Answer:
The period of motion is 0.5 second.
Explanation:
Given;
extension of the spring, x = 3.8 cm = 0.038 m
mass of the object, m = 13 g = 0.013 kg
Determine the force constant of the spring, k;
F = kx
k = F / x
k = mg / x
k = (0.013 x 9.8) / 0.038
k = 3.353 N/m
When the object is replaced with a block of mass 20 g, the period of motion is calculated as;
[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]
Therefore, the period of motion is 0.5 second.
A long, straight wire carries a 13.0-A current. An electron is fired parallel to this wire with a velocity of 250 km/s in the same direction as the current, 2.00 cm from the wire.
(a) Find the magnitude and direction of the electron’s initial acceleration.
(b) What should be the magnitude and direction of a uniform electric field that will allow the electron to continue to travel parallel to the wire?
(c) Is it necessary to include the effects of gravity? Justify your answer.
Answer:
A.5.7 x 10^-12m/s²
B. 32.5V/m
C. Pls. See attached file for explanations
Explanation:
A parallel plate air-filled capacitor is being charged.The circular plates have a radius of 4.00cm, and at a particular instant the conduction current in the wiresis 0.280A.
a. What is the displacement current density (jD) in theair space between the plates?
b. What is the rate at which the electric field between the platesis changing?
c. What is the induced magnetic field between the plates at adistance of 2.00 cm from the axis?
d. What is the induced magnetic field between the plates at adistance of 1.00 cm form the axis?
Answer:
Given that
capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴
displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m
( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2
( B ) . displacement current density j
D = ε( dE / dt )the rate at which the electric field between the plates is changing(
dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer
Suppose a 500 mb chart valid today at 12 Z indicates a large trough over the eastern US and a large ridge over the western US. An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will _____ altitude if the altimeter is not corrected. Group of answer choices
Answer:
An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected
A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which a wave travels on this string.
Answer:
Velocity of wave (V) = 5.2 × 10² m/s
Explanation:
Given:
Per unit length mass (U) = 4.80 × 10⁻³ kg/m
Tension (T)= 1,300 N
Find:
Velocity of wave (V)
Computation:
Velocity of wave (V) = √T / U
Velocity of wave (V) = √1300 / 4.80 × 10⁻³
Velocity of wave (V) = √ 270.84 × 10³
Velocity of wave (V) = 5.2 × 10² m/s
By what length will a slab of concrete that is originally 18 m long contract when the temperature drops from 24°C to -16°C? The coefficient of linear thermal expansion for this concrete is 1.0 × 10-5 C-1.
Explanation:
According to Thermal Expansion of solids:
[tex]dl = \alpha \times l \times dt[/tex]
[tex]dl = {10}^{ - 5} \times 18 \times 40 [/tex]
[tex]dl = 7.2 \times {10}^{ - 3} [/tex]
From a static hot air balloon, a 10kg projectile is launched at a speed of 10m / s upwards. If the balloon has a mass of 90kg. What is the final velocity of the latter? Select one:
a. 0.57m / s down
b. 2.56m / s down
c. 1.11m / s down
d. 2.03m / s down
e. 3.15m / s down
Answer:
c. 1.11 m/s down
Explanation:
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Assuming the balloon and projectile are originally at rest:
(90 kg) (0 m/s) + (10 kg) (0 m/s) = (90 kg) v + (10 kg) (10 m/s)
0 kg m/s = (90 kg) v + 100 kg m/s
v = -1.11 m/s
Two resistors connected in series have an equivalent resistance of 690 Ohms. When they are connected in parallel, their equivalent resistance is 118 Ohms. Find the resistance of each resistor.
Explanation:
Let [tex]R_1[/tex] and [tex]R_2[/tex] be the the resistances of the resistors. We are given that
[tex]R_1 + R_2 = 690\:Ω\:\:\:\:\:\:\:(1)[/tex]
and
[tex]\dfrac{1}{R_1} + \dfrac{1}{R_2} = \dfrac{1}{118\:Ω}\:\:\:\:\:(2)[/tex]
From Eqn(1), we can write
[tex]R_2 = 690\:Ω - R_1[/tex]
and then plug this into Eqn(2):
[tex]\dfrac{1}{R_1} + \dfrac{1}{690\:Ω - R_1} = \dfrac{1}{118\:Ω}[/tex]
or
[tex]\dfrac{690\:Ω}{(690\:Ω)R_1 - R_1^2}= \dfrac{1}{118\:Ω}[/tex]
[tex]\Rightarrow R_1^2 - (690\:Ω)R_1 + (690\:Ω)(118\:Ω)= 0[/tex]
or
[tex]R_1^2 - 690R_1 + 81420 = 0[/tex]
Using the quadratic formula, we find that the above equation has two roots:
[tex]R_1 = 151.1\:Ω,\:\:538.9\:Ω[/tex]
This means that if you choose one root value for [tex]R_1[/tex], the other root will be the value for [tex]R_2[/tex].
If a soap bubble is 130 nmnm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated normally by white light
Answer:
The question is not complete, here is the other part.
Assume that n = 1.36.
Express your answer to three significant figures and include the appropriate units.
λ = 707.2nm
Explanation:
n = 1.36
t = 130
2 n t= (m+1/2) λ
To solve for λ.
λ = 2 n t ÷ m+1/2
λ = 2 × 1.36 × 130 ÷ m +1/2
λ =
If m= 0
λ= 353.6 ÷ 0+ 1/2
λ = 353.6 × 2
= 707.2nm
A lamp has the shape of a parabola when viewed from the side. The lamp is centimeters wide and centimeters deep. How far is the light source from the bottom of the lamp if the light source is placed at the focus
The question is not complete so i have attached it.
Answer:
The light source is 2 cm from the bottom of the lamp
Explanation:
From the attached image, we can see that the parabola opens up with its vertex at the origin.
Now, the standard form of equation for a parabola is:
x² = 4ay
Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)
Thus, we have;
12² = 4a(18)
144 = 72a
a = 144/72
a = 2
Looking at the parabola again, the line of symmetry is at x = 0
Thus, axis of symmetry is at x = 0.
Thus, focus is at (0, 2)
So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm
The distance of the light source from the bottom of the lamp is 2 cm.
The given parameters;
the top right edge of the lamp has a coordinate of (12,18)Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;
[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]
Thus, the distance of the light source from the bottom of the lamp is 2 cm.
"Your question is not complete, it seems to be missing the following information";
the top right edge of the lamp has a coordinate of (12,18)
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an 1800 kg car travelling east at 30 m/s slows down to 10 m/s what is the work done by brakes
This was a question on my test do y’all know how to do it cuz I couldn’t figure it out
Answer:
Explanation:
The work done by brake friction changes the kinetic energy,
ASSUMING the road is level and other friction is ignored.
W = KEf - KEi = ½mvf² - ½mvi² = ½m(vi² - vf²)
W = ½(1800)(30² + 10²) = 900(900 - 100) = 720000 J = 720 KJ
The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume that the emissivity eee is equal to 1 for these surfaces.
Required:
a. Find the radius RRigel of the star Rigel, the bright blue star in the constellation Orion that radiates energy at a rate of 2.7 x 10^31 W and has a surface temperature of 11,000 K.
b. Find the radius RProcyonB of the star Procyon B, which radiates energy at a rate of 2.1 x 10^23 W and has a surface temperature of 10,000 K. Assume both stars are spherical. Use σ=5.67 x 10−8^ W/m^2*K^4 for the Stefan-Boltzmann constant.
Given that,
Energy [tex]H=2.7\times10^{31}\ W[/tex]
Surface temperature = 11000 K
Emissivity e =1
(a). We need to calculate the radius of the star
Using formula of energy
[tex]H=Ae\sigma T^4[/tex]
[tex]A=\dfrac{H}{e\sigma T^4}[/tex]
[tex]4\pi R^2=\dfrac{H}{e\sigma T^4}[/tex]
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}[/tex]
[tex]R=5.0\times10^{10}\ m[/tex]
(b). Given that,
Radiates energy [tex] H=2.1\times10^{23}\ W[/tex]
Temperature T = 10000 K
We need to calculate the radius of the star
Using formula of radius
[tex]R^2=\dfrac{H}{e\sigma T^4\times4\pi}[/tex]
Put the value into the formula
[tex]R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}[/tex]
[tex]R=5.42\times10^{6}\ m[/tex]
Hence, (a). The radius of the star is [tex]5.0\times10^{10}\ m[/tex]
(b). The radius of the star is [tex]5.42\times10^{6}\ m[/tex]
The work function of a certain metal is φ = 3.55 eV. Determine the minimum frequency of light f0 for which photoelectrons are emitted from the metal. (Planck's constant is: h = 4.1357×10-15 eVs.)
Answer:
Explanation:
Let f₀ be the frequency .
energy of photons having frequency of f₀
= hf₀ where h is plank's constant
for electron to get ejected , work function should be equal to energy of photon
hf₀ = 3.55
4.1357 x 10⁻¹⁵ x f₀ = 3.55
f₀ = 8.58 x 10¹⁴ Hz .
How are electricity and magnets connected
Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.
A cook preparing a meal for a group of people is an example of
O kinetic energy because he has the ability to make a meal
O potential energy because he has the ability to make a meal
O kinetic energy because he is making the meal
o potential energy because he is making the meal
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by:_____
a. a whole number of half-wavelengths.
b. a whole number of wavelengths.
c. an odd number of half-wavelengths.
Answer:
(B) a whole number of wavelengths.
Explanation:
Two beams of coherent light start out at the same point in phase and travel different paths to arrive at point P. If the maximum destructive interference is to occur at point P, the two beams must travel paths that differ by a whole number of wavelengths.
When the resultant effect of the combination of two identical waves result in their annihilation or complete cancellation of the effect of each other, destructive interference takes place. Hence to have two wave sources producing waves that have the same frequency wavelength and amplitude and which are always in phase with each other or have a constant phase difference are said to be Coherent source
A projectile is launched with speed of 128 m/s, at an angle of 60° with the horizontal. After 2.0 s, what is the vertical component of the projectile's velocity?
After 2.0s, what is the speed of the projectile?
Answer:
a) 91 m/s
b) 111 m/s
Explanation:
v = u + at
v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s
v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s
Just wondering if I did this right
Yeah
All they are all correct
Which is a dopant for a p-type semiconductor? arsenic indium phosphorus antimony
Answer:
As opposed to n-type semiconductors, p-type semiconductors have a larger hole concentration than electron concentration.
Explanation:
In p-type semiconductors, holes are the majority carriers and electrons are the minority carriers. A common p-type dopant for silicon is boron or gallium. hope this you :)
Answer:
IndiumHere are notes I took on semiconductor conductivity :
________________________________________________________
-A p-type semiconductor is made of a material in which electrical conduction is due to the movement of a positive charge.
-Examples of p-type dopants - boron, aluminum, gallium, indium, and thallium
Explanation:
In this case, indium only correct option being a dopant of a P-type semiconductor. Other options are N-type dopants.
Hopefully its correct !! <3
6)the speed of light is approximately 186,000 mi/sec. It takes light from a particular star approximately 9 yrs to reach Earth. How many miles away is the star from Earth? Express the answer in scientific notation. Use 365 days in 1 year. The star is nothing miles away from Earth.
Answer:
5.2791264*10¹³
Explanation:
Convert the 9 years to seconds and then multiple it by 186000
The star is 4.62 x 10¹⁶ miles away from Earth.
The speed of light is 186,000 miles per second. It takes light from a particular star approximately 9 years to reach Earth. There are 365 days in 1 year, so it takes 9 x 365 = 3285 days for light from the star to reach Earth.
The distance between the star and Earth is 3285 x 186,000 = 608,810,000 miles. In scientific notation, this is 4.62 x 10¹⁶ miles.
Here is the calculation:
distance = speed * time
distance = 186,000 miles/second * 3285 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute
distance = 608,810,000 miles
distance = 4.62 x 10¹⁶ miles
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A lens is made with a focal length of -40 cm using a material with index of refraction 1.50. A second lens is made with the SAME GEOMETRY as the first lens, but using a material having refractive index of 2.00. What is the focal length of the second lens
Answer:
f = - 20 cm
Explanation:
This exercise asks us for the focal length, which for a lens in air is
1 / f = (n₂-n₁) (1 / R₁ - 1 / R₂)
where n₂ is the refractive index of the material, n₁ is the refractive index of the medium surrounding the lens, R₁ and R₂ are the radii of the two surfaces.
In this exercise the medium that surrounds the lens is air n₁ = 1 and the lens material has an index of refraction n₂ = n = 1.50, let's substitute in the expression
- 1/40 = (n-1) (1 / R₁ -1 / R₂)
(1 / R₁ - 1 / R₂) = - 1/40 (n-1)
let's calculate
(1 / R₁ -1 / R₂) = - 1/40 (1.50 -1)
(1 / R₁ -1 / R₂) = -1/20
Now we change the construction material for one with refractive index
n = 2, keeping the radii,
1 / f = (n-1) (1 / R₁-1 / R₂)
1 / f = (n-1) (-1/20)
let's calculate
1 / f = (2.00-1) (-1/20)
1 / f = -1/20
f = - 20 cm
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are the pressure and velocity of the water after the contraction
Answer:
the pressure after contraction is 2×10^5 Pa
the speed after contraction is 15m/s
Explanation:
We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,
For us to calculate pressure we need to calculate the area first as;
Let initial Area = A₁
And Final area A₂
We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means
A₂= A₁/3.................
V₁ is the speed
the pressure and speed of the water after the contraction can be calculated using equation of continuity below
A₂V₂ = A₁V₁
But
If we substitute given value in the expresion we have
V₂ = (3A *5)/A
V₂ = 15m/s
Therefore, the speed after contraction is 15m/s
Now we can calculate the pressure using
Bernoulli's equation
p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂
But we know that the pipe is horizontal, then "h" terms cancel out then
p₁ + ½ρv₁² = p₂ + ½ρv₂²
Making P₂ subject of formula we have
p₂ = 0.5ρ( V ₁² - v₂² ) + P₁
P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5
=2×10^5 Pa
Therefore, the pressure after contraction is 2×10^5 Pa
(a) the final speed of the water after contraction is 15 m/s.
(b) The final pressure of the water after contraction is 2.5 x 10⁵ Pa.
The given parameters;
initial pressure, P₁ = 3.5 x 10⁵ Painitial speed, v₁ = 5 m/sdensity of water, ρ = 1000 kg/m³Let the initial area of the pipe = A₁
Apply the continuity equation to determine the final speed of the water after contraction as follows;
[tex]A_1 V_1 = A_2 V_2\\\\V_2 = \frac{A_1V_1}{A_2} \\\\V_2 = \frac{A_1 \times 5}{\frac{1}{3} A_1 } \\\\V_2 = 15 \ m/s[/tex]
The final pressure of the water after contraction is determined by applying Bernoulli's equation for horizontal pipe;
[tex]P_1 + \frac{1}{2} \rho V_1^2= P_2 + \frac{1}{2} \rho V_2^2\\\\P_2 = \frac{1}{2} \rho (V_1^2 - V_2^2) + P_1\\\\P_2 = \frac{1}{2} \times 1000(5^2 - 15^2) + 3.5 \times 10^5\\\\P_2 = 2.5 \times 10^5 \ Pa[/tex]
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A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answer:
Around the center of the mirror
Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields. Which of the following statements correctly describe the various applications listed above?
a) All these technologies use radio waves, including low-frequency microwaves.
b) All these technologies use radio waves, including high-frequency microwaves.
c) All these technologies use a combination of infrared waves and high-frequency microwaves.
d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f) All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m.
g) All the technologies emit waves with a wavelength in the range 0.01 to 10.0 km.
Answer:
d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
Explanation:
Microwave are radio waves of short wavelength, from about 10 centimetres to one millimetre, in the Super High Frequency and the Extremely High Frequency bands. Microwaves can penetrate into materials and deposit their energy below the surface which is why is is used in microwave heating found in microwave oven. Transmission of data sometimes involves the use of microwaves to send and receive information over a long distance. Microwaves are the mainly used in radar, used for satellite communication, and wireless networking technologies such as Wi-Fi.
Fast please!!. Serious answers only.
Considering that Susan is 38 years old, what is the probability that she lives to the age of 85?
Answer:
The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.
Explanation:
Edmentum
Answer:
45,436/97,825= 0.4644.
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a force F? =88i^?23j^ to the end of the wrench. What is the torque on the bolt?
Answer:
The torque on the wrench is 4.188 Nm
Explanation:
Let r = xi + yj where is the distance of the applied force to the origin.
Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,
r = 0.18i + 0.055j
The applied force f = 88i - 23j
The torque τ = r × F
So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j
= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j
= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0 since i × i = 0, j × j = 0, i × j = k and j × i = -k
= 0 - 4.14k + 0.0484(-k) + 0
= -4.14k - 0.0484k
= -4.1884k Nm
≅ -4.188k Nm
So, the torque on the wrench is 4.188 Nm
Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits
Answer:
λ = 6.61 x 10⁻⁷ m = 661 nm
Explanation:
From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:
Δx = λL/d
where,
Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m
L = Distance between slits and screen = 2.12 m
d = slit separation = 0.49 mm = 0.49 x 10⁻³ m
λ = wavelength of light = ?
Therefore,
2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)
(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ
λ = 6.61 x 10⁻⁷ m = 661 nm
A convex refracting surface has a radius of 12 cm. Light is incident in air (n = 1) and refracted into a medium with an index of refraction of 1.5. Light incident parallel to the central axis is focused at a point _____________
Answer:
36cm from the surfaceExplanation:
Equation of refraction of a lens is expression according to the formula given below;
[tex]\dfrac{n_2}{v} = \dfrac{n_1}{u}= \dfrac{n_2-n_1}{R}[/tex]
R is the radius of curvature of the convex refracting surface = 12cm
v is the image distance from the refracting surface
u is the object distance from the refracting surface
n₁ and n₂ are the refractive indices of air and the medium respectively
Given parameters
R = 12 cm
u = [tex]\infty[/tex] (since light incident is parallel to the axis)
n₁ = 1
n₂ = 1.5
Required
focus point of the light that is incident and parallel to the central axis (v)
Substituting this values into the given formula we will have;
[tex]\dfrac{1.5}{v} - \dfrac{1}{\infty}= \dfrac{1.5-1}{12}\\\\\dfrac{1.5}{v} -0= \dfrac{0.5}{12}\\\\\dfrac{1.5}{v}= \dfrac{0.5}{12}\\\\[/tex]
Cross multiply
[tex]1.5*12 = 0.5*v\\ \\18 = 0.5v\\\\v = \frac{18}{0.5}\\ \\v = 36cm[/tex]
Hence Light incident parallel to the central axis is focused at a point 36cm from the surface
A professor of physics cannot see clearly at a distance shorter than 1 m. What should be the focal length of an eyeglass lens that would assist him in reading a newspaper while holding it at a desired distance of 25 cm
Answer:
Explanation:
elastic wire extend by 1.ocm when a load on 20g range from It, what additional load will it be required Cause the futher extension of 2.0cm
Answer:
40g
Explanation:
20g range > 1.0cm
Therefore,
40g range > 2.0cm
Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for longer wavelengths of light. The work function of the surface is
Explanation:
Given: [tex]\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}[/tex]
[tex]\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}[/tex]
[tex]\:\:\:\:\:= 6×10^{14}\:\text{Hz}[/tex]
The work function [tex]\phi[/tex] is then
[tex]\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})[/tex]
[tex]\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}[/tex]
The work function of the surface is equal to 3.98 × 10⁻¹⁹J.
What are frequency and wavelength?The frequency can be explained as the number of oscillations of a wave in one second. The frequency has S.I. units of hertz.
The wavelength can be explained as the distance between the two adjacent points such as two crests or troughs on a wave.
The expression between wavelength (λ), frequency, and speed of light (c) is:
c = νλ
Given, the wavelength of the light, ν = 500 nm
The frequency of the light can determine from the above-mentioned relationship:
ν = c/λ= 3 × 10⁸/500 × 10⁻⁹ = 6 × 10¹⁴ Hz
The work function = h ν = 6 × 10¹⁴ × 6.626 × 10⁻³⁴
φ = 3.98 × 10⁻¹⁹J
Therefore, the work function of the surface is 3.98 × 10⁻¹⁹J.
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