B) Gravitational force between them-Gravitational force between two object with mass [tex]m_1[/tex] and [tex]m2[/tex] can be given as,
[tex]F_g=G\dfrac{m_1m_2}{r^2}[/tex]
Here [tex]G=6.67\times 10^{-11}[/tex] N-m squared per kg squared.
Put the values,
[tex]F_g=6.67\times 10^{-11}\times\dfrac{6\times17}{3^2}\\F_g=7.56\times10^{-10}[/tex]
Thus, the gravitational force between them is [tex]7.56\times10^{-10}[/tex] N.
Hence,
The electric force that these objects exert on one another is [tex]7.28\times 10^5[/tex] N.The gravitational force between them is [tex]7.56\times10^{-10}[/tex] N.Learn more about the electric force here;
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Final answer:
The electric force and gravitational force between two objects can be calculated using specific laws involving charge, mass, and distance.
Explanation:
The electric force between two objects can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The gravitational force between the two objects can be determined using Newton's Law of Universal Gravitation, where the force is proportional to the product of the masses and inversely proportional to the square of the distance between them.
Which of the following is not a fossil fuel?
a.
oil
c.
coal
b.
natural gas
d.
hydrogen
what is showed in the diagram /Users/jacinta/Desktop/NNNNNN.png
A vertical polarizing filter is used on the lens of a camera. Which waves do not strike the lens?
A light source emits a beam of photons, each of which has a momentum of 2.7 × 10-29 kg·m/s. (a) what is the frequency of the photons? (b) to what region of the electromagnetic spectrum do the photons belong?
The frequency of the photons is equal to 1.22 ×10¹³ Hz and lies in the infrared region of the electromagnetic spectrum.
What is the frequency?The frequency of the photons or light can be described as the number of oscillations in one second. The frequency possesses S.I. units per second or Hertz.
The relationship between momentum (p), frequency (ν), and speed of light (c) is:
p = hν/c
ν = pc/h
Given, the momentum of the photons, p = 2.7 ×10⁻²⁹ Kg.m/s
The speed of light, c = 3×10⁸ m/s
The plank's constant, h = 6.626 ×10⁻³⁴ Js
The frequency of the photons can determine from the above-mentioned relationship:
ν = (2.7 × 10⁻²⁹).( 3 × 10⁸)/ 6.626 × 10⁻³⁴
ν = 1.22 × 10¹³ Hz
Therefore, the frequency of the photons is 1.22 × 10¹³ Hz and lies in the infrared region of the spectrum.
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A crying baby emits sound with an intensity of 8.0 × 10-8 w/m2. calculate a reasonable estimate for the intensity level from a set of quintuplets (five babies), all crying simultaneously at the same place? the lowest detectable intensity is 1.0 × 10-12 w/m2.
The intensity level from a set of quintuplets (five babies) : 56 dB
Further explanationWave intensity is the power of a wave that is moved through a plane of one unit that is perpendicular to the direction of the wave
Can be formulated
[tex]\rm I=\dfrac{P}{A}[/tex]
I = intensity, W m⁻²
P = power, watt
A = area, m²
The farther the distance from the sound source, the smaller the intensity
[tex]\rm \dfrac{I_2}{I_1}=\dfrac{(r_1)^2}{(r_2)^2}[/tex]
So the intensity is inversely proportional to the square of the distance from the source
[tex]\rm I\approx \dfrac{1}{r^2}[/tex]
Intensity level (LI) can be formulated
[tex]\rm LI=10\:log\dfrac{I}{I_o}[/tex]
Io = 10⁻¹²
For the level of intensity of several sound sources as many as n pieces can be formulated:
LIn = LI1 + 10 log n
The intensity level of 1 baby is
[tex]\rm LI=10\:log\dfrac{8.10^{-8}}{10^{-12}}[/tex]
LI = 10 log 8.10⁴
LI = 49
The intensity level of 5 babies :
LI5 = LI + 10 log n
LI5 = 49 + 10 log 5
LI5 = 49 + 7
LI5 = 56
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which change is the best example of a physical change 1) a cookie baking 2) paper burning 3) ice cream melting 4) a nail rusting
Nickel is a common ferromagnetic material. t/f
In a typical lightning strike, 2.9 c flows from cloud to ground in 0.18 ms. what is the current during the strike?
The pressure exerted by a phonograph needle on a record is surprisingly large. if the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.210 mm in radius, what pressure is exerted on the record in n/m2?
The pressure exerted by a phonograph needle on a record, if the equivalent of 1g is supported by the needle with a radius of 0.210 mm, is approximately 7.05*10^7 Pa or N/m².
Explanation:
The pressure exerted by a phonograph needle on a record is calculated using the formula for pressure: P = F / A . To obtain the force (F), we multiply the mass of the needle (1g or 0.001 kg) by the acceleration due to gravity (9.8 m/s²). This yields a force of 0.0098 N. The area (A) is calculated using the formula for the area of a circle, A=πr², where r is the radius of the needle tip (0.210 mm or 0.00021 m). So, the area amounts to roughly 0.000000139 m². The resulting pressure (P), when calculated comes out to be approximately 7.05*10^7 Pa or N/m².
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If a hydrogen atom in the ground state absorbs a photon of energy 12.09 ev, to which state will the electron make a transition?
Light of wavelength 520 nm passes through a slit of width 0.220 mm. (a) the width of the central maximum on a screen is 8.30 mm. how far is the screen from the slit?
What is the escape speed of an electron launched from the surface of a 1.1-cm-diameter glass sphere that has been charged to 8.0 nc ?
The calculation of escape speed from a charged glass sphere involves principles from electrostatics and classical mechanics, requiring an understanding of how kinetic and electric potential energies equate as an electron moves away from the sphere. Unfortunately, without detailed formulae specific to this electrostatic scenario, a precise answer cannot be provided here.
Explanation:The question involves calculating the escape speed of an electron from a charged glass sphere, which is a task that falls under the domain of electrostatics and classical mechanics in physics. Generally, the escape speed from a celestial body like Earth is determined by its mass and the gravitational forces involved. However, in this electrostatic context, the key forces are electrical, not gravitational. The escape speed in this scenario would depend on the electric potential energy and kinetic energy equivalence. Given the unique nature of the question which combines concepts from electrostatics with classical escape velocity calculations, a straightforward formula application from physics textbooks might not directly apply without considering the electric force on the electron due to the charged sphere. Nonetheless, the principle remains that to calculate escape speed, one would need to equate the kinetic energy of the electron with the work done against the electric force as it moves to infinity (where the electrical potential energy becomes zero).
The escape speed of an electron from the surface of a charged sphere can be found using the formula, yielding approximately 162 meters per second. This is derived considering the charge of the sphere, its radius, and Coulomb's constant.
To find the escape speed of an electron from a charged sphere, you need to use the concept of electric potential energy and kinetic energy. The escape speed, vesc, is given by:
vesc = √((2 * k * Q) / r)where Q is the charge of the sphere, r is the radius of the sphere, and k is Coulomb's constant (k = 8.99 × [tex]10^9[/tex] N [tex]m^2[/tex]/[tex]C^2[/tex]).
Given:
Diameter of the sphere = 1.1 cm → radius, r = 0.55 cm = 0.0055 mCharge of the sphere, Q = 8.0 nC = 8.0 × [tex]10^{-9}[/tex] CUsing the formula:
vesc = √((2 * 8.99 × [tex]10^9[/tex] N [tex]m^2[/tex]/[tex]C^2[/tex] * 8.0 × [tex]10^{-9}[/tex] C) / 0.0055 m)vesc = √((1.4384 × [tex]10^2[/tex] ) / 0.0055 m)vesc ≈ √(2.6145 × [tex]10^4[/tex])vesc ≈ 162 m/sThus, the escape speed of the electron from the surface of the charged sphere is approximately 162 meters per second.
Explain why wet clothes that are hung on a washing line dry best