Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
A diffraction grating 19.2 mm wide has 6010 rulings. Light of wavelength 337 nm is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of θ at which maxima appear on a distant viewing screen?
Answer:
(a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
Explanation:
Given that,
Width of diffraction grating [tex]w= 19.2\ mm[/tex]
Number of rulings[tex]N=6010[/tex]
Wavelength = 337 nm
We need to calculate the distance between adjacent rulings
Using formula of distance
[tex]d=\dfrac{w}{N}[/tex]
Put the value into the formula
[tex]d=\dfrac{19.2\times10^{-3}}{6010}[/tex]
[tex]d=3.19\times10^{-6}\ m[/tex]
We need to calculate the value of m
Using formula of constructive interference
[tex]d \sin\theta=m\lambda[/tex]
[tex]\sin\theta=\dfrac{m\lambda}{d}[/tex]
Here, m = 0,1,2,3,4......
[tex]\lambda[/tex]=wavelength
For largest value of θ
[tex]\dfrac{m\lambda}{d}>1[/tex]
[tex]m>\dfrac{d}{\lambda}[/tex]
Put the value into the formula
[tex]m>\dfrac{3.19\times10^{-6}}{337\times10^{-9}}[/tex]
[tex]m>9.46[/tex]
[tex]m = 9[/tex]
(a). We need to calculate the largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{9\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=71.9^{\circ}[/tex]
(b). We need to calculate the second largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{8\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=57.7^{\circ}[/tex]
(c). We need to calculate the third largest value of θ
Using formula of constructive interference
[tex]\theta=\sin^{-1}(\dfrac{m\lambda}{d})[/tex]
Now, put the value of m in to the formula
[tex]\theta=\sin^{-1}(\dfrac{7\times337\times10^{-9}}{3.19\times10^{-6}})[/tex]
[tex]\theta=47.7^{\circ}[/tex]
Hence, (a). The largest value of θ is 71.9°.
(b). The second largest value of θ is 57.7°.
(c). The third largest value of θ is 47.7° .
_____ is a distinct element found in Balinese Gamelan music in which a pair of gongs, one of them de-tuned, is played together, creating a sound similar to a tremolo.
its music btw
Answer:
best song ever will you follow me I will also follow you
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by in SI units. What is the frequency of the wave
Complete Question
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex] in SI units.
Answer:
The value is [tex]f = 1.98918*10^{5}\ Hz[/tex]
Explanation:
From the question we are told that
The magnetic field is [tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ][/tex]
This above equation can be modeled as
[tex]B=1.2* 10^{-6} sin [2\pi[(\frac{z}{240} ) - ( \frac{t * 10^7}{8} ) ] ] \equiv A sin ( kz -wt )[/tex]
So
[tex]w = \frac{10^7}{8}[/tex]
Generally the frequency is mathematically represented as
[tex]f = \frac{w}{2 \pi}[/tex]
=> [tex]f = \frac{ \frac{10^7}{8} }{2 \pi}[/tex]
=> [tex]f = 1.98918*10^{5}\ Hz[/tex]
UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".
Kinetic energyAccording to the question,
Speed of light, c = 3 × 10⁸ m/s
Wavelength, λ = 120 nm or,
= 1.2 × 10⁻⁷ m
Plank's Constant, h = 6.626 × 10⁻³⁴ J.s
Now,
The energy of photon will be:
→ E = [tex]\frac{hc}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]
= [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]
= 10.35 eV
By using Einstein's Photoelectric equation,
Energy of Photon = Work function + K.E
10.35 = 4.82 + K.E
K.E = 10.35 - 4.82
= 5.53 eV or,
= 8.85 × 10⁻¹⁹ J
Thus the response above is correct.
Find out more information about Kinetic energy here:
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What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.630 m long and has a mass of 5.69 g.
Answer:
220.698HzExplanation:
The fundamental frequency f₀ is expressed as f₀ =V/2L where;
V is the speed of the string = [tex]\sqrt{\frac{T}{M} }[/tex]
m is the mass of the string
L is the length of the string
T is the tension in the string
f₀ = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]
Given datas
m = 5.69g = 0.00569 kg
T = 440N
L = 0.630 m
Required
Fundamental frequency of the steel piano wire f₀
[tex]f_0 = \frac{1}{2(0.630)}\sqrt{\frac{440}{0.00569} } \\ \\f_0 = \frac{1}{1.26}\sqrt{77,328.65 } \\\\f_0 = \frac{1}{1.26} * 278.08\\\\f_0 = 220.698Hz[/tex]
Hence the frequency of the fundamental mode of vibration of the steel piano wire stretched to a tension of 440N is 220.698Hz
A square coil of wire with 15 turns and an area of 0.40 m2 is placed parallel to a magnetic field of 0.75 T. The coil is flipped so its plane is perpendicular to the magnetic field in 0.050 s. What is the magnitude of the average induced emf
Answer:
The magnitude of the average induced emf is 90V
Explanation:
Given;
area of the square coil, A = 0.4 m²
number of turns, N = 15 turns
magnitude of the magnetic field, B = 0.75 T
time of change of magnetic field, t = 0.05 s
The magnitude of the average induced emf is given by;
E = -NAB/t
E = -(15 x 0.4 x 0.75) / 0.05
E = -90 V
|E| = 90 V
Therefore, the magnitude of the average induced emf is 90V
A parachutist relies on air resistance (mainly on her parachute) to decrease her downward velocity. She and her parachute have a mass of 58.0 kg and air resistance exerts a total upward force of 670 N on her and her parachute. What is the combined weight of the parachutist and parachute? Calculate the net force on the parachutist. What is the magnitude of the acceleration of the parachutist?
Answer:
1. F = 569 N
2. F_net = 101N
3. a = 1.74m/s²
Explanation:
Weight is a force measurement.
F = m*a
F = 58.0kg*9.81m/s²
F = 568.98 N
F_net = 670N+(-569N)
F_net = 101N
a = F/m
a = 101N/58.0kg
a = 1.74m/s²
The combined weight of the parachutist and parachute is equal to 568.98 N and the net force due to air resistance exerts a total upward force of 101.02N on her and her parachute, then the magnitude of the acceleration of the parachutist is 1.74 m/s².
What is Newton's second law?Newton's second law states that the resultant force acting on a body is proportional to the rate of change of momentum of that body.
If a parachutist relies on air resistance to decrease her downward velocity.
The mass of the parachutist and her parachute is 58 kg
The air resistance exerts a total upward force of 670 N
The combined weight of the parachutist and parachute, W = mg
W = 58 × 9.81
W = 568.98 N
The net force on the parachutist = 670 - 568.98 = 101.02 N
The acceleration of the Parachutist = Net force/mass
a = F/m
a = 101.02/58
a = 1.74 m/s²
Thus, the magnitude of the acceleration of the parachutist would be 1.74m/s².
Learn more about Newton's second law, here:
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A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated
We know
[tex]\boxed{\sf m=-\dfrac{v}{u}}[/tex]
[tex]\\ \sf\longmapsto 3=-\dfrac{-40}{u}[/tex]
[tex]\\ \sf\longmapsto 3=\dfrac{40}{u}[/tex]
[tex]\\ \sf\longmapsto u=\dfrac{40}{3}[/tex]
[tex]\\ \sf\longmapsto u=13.3cm[/tex]
What is the momentum of a 36.9 N bowling ball with a velocity of 7.56 m/s?
Answer:
momentum (m)=36.9N
velocity (v)=7.56m/s
now,
momentum (m)=m×v
36.9=m×7.56
36.9÷7.56=m
m=4.89kg
1) True or False:
Atomic mass number is the number of neutrons and protons.
2) True or False:
Fe (iron) has 26 protons. Hint: protons equal what number?
3) True or False:
A photon of infrared light has less energy than a photon of red light.
Answer:
1.true
Explanation:
Answer:
1. True
2. True
3. True
Explanation for Question 1.
A nucleus consists of a bunch of protons and neutrons; these are known as nucleons. The atomic mass number, which is the total number of nucleons;
So, this sentence says that the atomic number is the number of protons and neutrons.
Explanation for Question 2.
Iron has 26 protons.
The number of protons = the atomic number.
So, the atomic number should be 26 also,
When we see the periodic table, Iron's atomic number is 26, so the statement is true.
Explanation for Question 3.
Red photons of light carry about 1.8 electron volts of energy. Infrared radiation has longer waves than red light, and thus oscillates at a lower frequency and carries less energy.
So, the above statement proves that the photon of infrared light has less energy than the photon of red light.
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it is true to say that
Answer:
A) the moment of inertia of the system decreases and the angular speed increases.
Explanation:
The complete question is
A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that
A) the moment of inertia of the system decreases and the angular speed increases.
B) the moment of inertia of the system decreases and the angular speed decreases.
C) the moment of inertia of the system decreases and the angular speed remains the same.
D) the moment of inertia of the system increases and the angular speed increases.
E) the moment of inertia of the system increases and the angular speed decreases
In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as
[tex]I_{1} w_{1} = I_{2} w_{2}[/tex] ....1
where [tex]I_{1}[/tex] and [tex]I_{2}[/tex] are the initial and final moment of inertia respectively.
and [tex]w_{1}[/tex] and [tex]w_{2}[/tex] are the initial and final angular speed respectively.
Also, we know that the moment of inertia of a rotating body is given as
[tex]I = mr^{2}[/tex] ....2
where [tex]m[/tex] is the mass of the rotating body,
and [tex]r[/tex] is the radius of the rotating body from its center.
We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.
From equation 1, we see that in order for the angular momentum to be conserved, the decrease from [tex]I_{1}[/tex] to [tex]I_{2}[/tex] will cause the angular speed of the system to increase from [tex]w_{1}[/tex] to [tex]w_{2}[/tex] .
From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Complete Question
A voltaic cell utilizes the following reaction and operates at 298 K:
3Ce4+(aq)+Cr(s)→3Ce3+(aq)+Cr3+(aq).
What is the emf of this cell under standard conditions? Express your answer using three significant figures.
Answer:
The value is [tex]E^o_{cell} = 2.35 V[/tex]
Explanation:
From the question we are told that
The ionic equation is
[tex]3 Ce^{4 +} _{(aq)} + Cr _{(s)} \to 3 Ce^{3+} _{(aq)} + Cr^{3r} _{(aq)}[/tex]
Now under standard conditions the reduction half reaction is
[tex]Ce^{4+} + e \to Ce^{3+} ; \ \ E^o_r = 1.61 V[/tex]
And the oxidation half reaction is
[tex]Cr^{3+} + 3e^{-} \to Cr ; \ \ \ E^o_o = - 0.74 V[/tex]
The emf of this cell under standard conditions is mathematically represented as
[tex]E^o_{cell} = E^o _r - E^o _o[/tex]
substituting values
[tex]E^o_{cell} = 1.61 - (- 0.74)[/tex]
[tex]E^o_{cell} = 2.35 V[/tex]
Alpha particles (charge = +2e, mass = 6.68 × 10-27 kg) are accelerated in a cyclotron to a final orbit radius of 0.30 m. The magnetic field in the cyclotron is 0.80 T. The period of the circular motion of the alpha particles is closest to: A. 0.25 μs B. 0.16 μs C. 0.49 μs D. 0.40 μs E. 0.33 μs
Answer:
Option B: T ≈ 0.16 μs
Explanation:
We are given;
Mass; m = 6.68 × 10^(-27) kg
Magnetic field;B = 0.80 T
Charge;q = 2e
Now, e is the charge on an electron and it has a value of 1.6 × 10^(-19) C
So, q = 2 × 1.6 × 10^(-19)
q = 3.2 × 10^(-19) C
The period of the circular motion of the alpha particles moving along a in the presence of the magnetic field is given by;
T = 2πm/qB
Where ;
m, q and B are as stated earlier.
Plugging in the relevant values, we have;
T = (2π × 6.68 × 10^(-27))/(3.2 × 10^(-19) × 0.8)
T = 0.16395 × 10^(-6) s
This can also be written as;
T ≈ 0.16 μs
A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)
Answer:
Change in the frequency (in Hz) = 104.96 Hz
Explanation:
Given:
Speed of sound in air (v) = 343 m/s
Speed of car (v1) 36 m/s
Frequency(f) = 500 Hz
Find:
Change in the frequency (in Hz)
Computation:
Frequency hear by the observer(before)(f1) = [f(v+v1)] / v
Frequency hear by the observer(f1) = [500(343+36)] / 343
Frequency hear by the observer(f1) = 552.48 Hz
Frequency hear by the observer(after)(f2) = [f(v-v1)] / v
Frequency hear by the observer(f2) = [500(343-36)] / 343
Frequency hear by the observer(f2) = 447.52 Hz
Change in the frequency (in Hz) = f1 - f2
Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz
Change in the frequency (in Hz) = 104.96 Hz
A sinusoidal voltage is displayed on an oscilloscope screen. The separation on the screen between two adjacent peaks is 5.5 divisions, and you notice that the sweep speed is set to 1 ms per division. What is the frequency of the sinusoidal voltage
Answer:
f = 1.8 10² Hz
Explanation:
With the readings of the oscilloscope screen we can calculate the period of the wave
T = #_divisions time_base
T = 5.5 1 10⁻³
T = 5.5 10⁻³ s
the period and frequency are related
f = 1 / T
f = 1 / 5.5 10⁻³
f = 1.8 10² Hz
a coach is travelling east wards at 12.6 m/s after 12 second its velocity is 9.5 m/s in the same direction. what is the acceleration and direction of its acceleration?
pls do it with the formula
thx mates :)
[tex]\\ \rm\longmapsto a=\dfrac{v-u}{t}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{12.6-9.5}{12}[/tex]
[tex]\\ \rm\longmapsto a=\dfrac{3.1}{12}[/tex]
[tex]\\ \rm\longmapsto \overrightarrow{a}=0.25m/s^2[/tex]
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?
a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
Find the minimum thickness (in nm) of a soap bubble that appears green when illuminated by white light perpendicular to its surface. Take the wavelength to be 549 nm, and assume the same index of refraction as water (nw
Answer:
103nm
Explanation:
Pls see attached file
summarize the steps a scientist might take to determine if a new drug Works in cancer patients
blah help
1. Identification of the problem whether the new drug works in cancer patients.
2. Create a hypothesis like, if the new drugs works on all types of cancers.
3. Variables like placebo effect of the drug and its dosages to be administered.
4. Creating experiments to test the viability of the drug.
5. Analyzing results of the experimentation.
6. Form a conclusion and test further depending on the result of the experiments.
I hope this answer helps you out. Brainliest would be appreciated :)A resistance heater having 20.7 kW power is used to heat a room having 16 m X 16.5 m X 12.3 m size from 13.5 to 21 oC at sea level. The room is sealed once the heater is turned on. Calculate the amount of time needed for this heating to occur in min. (Write your answer in 3 significant digits. Assume constant specific heats at room temperature.)
Answer:
t = 23.6 min
Explanation:
First we need to find the mass of air in the room:
m = ρV
where,
m = mass of air in the room = ?
ρ = density of air at room temperature = 1.2041 kg/m³
V = Volume of room = 16 m x 16.5 m x 12.3 m = 3247.2 m³
Therefore,
m = (1.2041 kg/m³)(3247.2 m³)
m = 3909.95 kg
Now, we find the amount of energy consumed to heat the room:
E = m C ΔT
where,
E = Energy consumed = ?
C = Specific Heat of air at room temperature = 1 KJ/kg.⁰C
ΔT = Change in temperature = 21 °C - 13.5 °C = 7.5 °C
Therefore,
E = (3909.95 kg)(1 KJ/kg.°C)(7.5 °C)
E = 29324.62 KJ
Now, the time period can be calculated as:
P = E/t
t = E/P
where,
t = Time needed = ?
P = Power of heater = 20.7 KW
Therefore,
t = 29324.62 KJ/20.7 KW
t = (1416.65 s)(1 min/60 s)
t = 23.6 min
A5 kg box slides 3 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3 m / s?
Answer:
Coefficient of kinetic friction (Cof. KE) = 0.153
Explanation:
Given:
Mass of box (M) = 5 kg
Distance = 3 m
Initial speed (v) = 3 m/s
Find:
Coefficient of kinetic friction (Cof. KE)
Computation:
v² = u² + 2as
a = v² / 2s
a = 9 / 2(3)
a = 1.5 m/s²
Coefficient of kinetic friction (Cof. KE) = a / g
Coefficient of kinetic friction (Cof. KE) = 1.5 / 9.8
Coefficient of kinetic friction (Cof. KE) = 0.153
Which is produced around a wire when an electrical current is in the wire? magnetic field solenoid electron flow electromagnet
Answer:
A. magnetic field
Explanation:
The magnetic field is produced around a wire when an electrical current is in the wire because of the magnetic effect of the electric current therefore the correct answer is option A .
What is a magnetic field ?A magnetic field could be understood as an area around a magnet, magnetic material, or an electric charge in which magnetic force is exerted.
As given in the problem statement we have to find out what is produced around a wire when an electrical current is in the wire.
The magnetic field is produced as a result when an electrical current is passed through the conducting wire .
Option A is the appropriate response because a wire's magnetic field is created when an electrical current flows through it due to the magnetic influence of the electric current .
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(a) If electrons were used (electron microscope), what minimum kinetic energy would be required for the electrons
Answer:
K = 1.6 10⁻¹⁵ J
Explanation:
In an electron microscope, electrons are used to form images, these electrons are accelerated in electric fields so that they have a kinetic energy that allows obtaining a good amplification with the microscope.
electrical potential energy is converted to kinetic energy
U = K
e V = ½ m v²
v = √2eV /m
the wavelength of these electrons we obtain from the de Broglie equation
λ = h / p
p = mv
λ = h / mv
λ = h / mra 2eV / m
λ = h / ra 2eVm
where we can see that as the potential energy increases, it electrifies the shorter the wavelength of the electrons and consequently the greater the magnification of the microscope
in general these microscopes use from 10000X onwards therefore for this saponification
K = e V
K = 1.6 10⁻¹⁹ 10000
K = 1.6 10⁻¹⁵ J
Ultraviolet light having a wavelength of 97 nm strikes a metallic surface. Electrons leave the surface with speeds up to 3.48 × 105 m/s. What is the work function, in eV of the metal?
Answer:
12.45eVExplanation:
Before calculating the work function, we must know the formula for calculating the kinetic energy of an electron. The kinetic energy of an electron is the taken as the difference between incident photon energy and work function of a metal.
Mathematically, KE = hf - Ф where;
h is the Planck constant
f is the frequency = c/λ
c is the speed of light
λ is the wavelength
Ф is the work function
The formula will become KE = hc/λ - Ф. Making the work function the subject of the formula we have;
Ф = hc/λ - KE
Ф = hc/λ - 1/2mv²
Given parameters
c = 3*10⁸m/s
λ = 97*10⁻⁹m
velocity of the electron v = 3.48*10⁵m/s
h = 6.62607015 × 10⁻³⁴
m is the mass of the electron = 9.10938356 × 10⁻³¹kg
Substituting the given parameters into the formula Ф = hc/λ - 1/2mv²
Ф = 6.63 × 10⁻³⁴*3*10⁸/97*10⁻⁹ - 1/2*9.11*10⁻³¹(3.48*10⁵)²
Ф = 0.205*10⁻¹⁷ - 4.555*10⁻³¹*12.1104*10¹⁰
Ф = 0.205*10⁻¹⁷ - 55.163*10⁻²¹
Ф = 0.205*10⁻¹⁷ - 0.0055.163*10⁻¹⁷
Ф = 0.1995*10⁻¹⁷Joules
Since 1eV = 1.60218*10⁻¹⁹J
x = 0.1995*10⁻¹⁷Joules
cross multiply
x = 0.1995*10⁻¹⁷/1.60218*10⁻¹⁹
x = 0.1245*10²
x = 12.45eV
Hence the work function of the metal in eV is 12.45eV
A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples does it produce when a 3.25 m focal length eyepiece is used? ✕
Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
A person starts at position zero, walks to position 8, then walks to position 5. Which answer correctly identifies the person's distance traveled? *
Answer:
Distance = 13 units
Explanation:
The overall path covered by an object during its journey is called distance covered.
In this problem, a person starts at position zero, walks to position 8, then walks to position 5.
We need to find the person's distance traveled. It can be calculated simply by adding all the positions i.e.
Distance = 0+8+5
Distance = 13
Hence, the distance covered by the person is 13 units.
The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.70 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface.
A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
B) Find the height of the image formed by paraxial rays incident on the convex surface.
C) Is the image erect or inverted?
Answer:
A) 0.1477
B) 0.65388 mm
C) object is inverted
Explanation:
The formula for object - image relationships for spherical reflecting surface is given as;
n1/s + n2/s' = = (n2 - n1)/R
Where;
n1 & n2 are the Refractive index of both surfaces
s is the object distance from the vertex of the spherical surface
s' is the image distance from the vertex of the spherical surface
R is the radius of the spherical surface
We are given;
index of refraction of glass; n2 = 1.60
s = 24 cm = 0.24 m
R = 4 cm = 0.04 m
index of refraction of air has a standard value of 1. Thus; n1 = 1
a) So, making s' the subject from the initial equation, we have;
s' = n2/[((n2 - n1)/R) - n1/s]
Plugging in the relevant values, we have;
s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]
s' = 0.1477
b) The formula for lateral magnification of spherical reflecting surfaces is;
m = -(n1 × s')/(n2 × s) = y'/y
Where;
m is the magnification
n1, n2, s & s' remain as earlier explained
y is the height of the object
y' is the height of the image
Making y' the subject, we have;
y' = -(n1 × s' × y)/(n2 × s)
We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.
Thus;
y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)
y' = - 0.00065388021 m = -0.65388 mm
C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.
Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.
Thus, the object is inverted since m is negative.
Ellen says that whenever the acceleration is directly proportional to the displacement of an object from its equilibrium position, the motion of the object is simple harmonic motion. Mary says this is true only if the acceleration is opposite in direction to the displacement. Which one, if either, is correct
Answer:
Both Ellen and Mary are correct.
Explanation:
Both are correct, it's just different ways of saying the same thing.
When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position
A bicycle has wheels that are 60 cm in diameter. What is the angular speed of these wheels when it is moving at 4.0 m/s
Answer:
13.33 rad/s
Explanation:
Applying,
v = ωr......................... Equation 1
Where v = linear speed, ω = angular speed and r = radius.
Note that,
r = d/2................. Equation 2
Where d = diameter of the wheel.
Substitute equation 2 into equation 1
v = ωd/2............... Equation 3
make ω the subject of the equation
ω = 2v/d................ Equation 4
Given: v = 4 m/s, d = 60 cm = 0.6 m
Substitute these values into equation 4
ω = 2(4)/0.6
ω = 13.33 rad/s
In the lab, you shoot an electron towards the south. As it moves through a magnetic field, you observe the electron curving upward toward the roof of the lab. You deduce that the magnetic field must be pointing:_______.
a. to the west.
b. upward.
c. to the north.
d. to the east.
e. downward.
Answer:
a. to the west.
Explanation:
An electron in a magnetic field always experience a force that tends to change its direction of motion through the magnetic field. According to Lorentz left hand rule (which is the opposite of Lorentz right hand rule for a positive charge), the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.
In this case, if we point the thumb (which shows the direction we shot the electron) to the south (towards your body), with the palm (shows the direction of the force) facing up to the roof, then the fingers (the direction of the field) will point west.