Answer:
A
[tex]\lambda = 361.45 \ m[/tex]
B
[tex]k = 0.01739 \ rad/m[/tex]
C
[tex]w = 5.22 *10^{6} \ rad/s[/tex]
D
[tex]E = 0.01446 \ N/C[/tex]
Explanation:
From the question we are told that
The frequency is [tex]f = 83 0 \ kHz = 830 *10^{3} \ Hz[/tex]
The magnetic field amplitude is [tex]B = 4.82*10^{-11} \ T[/tex]
Generally wavelength is mathematically represented as
[tex]\lambda = \frac{c}{f}[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]\lambda = \frac{3.0*10^{8}}{ 830 *10^{3}}[/tex]
=> [tex]\lambda = 361.45 \ m[/tex]
Generally the wave number is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]k = \frac{2 * 3.142 }{ 361.45 }[/tex]
=> [tex]k = 0.01739 \ rad/m[/tex]
Generally the angular frequency is mathematically represented as
[tex]w = 2 * \pi * f[/tex]
=> [tex]w = 2 * 3.142 * 830*10^{3}[/tex]
=> [tex]w = 5.22 *10^{6} \ rad/s[/tex]
The the electric-field amplitude is mathematically represented as
[tex]E = B * c[/tex]
=> [tex]E = 4.82 *10^{-11} * 3.0*10^{8}[/tex]
=> [tex]E = 0.01446 \ N/C[/tex]
This question involves the concepts of wavelength, frequency, wave number, and electric field.
a) The wavelength is "361.44 m".
b) The wave number is "0.0028 m⁻¹".
c) The angular frequency is "5.22 x 10⁶ rad/s".
d) The electric field amplitude is "0.0145 N/C".
a)
The wavelength can be given by the following formula:
[tex]c=f\lambda[/tex]
where,
c = speed of light = 3 x 10⁸ m/s
f = frequency = 830 KHz = 8.3 x 10⁵ Hz
λ = wavelength = ?
Therefore,
[tex]3\ x\ 10^8\ m/s=(8.3\ x\ 10^5\ Hz)\lambda\\\\\lambda=\frac{3\ x\ 10^8\ m/s}{8.3\ x\ 10^5\ Hz}\\\\[/tex]
λ = 361.44 m
b)
The wave number can be given by the following formula:
[tex]wave\ number = \frac{1}{\lambda} = \frac{1}{361.44\ m}[/tex]
wave number = 0.0028 m⁻¹
c)
The angular frequency is given as follows:
[tex]\omega = 2\pi f = (2)(\pi)(8.3\ x\ 10^5\ Hz)[/tex]
ω = 5.22 x 10⁶ rad/s
d)
The electric field amplitude can be given by the following formula:
[tex]\frac{E}{B} = c\\\\c(B)=E\\\\E = (3\ x\ 10^8\ m/s)(4.82\ x\ 10^{-11}\ T)\\[/tex]
E = 0.0145 N/C
Learn more about wavelength and frequency here:
https://brainly.com/question/12924624?referrer=searchResults
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?
Answer:
The mas unit is the the 'Kilogram' written as 'kg'
Volume is 10 L
Explanation:
The complete question is
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?
Second, what is the volume
mass units is the 'Kilogram', written as 'kg'
density = mass/volume = 100 kg/L
the mass = 1000 kg
volume = mass/density = 1000/100 = 10 L
A small helium-neon laser emits red visible light with a power of 5.40 mW in a beam of diameter 2.30 mm.
Required:
a. What is the amplitude of the electric field of the light? Express your answer with the appropriate units.
b. What is the amplitude of the magnetic field of the light?
c. What is the average energy density associated with the electric field? Express your answer with the appropriate units.
d. What is the average energy density associated with the magnetic field? Express your answer with the appropriate units.
E) What is the total energy contained in a 1.00-m length of the beam? Express your answer with the appropriate units.
Answer:
A. 990v/m
B.330x10^-8T
C.2.19x10^-6J/m³
D.1.45x10^-11J
Explanation:
See attached file
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days? Also find its Standard Deviation
Answer:
so the probability will be = 0.062
Standard deviation = 0.8925
Explanation:
The probability of rain = 15% = 15/100= 0.15
and the probability of no rain=q = 1-p= 1-0.15= 0.85
The number of trials = 7
so the probability will be
7C3 * ( 0.15)^3 (0.85)^4= 35* 0.003375 * 0.52200 =0.06166= 0.062
Taking this as binomial as the p and q are constant and also the trials are independent .
For a binomial distribution
Standard deviation = npq= 0.15 *0.85 *7= 0.8925
The advantage of a hydraulic lever is A : it transforms a small force acting over a large distance into a large force acting over a small distance. B : it transforms a small force acting over a small distance into a large force acting over a large distance. C : it allows you to exert a larger force with less work. D : it transforms a large force acting over a large distance into a small force acting over a small distance. E : it transforms a large force acting over a small distance into a small force acting over a large distance.
Answer:
A) it transforms a small force acting over a large distance into a large force acting over a small distance.
Explanation:
The hydraulic lever works based on Pascal's law of transmission of pressure through a fluid. In the hydraulic lever, the pressure transmitted is the same.
Pressure transmitted P = F/A
where F is the force applied
and A is the area over which the force is applied.
This pressure can be manipulated on the input end as a small force applied over a small area, and then be transmitted to the output end as a large force over a large area.
F/A = f/a
where the left side of the equation is for the output, and the right side is for the input.
The volume of the displaced fluid will be the same on both ends of the hydraulic lever. Since we know that
volume V = (area A) x (distance d)
this means that the the piston on the input smaller area of the hydraulic lever will travel a greater distance, while the piston on the larger output area of the lever will travel a small distance.
From all these, we can see that the advantage of a hydraulic lever is that it transforms a small force acting over a large distance into a large force acting over a small distance.
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.
Answer:
b. approaches infinity
Explanation:
Because Capacitive reactance is given as Xc = 1/ωC
So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.
Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC
Question 18(Multiple Choice Worth 2 polnis)
When riding your skateboard you crash into a curb, the skateboard stops, and you continue moving forward. Which law of
motion is being described in this scenario?
O Law of Universal Gravitation
o Newton's Second Law of Motion
o Law of Conservation of Energy
o Newton's First Law of Motion
A fireworks rocket is launched vertically upward at 40 m/s. At the peak of its trajectory, it explodes into two equal-mass fragments. One reaches the ground t1 = 2.71s after the explosion.When does the second reach the ground?t=?
Answer:
6.13 seconds
Explanation:
At the peak of the fireworks trajectory, the velocity of the firework would be zero. Using equation of motion, we have:
v² = u² + 2gh
0 = 40² - (2)(9.81)(h)
0 = 1600 - 19.62h
19.62h = 1600
h = 1600/19.62
h = 81.55 m
Now during the process of explosion, the two parts gained equal vertical momentum but in opposite directions.
We are told the first piece lands in a time of 2.71 s,
Using 3rd equation of motion, we have;
h = ut + ½gt²
81.55 = u(2.71) + ½(9.81 × 2.71²)
81.55 = 2.71u + 36.0228
2.71u = 81.55 - 36.0228
2.71u = 45.5272
u = 45.5272/2.71
u = 16.8 m/s
The time it takes a projectile to return back to its original launch point assuming the projectile was launched
vertically with speed u = 16.8 m/s is;
t = 2u/g
t = (2 × 16.8)/9.81
t = 3.43 s
Thus total time it takes the second mass to reach the ground = 3.43 + 2.71 = 6.13 seconds
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
An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______
Answer:
-z axis
Explanation:
According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
"A thin film with an index of refraction of 1.50 is placed in one of the beams of a Michelson interferometer. If this causes a shift of 8 bright fringes in the pattern produced by light of wavelength 540 nm, what is the thickness of the film?"
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder
Answer:
692 J/kg/°C
Explanation:
Electric energy added = amount of heat
Power × time = mass × SHC × increase in temperature
Pt = mCΔT
(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)
C = 692 J/kg/°C
Equipotential lines are lines with equal electric potential (for example, all the points with an electric potential of 5.0 V). Using the plot tool that comes with voltmeter (pencil icon) make two equipotential lines at r = 0.5 m and r = 1.5 m. Enable electric field vectors in the simulation. Put an electric field sensor at different points on the equipotential line and note the direction of the electric field vector. What can you conclude about the direction of the electric field vector in relation to the equipotential lines?
The direction for each field vector is perpendicular to equipotential lines.
Take a snapshot of the simulation showing equipotential lines and paste to a word document.
....................
3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.
Answer:
Option (c)
Explanation:
In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.
(C.) is the only correct statement in the list of choices.
In a series circuit, the current can flow through only one path from start to finish.
Charge of uniform linear density (6.7 nC/m) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y
Thw question is not complete. The complete question is;
Charge of uniform linear density (6.7 nCim) is distributed along the entire x axis. Determine the magnitude of the electric field on the y axis at y = 1.6 m. a. 32 N/C b. 150 NC c 75 N/C d. 49 N/C e. 63 NC
Answer:
Option C: E = 75 N/C
Explanation:
We are given;
Uniform linear density; λ = 6.7 nC/m = 6.7 × 10^(-9) C/m
Distance on the y-axis; d = 1.6 m
Now, the formula for electric field with uniform linear density is given as;
E = λ/(2•π•r•ε_o)
Where;
E is electric field
λ is uniform linear density = 6.7 × 10^(-9) C/m
r is distance = 1.6m
ε_o is a constant = 8.85 × 10^(-12) C²/N.m²
Thus;
E = (6.7 × 10^(-9))/(2π × 1.6 × 8.85 × 10^(-12))
E = 75.31 N/C ≈ 75 N/C
A girl is sitting on the edge of a pier with her legs dangling over the water. Her soles are 80.0 cm above the surface of the water. A boy in the water looks up at her feet and wants to touch them with a reed. (nwater =1.333). He will see her soles as being:____
a. right at the water surface.
b. 53.3 cm above the water surface.
c. exactly 80.0 cm above the water surface.
d. 107 cm above the water surface.
e. an infinite distance above the water surface.
Answer:
d. 107 cm above the water surface.
Explanation:
The refractive index of water and air = 1.333
The real height of the girl's sole above water = 80.0 cm
From the water, the apparent height of the girl's sole will be higher than it really is in reality by a factor that is the refractive index.
The boy in the water will therefore see her feet as being
80.0 cm x 1.333 = 106.64 cm above the water
That is approximately 107 cm above the water
A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exerts a resistive force with magnitude proportional to the square of the speed with k
Answer:
The velocity is 40 ft/sec.
Explanation:
Given that,
Force = 3200 lb
Angle = 30°
Speed = 64 ft/s
The resistive force with magnitude proportional to the square of the speed,
[tex]F_{r}=kv^2[/tex]
Where, k = 1 lb s²/ft²
We need to calculate the velocity
Using balance equation
[tex]F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}[/tex]
Put the value into the formula
[tex]3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}[/tex]
Put the value of k
[tex]3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}[/tex]
[tex]1600-v^2=m\dfrac{d^2v}{dt^2}[/tex]
At terminal velocity [tex]\dfrac{d^2v}{dt^2}=0[/tex]
So, [tex]1600-v^2=0[/tex]
[tex]v=\sqrt{1600}[/tex]
[tex]v=40\ ft/sec[/tex]
Hence, The velocity is 40 ft/sec.
Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J = cr2 = 9.00 ✕ 106 A/m4 r2. What is the current (in A) through the inner section of the wire from the center to r = 0.5R?
Answer:
The current is [tex]I = 8.9 *10^{-5} \ A[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 3.17 \ mm = 3.17 *10^{-3} \ m[/tex]
The current density is [tex]J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2[/tex]
The distance we are considering is [tex]r = 0.5 R = 0.001585[/tex]
Generally current density is mathematically represented as
[tex]J = \frac{I}{A }[/tex]
Where A is the cross-sectional area represented as
[tex]A = \pi r^2[/tex]
=> [tex]J = \frac{I}{\pi r^2 }[/tex]
=> [tex]I = J * (\pi r^2 )[/tex]
Now the change in current per unit length is mathematically evaluated as
[tex]dI = 2 J * \pi r dr[/tex]
Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows
[tex]I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr[/tex]
[tex]I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.[/tex]
[tex]I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ][/tex]
substituting values
[tex]I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ][/tex]
[tex]I = 8.9 *10^{-5} \ A[/tex]
A uniform narrow tube 1.90 m long is open at both ends. It resonates at two successive harmonics of frequencies 280 Hz and 294 Hz.(a) What is the fundamental frequency?_____Hz(b) What is the speed of sound in the gas in the tube?________ m/s
Answer:
a)14Hz
b)26.6m/s
Explanation:
a)we were given
the first harmonics frequencies as 280 Hz
The second harmonic frequency as 294 Hz.
The fundamental frequency is equal to the gap which means the distance that exist between the harmonics, then
the fundamental frequency=(294 - 280 = 10 Hz)
= 14Hz
b) We know the frequency and the wavelength of the sound wave (
We were told that the wavelength must be twice the length of the tube then, velocity can be calculated as
And fundamental frequency= 14Hz, and distance of 1.90 m then
v = f*2L = (14Hz)*2*(1.90 m) = 26.6m/s
Therefore, the speed of sound in the gas in the tubes is 26.6m/s
¿Qué resistencia debe ser conectada en paralelo con una de 20 Ω para hacer una
resistencia combinada de 15 Ω?
Answer:
60 Ω
Explanation:
R(com) = 15 Ω
1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn
1/15 = 1/20 + 1/R2
1/R2 = 1/15 - 1/20
1/R2 = (4 - 3) / 60
1/R2 = 1/60
R2 = 60 Ω
así, la combinada de resistencia necesaria es 60 Ω
The speed of sound through air is 340 m/s. If a person hears the clap of thunder 9.6 s after seeing the bolt of lightning, how far away is the lightning?
Explanation:
Distance = speed × time
d = (340 m/s) (9.6 s)
d = 3264 m
A swimmer is treading water with their head above the surface of a pool and sees a penny at the bottom of the pool 5.0 mm below. How deep does the coin appear to be? (Index of refraction of water = 1.33) [Conceptual note: Does the coin appear to be shallower or deeper?]
Answer:
The apparent depth is [tex]D' = 0.00376 \ m[/tex]
Explanation:
From the question we are told that
The depth of the water is [tex]D = 5.0 \ mm = 5.0 *10^{-3} \ m[/tex]
The refractive index of water is [tex]n = 1.33[/tex]
Generally the apparent depth of the coin is mathematically represented as
[tex]D' = D * [\frac{ n_a}{n} ][/tex]
Here [tex]n_a[/tex] is the refractive index of air the value is [tex]n_a = 1[/tex]
So
[tex]D' = 5.0 *10^{-3} * [\frac{1}{1.33} ][/tex]
[tex]D' = 0.00376 \ m[/tex]
The apparent depth will be 0.00376 m.
What is an index of refraction?
The index of refraction of a substance also known as the refraction index is a dimensionless quantity that specifies how quickly light passes through it in optics.
d is the depth of the water =5.0 mm =5.0 ×10⁻³
n is the refractive index of water =1.33
[tex]\rm n_a[/tex] is the refractive index of wire=1
The apparent depth of the coin is given as;
[tex]\rm D'=D \times \frac{n_a}{n} \\\\ \rm D'=5.0 \times 10^{-3} \times \frac{1}{1.33} \\\\ \rm D'=0.00376 \ m[/tex]
Hence the apparent depth will be 0.00376 m.
To learn more about the index of refraction refer to the link;
https://brainly.com/question/23750645
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
A pulley 326 mm in diameter and rotating initially at 4.00 revolutions per second receives a constant angular acceleration of 2.25 radians per second squared by a drive belt. What is the linear velocity of the belt after 5.00 seconds
Answer:
The linear velocity, v = 5.93 m/s
Explanation:
To find the linear velocity after 5 seconds, we find its angular velocity after 5 seconds using
ω' = ω + αt where ω = initial angular speed = 4.00 rev/s = 4.00 × 2π rad/s = 25.13 rad/s, ω' = = final angular speed, α = angular acceleration = 2.25 rad/s² and t = time = 5.00 s
ω' = ω + αt
= 25.13 rad/s + 2.25 rad/s² × 5.00 s
= 25.13 rad/s + 11.25 rad/s
= 36.38 rad/s
The linear velocity v is gotten from v = rω' where r = radius of pulley = 326 mm/2 = 163 mm = 0.163 m
v = rω'
= 0.163 m × 36.38 rad/s
= 5.93 m/s
So, the linear velocity v = 5.93 m/s
The target variable is the speed of light v in the glass, which you can determine from the index of refraction n of the glass. Which equations will you use to find n and v?
Answer:
n= speed of light in vacuum/ speed of light in the other medium.
Explanation:
If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;
n= speed of light in vacuum/ speed of light in the other medium.
Hence;
n= c/v
Where;
n= refractive index of the material
c= speed of light in vacuum
v = speed of light in another medium.
Note that the refractive index is the amount by which a transparent medium decreases the speed of light.
The accommodation limits for a nearsighted person's eyes are 20.0 cm and 82.0 cm. When he wears his glasses, he can see faraway objects clearly. At what minimum distance is he able to see objects clearly
Answer;
26.45cm
See attached file for explanation
Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, what happens to the refracted ray?
Answer:
It bends away from the normal
Explanation:
From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.
Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.
This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal
The refracted ray is seen to bend away from the normal.
Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of 1.3 is the denser medium and the medium with a refractive index of 1.2 is the less dense medium.
From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.
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A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Answer:
Explanation:
According to Equations of Projectile motion :
[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]
vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec
[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]
(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m
[tex]Horizontal Range = vcosx * t[/tex]
(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
A semi-circular loop consisting of one turn of wire is place in the x-y plane. A constant magnetic field B=1.7T points along the negative z-axis(into the page), and a current I=0.7A flows counterclockwisefrom the positive z-axis. The net magnetic force on the circular section of the loop points in what direction? What is the net magnetice force on the circular section of the loop?
Answer:
The direction of net magnetic force on the circular section of the loop is in the positive y-axis
The net magnetic force on the circular section of the loop is 3.74 N
Explanation:
The magnetic field strength [tex]B[/tex] = 1.7 T
the current [tex]I[/tex] = 0.7 A
The diameter of the loop = 2 m
the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2
==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m
The force on the semi-circular is given as
F = [tex]BIl[/tex] sin ∅
but the loop is perpendicular to the field, therefore
sin ∅ = sin 90° = 1
F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N
The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".
According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis
