Answer:
Explanation:
a )
Amplitude A = 14 mm , angular frequency ω = 2π / T
= 2π / .5
ω = 4π rad /s
φ₀ = initial phase
Putting the given values in the equation
14 = 25 cos(ωt + φ₀ )
14/25 = cosφ₀
φ₀ = 56 degree
x(t) = 25cos(4πt + 56° )
b )
maximum velocity = ω A
= 4π x 25
100 x 3.14 mm /s
= 314 mm /s
At x = 0 ( equilibrium position or middle point , this velocity is achieved. )
maximun acceleration = ω² A
= 16π² x A
= 16 x 3.14² x 25
= 3943.84 mm / s²
3.9 m / s²
It occurs at x = A or at extreme position.
Suppose that a parallel-plate capacitor has circular plates with radius R = 26 mm and a plate separation of 4.0 mm. Suppose also that a sinusoidal potential difference with a maximum value of 220 V and a frequency of 76 Hz is applied across the plates; that is, V = (220 V) sin[2π(76 Hz)t]. Find Bmax(R), the maximum value of the induced magnetic field that occurs at r = R.
Answer:
B(max) = 3.7971 × [tex]10^{-12}[/tex] T
Explanation:
given data
radius R = 26 mm
plate separation d = 4.0 mm
potential difference Vm = 220 V
frequency f = 76 Hz
V = (220 V) sin[2π(76 Hz)t]
solution
we know that E will be
E = V ÷ d ............1
put here value
E = [tex]\frac{220 \times sin(2\pi 76\times t)}{d}[/tex]
and here we take as given r = R
so A = π R² .................2
and
ФE = E × A
ФE = [tex]\frac{\pi R^2 \times 220 \times sin(2\pi 76 \times t)}{d}[/tex] .....................3
so use use here now Ampere's Law that is
∫ B ds = [tex]\mu_o \times \epsilon_o \times \frac{d\Phi E}{dt} + \mu_o \times I_{encl}[/tex] .....................4
and
here [tex]I_{encl}[/tex] is = 0 and r = R
so
[tex]2B \times \pi \times R = \mu_o \times \epsilon_o \times \frac{d\Phi E}{dt}[/tex] .....................5
and put here value we get
B = [tex]\frac{\mu_o \times \epsilon_o \times \pi \times f \times R \times V_m cos(2\pi f t)}{d}[/tex] .....................6
put here value for B maximum cos(2πft) = 1
and we get B (max)
B(max) = [tex]\frac{\mu_o\times \epsilon_o\times \pi \times f\times R\times V_m}{d}[/tex] ....................7
put here all value
B(max) = [tex]\frac{4\pi \times10^{-7} \times 8.85\times 10^{-12}\times \pi \times 76 \times 0.026\times 220 }{4\times 10^{-3}}[/tex]
solve it we get
B(max) = 3.7971 × [tex]10^{-12}[/tex] T
When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is 1.10 eV.
What is the maximum kinetic energy K_0 of the photoelectrons when light of wavelength 330 nm falls on the same surface?
Use h = 6.63×10-34 J*s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
1.76 eV
Explanation:
Maximum kinetic energy of the emitted photo electrons = energy of the photon- work function of the metal
K.E' = (hc/λ)-∅.................. Equation 1
Where K.E' = maximum kinetic energy of the emitted photo electrons, h = Planck's constant, c = speed, λ = wave length, ∅ = work function of the metal.
make ∅ the subject of the equation
∅ = (hc/λ)-K.E'.................. Equation 2
Given: h = 6.63×10⁻³⁴ J.s, c = 3×10⁸ m/s, λ = 400 nm = 400×10⁻⁹ m, K.E' = 1.1 eV = 1.1(1.602×10⁻¹⁹) J = 1.7622×10⁻¹⁹ J
Substitute into equation 2
∅ = [6.63×10⁻³⁴(3×10⁸)/400×10⁻⁹ ]-1.7622×10⁻¹⁹
∅ = (4.973×10⁻¹⁹)-(1.7622×10⁻¹⁹)
∅ = 3.21×10⁻¹⁹ J.
The maximum kinetic energy of the photo electrons when the wave length is 330 nm is
K.E' = [6.63×10⁻³⁴( 3×10⁸ )/330×10⁻⁹]-(3.21×10⁻¹⁹)
K.E' = (6.03×10⁻¹⁹)-(3.21×10⁻¹⁹)
K.E' = 2.82×10⁻¹⁹ J
K.E' = 2.82×10⁻¹⁹/1.602×10⁻¹⁹
K.E' = 1.76 eV
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 11.8 cm^2 is rotated during the time interval 4.90×10^-2 s from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.60×10-5 T.
a) What is the total magnitude of the magnetic flux through the coil before it is rotated?
b) What is the magnitude of the total magnetic flux through the coil after it is rotated?
c) What is the magnitude of the average emf induced in the coil?
Answer:
Explanation:
Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli
= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴
= 13216 x 10⁻⁹ weber .
b ) When the coil becomes parallel to magnetic field , flux through it will become zero.
c ) e m f induced = change in flux / time
= 13216 x 10⁻⁹ / 4.9 x 10⁻²
= 2697.14 x 10⁻⁷ V
= 269.7 x10⁻⁶
269.7 μV.
Which statement best explains the pattern that causes people on Earth to see only one side of the moon?
The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
The moon orbits on its axis at the same rate at which Earth orbits the sun so that the side of the moon that faces Earth remains the same as it orbits.
The moon rotates on its axis at the same rate at which Earth orbits the sun so that the side of the moon that faces Earth remains the same as it orbits.
The correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
Explanation
The moon is the most popular and well-known natural satellite on planet earth, it is a satellite widely studied by humans, they have even walked on the moon. However, from Earth, at night when you see the moon you always see the same face of the moon, which has caused people to wonder why this phenomenon. The answer to this phenomenon is that we always see the same face of the moon because it takes the same time to rotate once on itself as it does to go around the Earth (about 27 days). The result is that the same part of the moon always points towards the earth. According to the above, the correct answer is A. The moon rotates on its axis at the same rate at which it orbits Earth so that the side of the moon that faces Earth remains the same as it orbits.
At what depth of lake water is the pressure equal to 201kpa?
A. 10m
B. 20m
C. 5.1m
D. none
Answer:
20m
Explanation:
Pressure = pgh
p = density of water 1000
kg/m^3
g = acceleration due to gravity 9.81 m/s^2
h is the depth of water
Pressure = 201 kPa = 201 x 10^3 Pa
201 x 10^3 = 1000 x 9.81 x h
201 x 10^3 = 9810h
h = 20.49 m
Approximately 20 m
How are the electric field lines around a positive charge affected when a second positive charge is rear it?
Answer:they repel
Explanation:
Like charges repels, unlike charges attracts
Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55
min
min
if she is not going to arrive late. Her exit is 43
mi
miHow much time would it take at the posted 60 mph speed?
away.
Complete Question
Alberta is going to have dinner at her grandmother's house, but she is running a bit behind schedule. As she gets onto the highway, she knows that she must exit the highway within 55 min if she is not going to arrive late. Her exit is 43 mi away. How much time would it take at the posted 60 mph speed?
Answer:
The time it would take at the given speed is [tex]x = 43.00 \ minutes[/tex]
Explanation:
From the question we are told that
The time taken to exist the highway is [tex]t = 55 min[/tex]
The distance to the exist is [tex]d = 43\ mi[/tex]
Alberta speed is [tex]v = 60 mph[/tex]
The time it would take travelling at the given speed is mathematically represented as
[tex]t_z = \frac{d}{v}[/tex]
substituting values
[tex]t_z = \frac{43}{60}[/tex]
[tex]t_z = 0.71667\ hrs[/tex]
Converting to minutes
1 hour = 60 minutes
So 0.71667 hours = x minutes
Therefore
[tex]x = 0.71667 * 60[/tex]
[tex]x = 43.00 \ minutes[/tex]
At a speed of 60 mph, it would take Alberta approximately 43 minutes to travel the 43 miles to her grandmother's house, which is within the 55 minutes time frame she has to avoid being late.
Explanation:To determine how long it will take Alberta to reach her grandmother's house if she travels at a constant speed of 60 mph, we need to use the formula for time which is time = distance ÷ speed. Alberta's exit is 43 miles away and the speed limit is 60 mph.
First, we calculate the time it would take her to travel 43 miles at 60 mph:
Time = Distance ÷ Speed
= 43 miles ÷ 60 mph
= 0.7167 hours
Since time in hours is not always intuitive, let's convert it to minutes by multiplying by 60 (since there are 60 minutes in one hour):
Time in minutes = 0.7167 hours × 60 minutes/hour
= 43 minutes
Thus, it will take Alberta approximately 43 minutes to reach her exit at the posted speed of 60 mph.
If 10 cal of energy are added to 2 g of ice at -30°C calculate the final temperature of the ice
Answer:
The final temperature of the ice is [tex]-40^{\circ}\ C[/tex].
Explanation:
It is given that,
Energy, Q = 10 cal
Mass of ice, m = 2 g
Initial temperature, [tex]T_1=-30^{\circ}\ C[/tex]
We need to find the final temperature of the ice. We know that the specific heat of ice is [tex]0.5\ cal\ g^{-1} ^{\circ} C^{-1}[/tex]
The heat added in terms of specific heat is given by :
[tex]Q=mc(T_1-T_2)[/tex]
[tex]T_2[/tex] final temperature of the ice
c is specific heat of ice
[tex]T_2=T_1-\dfrac{Q}{mc}\\\\T_2=(-30)-\dfrac{10}{2\times 0.5}\\\\T_2=-40^{\circ}[/tex]
So, the final temperature of the ice is [tex]-40^{\circ}\ C[/tex].
Answer: the correct answer is -20
Explanation:
78kg*9.8m/s2 what is the force
Answer:764.4N
Explanation:
Mass=78kg
Acceleration=9.8m/s^2
force=mass x acceleration
Force=78 x 9.8
Force=764.4
Force=764.4N
Monochromatic light of variable wavelength is incident normally on a thin sheet of plastic film in air. The reflected light is a maximum only for 477.1 nm and 668.0 nm in the visible spectrum.What is the minimum thickness of the film (n=1.58)?
Answer:
thickness t = 528.433 nm
Explanation:
given data
wavelength λ1 = 477.1 nm
wavelength λ2 = 668.0 nm
n = 1.58
solution
we know for constructive interference condition will be
2 × t × μ = (m1+0.5) × λ1 ....................1
2 × t × μ = (m2+0.5) × λ2 ....................2
so we can say from equation 1 and 2
(m1+0.5) × λ1 = (m2+0.5) × λ2
so
[tex]\frac{\lambda 2}{\lambda 1} = \frac{m1+0.5}{m2+0.5}[/tex] ..............3
put here value and we get
[tex]\frac{668.0}{477.1} = \frac{m1+0.5}{m2+0.5}[/tex]
[tex]\frac{m1+0.5}{m2+0.5}[/tex] = 1.4
[tex]\frac{m1+0.5}{m2+0.5} = \frac{7}{5}[/tex] ...................4
so we here from equation 4
m1+0.5 = 7
m1 = 3 .................5
m2+0.5 = 4
m2 = 2 .................6
so now put value in equation 1
2 × t × μ = (m1+0.5) × λ1
2 × t × 1.58 = (3+0.5) × 477.1
solve it we get
thickness t = 528.433 nm
Using the formula for thin film interference, the minimum thickness of the plastic film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.
Explanation:The question is asking for the minimum thickness of the film for which maximum light is reflected for the provided wavelengths. This is a classic example of thin film interference.
For constructive interference (maximum reflection), the thickness of the film (t) is given by the formula: t = mλ/2n. Here 'm' is the order of the bright fringe, 'λ' is wavelength, and 'n' is the index of refraction.
Considering the first-order maximum, we find the thickness for each wavelength:
t1 = (1)(477.1 x 10^-9 m)/2(1.58) = 1.51 x 10^-7 m (or 151 nm for 477.1 nm light)
t2 = (1)(668.0 x 10^-9 m)/2(1.58) = 2.12 x 10^-7 m (or 212 nm for 668.0 nm light)
Therefore, the minimum thickness of the film that would create a condition of maximum reflection in the visible spectrum is approximately 151 nm.
Learn more about Thin Film Interference here:https://brainly.com/question/33710977
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Why is it easier to slide a heavy box over a floor that it is to start it sliding in the first place?
Answer:
This is because of the inertia of the object.
Explanation:
This is because of the inertia of the object.
When you push a static object you must overcome the static friction force of the object. Once you have overcome the static friction the inertia law demands that the object tend to conserve its motion. That is the reason why you need less force when the object is already in motion. In other words, the inertia gained by the object with the initial force, "helps" you with the work of moving the object. This is also the reason why the kinetic friction of an object in motion over a surface is lower than the static friction.