Answer:
The blade undergoes 40 revolutions, so neither of the given options is correct!
Explanation:
The revolutions can be found using the following equation:
[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
Where:
α is the angular acceleration
t is the time = 2.5 s
[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min
First, we need to find the angular acceleration:
[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]
Now, the revolutions that the blade undergo are:
[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]
[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]
Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!
I hope it helps you!
The theory of the origin of the universe that is most popular among space scientists suggests that our universe originated how long ago
10-20 million years ago
10-20 trillion years ago,
10-20 billion years ago
10-20 thousand years ago
Answer:
10 - 20 million years ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled?
a. It will become four times as large.
b. It will double.
c. It will be cut in half.
d. It will become eight times as large.
e. It will be cut to one-quarter its original size.
Answer:
c : it wil be cut in half.
The pattern is formed on a distant screen so we can use the Fraunhofer difracction for a single slit. The formula of the width of the central bright spot is given by Δx = (2λz)/a, where λ is the wavelength and a is the width of the slit. So if the inicial width (a_1) is doubled (a_2= 2 x a_1),the width of the central spot will be cut in half Δx = (2λz)/a_2 = (2λz)/2xa_1 .
If a nucleus decays by successive b, a, a emissions, its mass number will Group of answer choices decrease by seven. decrease by two. decrease by four. decrease by eight. increase by four.
Answer:
The mass number will decrease by eight (8).
Explanation:
Given;
successive beta (b), alpha (a), alpha (a) emissions.
Generally, when a radioactive element emits a beta-particle (b), its mass number doesn't increase but its atomic number increases by 1 . [tex](^{0}_{-1}\beta )[/tex]
Also when a radioactive element emits an alpha-particle (a), its mass number decreases by 4, while its atomic number decrease by 2. [tex](^4_2\alpha)[/tex]
For the given question, a successive beta (b), alpha (a), and alpha (a) emissions = (0) + (-4) + (-4) = -8
Thus, when a radioactive element emit a successive beta (b), alpha (a), alpha (a) particles, the mass number will decrease by eight (8).
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A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in a time T. The heat of fusion of this metal is
Answer:
L = Pt/M
Explanation:
Power, P= Q/t = mL/t
we know that, (Q=m×l)
Now ⇒l= Pt/M
Thus l= Pt/M
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
B. A chemical change
Explanation:
I'm guessing ?
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
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10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ????? How far will this pulse travel in the same time if the tension is doubled?
Answer: Tension = 47.8N, Δx = 11.5×[tex]10^{-6}[/tex] m.
Tension = 95.6N, Δx = 15.4×[tex]10^{-5}[/tex] m
Explanation: A speed of wave on a string under a tension force can be calculated as:
[tex]|v| = \sqrt{\frac{F_{T}}{\mu} }[/tex]
[tex]F_{T}[/tex] is tension force (N)
μ is linear density (kg/m)
Determining velocity:
[tex]|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{0.00874 }[/tex]
[tex]|v| =[/tex] 0.0935 m/s
The displacement a pulse traveled in 1.23ms:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 9.35.10^{-2}*1.23.10^{-3}[/tex]
Δx = 11.5×[tex]10^{-6}[/tex]
With tension of 47.8N, a pulse will travel Δx = 11.5×[tex]10^{-6}[/tex] m.
Doubling Tension:
[tex]|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }[/tex]
[tex]|v| = \sqrt{2.0.00874 }[/tex]
[tex]|v| = \sqrt{0.01568}[/tex]
|v| = 0.1252 m/s
Displacement for same time:
[tex]\Delta x = |v|.t[/tex]
[tex]\Delta x = 12.52.10^{-2}*1.23.10^{-3}[/tex]
[tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex]
With doubled tension, it travels [tex]\Delta x =[/tex] 15.4×[tex]10^{-5}[/tex] m
The two metallic strips that constitute some thermostats must differ in:_______
A. length
B. thickness
C. mass
D. rate at which they conduct heat
E. coefficient of linear expansion
Answer:
E. Coefficient of linear expansion
The block moves up an incline with constant speed. What is the total work WtotalWtotalW_total done on the block by all forces as the block moves a distance LLL
Answer:
External force W₁ = F L
Friction force W₂ = - fr L
weight component W₃ = - mg sin θ L
Y Axis Force W=0
Explanation:
When the block rises up the plane with constant velocity, it implies that the sum of the forces is zero.
For these exercises it is indicated to create a reference system with the x axis parallel to the plane and the y axis perpendicular
let's write the equations of translational equilibrium in given exercise
X axis
F - fr -Wₓ = 0
F = fr + Wₓ
the components of the weight can be found using trigonometry
Wₓ = W sin θ
[tex]W_{y}[/tex] = W cos θ
let's look for the work of these three forces
W = F x cos θ
External force
W₁ = F L
since the displacement and the force have the same direction
Friction force
W₂ = - fr L
since the friction force is in the opposite direction to the displacement
For the weight component
W₃ = - mg sin θ L
because the weight component is contrary to displacement
Y Axis
N- Wy = 0
in this case the forces are perpendicular to the displacement, the angle is 90º and the cosine 90 = 0
therefore work is worth zero
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
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Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion
Answer:
C. planetary accretion
Explanation:
Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.
Answer:
[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]
Explanation:
Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.
Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.
The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would you walk(km)?
Answer:
2.5 km
Explanation:
Answer:
2.5 km
Explanation:
Distance = speed x time
So =5 x 0.5
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.
Answer:
8 seconds
160 meters
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 40 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (40 m/s) t + ½ (-10 m/s²) t²
0 = 40t − 5t²
0 = 5t (8 − t)
t = 0 or 8
Given in the x direction:
v₀ = 20 m/s
a = 0 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²
Δx = 160 m
Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away (in m) is the planet Venus if the echo time is 900 s? m (b) What is the echo time (in µs) for a car 80.0 m from a Highway Patrol radar unit? µs (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 11.5 m? ns
Answer:
a) 1.35 x 10^11 m
b) 0.53 µs
c) 8 ns
Explanation:
Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s
a) for 900 s,
The distance for a round trip = v x t
==> (3 x 10^8) x 900 = 2.7 x 10^11 m
The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m
b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of
d = 2 x 80 = 160 m
the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs
c) accuracy in distance Δd = 11.5 m
Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns
From a hot air balloon 2 km high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks west to see another town with angle of depression of 84 degrees. What is the distance between the two towns?
Answer:
7km
Explanation:
The angle of depression is congruent to the angle of elevation and can be explained as angle below horizontal in which the person observing an object must view for him/her to view object's that are lower than him/her.
In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.
hot air balloon 2 km high,
there exist two triangles
From trigonometry
Tanx= opposite/adjacent
Opp= 2km
Adj= X1
first triangle have base length of
Tan(16)=2/X1
X1=2/ tan(16)
X1=6.97
For Second triangle
Tanx= opposite/adjacent
Opp= 2km
Adj= X2
the other with a base length of
X2=2/tan(84)
X2=0.21
Therefore,, the total distance between the two towns is
x1+x2=6.97+0.21=7.18km
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 587 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 11th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source?
Answer:
The wavelength is [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the first light is [tex]\lambda _ 1 = 587 \ nm[/tex]
The order of the first light that is being considered is [tex]m_1 = 10[/tex]
The order of the second light that is being considered is [tex]m_2 = 11[/tex]
Generally the distance between the fringes for the first light is mathematically represented as
[tex]y_1 = \frac{ m_1 * \lambda_1 * D}{d}[/tex]
Here D is the distance from the screen
and d is the distance of separation of the slit.
For the second light the distance between the fringes is mathematically represented as
[tex]y_2 = \frac{ m_2 * \lambda_2 * D}{d}[/tex]
Now given that both of the light are passed through the same double slit
[tex]\frac{y_1}{y_2} = \frac{\frac{m_1 * \lambda_1 * D}{d} }{\frac{m_2 * \lambda_2 * D}{d} } = 1[/tex]
=> [tex]\frac{ m_1 * \lambda _1 }{ m_2 * \lambda_2} = 1[/tex]
=> [tex]\lambda_2 = \frac{m_1 * \lambda_1}{m_2}[/tex]
=> [tex]\lambda_2 = \frac{10 * 587 *10^{-9}}{11}[/tex]
=> [tex]\lambda_2 = 534 *10^{-9} \ m[/tex]
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.
A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 2.5 kg box sliding across a frictionless floor hits the end of the spring and compresses it 6.5 cm before the spring expands and shoots the box back out. How fast was the box going when it hit the spring
As the box compresses the spring, the spring performs
-1/2 (85 N/m) (0.065 m)² ≈ -0.18 J
of work on the box. By the work energy theorem, the total work performed on the box (which is done only by the spring since there's no friction) is equal to the change in the box's kinetic energy. At full compression, the box has zero instantaneous speed, so
W = ∆K ==> -0.18 J = 0 - 1/2 (2.5 kg) v ²
where v is the box's speed when it first comes into contact with the spring. Solve for v :
v ² ≈ 0.14 m²/s² ==> v ≈ 0.38 m/s
A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses in diopters he must use in his contacts to allow him to focus on distant objects?
Answer:
-0.24diopters
Explanation:
The lens is intended that makes an object at infinity appear to be 4.2 m away, so do=infinity, dI = - 4.2m (minus sign because image is on same side of lens as object)
So 1/do +1/di = 1/f
1/infinity + 1/-4.2 = 1/f
1/f = 1/-4.2 = -0.24diopters
You need to repair a broken fence in your yard. The hole in your fence is
around 3 meters in length and for whatever reason, the store you go to
has oddly specific width 20cm wood. Each plank of wood costs $16.20,
how much will it cost to repair your fence? (Hint: 1 meter = 100 cm) *
Answer:
cost = $ 243.00
Explanation:
This exercise must assume that it uses a complete table for each piece, we can use a direct ratio of proportions, if 1 table is 0.20 m wide, how many tables will be 3.00 m
#_tables = 3 m (1 / 0.20 m)
#_tables = 15 tables
Let's use another direct ratio, or rule of three, for cost. If a board costs $ 16.20, how much do 15 boards cost?
Cost = 15 (16.20 / 1)
cost = $ 243.00
the acceleration due to gravity jupiter is 25m/s Square . what does it mean
Answer:
The acceleration due to gravity of Jupiter is 25 m/s^2. means that, Any object dropped near Jupiter's Surface will accelerate downward (towards the Jupiter's surface) at the rate of 25 m/s^2 due to the gravity of Jupiter
Explanation:
hope it's help
Answer:
Explanation:
If any object is dropped from a height above the Jupiter's surface the object will fall towards jupiter's surface with a constant acceleration of 25m/s^2.
The only force acting on a 3.4 kg canister that is moving in an xy plane has a magnitude of 3.0 N. The canister initially has a velocity of 2.5 m/s in the positive x direction, and some time later has a velocity of 4.8 m/s in the positive y direction. How much work is done on the canister by the 3.0 N force during this time
Answer:
16.79JExplanation:
Given data
mass of canister= 3.4 kg
force acting on canister= 3 N
initial velocity u= 2.5 m/s
final velocity v= 4.8 m/s
The work done on the canister is the change in kinetic energy on the canister
change in [tex]KE= Kfinal- Kinitial[/tex]
K.E initial
[tex]Kintial= \frac{1}{2} mv^2\\\\Kintial= \frac{1}{2}*2*2.5^2\\\\KInitial= \frac{1}{2} *2*6.25\\\\Kinitial= 6.25J[/tex]
K.E final
[tex]Kfinal= \frac{1}{2} mv^2\\\\ Kfinal= \frac{1}{2}*2*4.8^2\\\\ Kfinal= \frac{1}{2} *2*23.04\\\\ Kfinal= 23.04J[/tex]
The net work done is [tex]KE= Kfinal- Kinitial[/tex]
[tex]W net= 23.04-6.25= 16.79J[/tex]
what conventions are used in SI to indicate units
Answer:
Conventions used in SI to indicate units are as follows:
Only singular form of units are used. for example: use kg and not kgs.Do not use full stop after the abbreviations of any unit. for example: do not use kg. or cm.Use one space between last numeric digit and SI unit. for example: 10 cm, 9 km.Symbols and words should not be mixed. for example: use Kilogram per cubic and not kilogram/m3.While writing numerals, only the symbols of the units should be written. for example: use 10 cm and not Ten cm.Units named after a scientist should be written in small letters. for example: newton, henry.Degree sign should not be used when the kelvin unit is used. for exmaple: use 37° and not 37°kAn expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answer:
The value is [tex]N = 3.708*10^{7} \ \ atoms[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]
The temperature is [tex]T = 26 + 273 = 299 \ K[/tex]
The volume is 1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]
Generally according to the ideal gas law we have that
[tex]PV = NkT[/tex]
here k is the Boltzmann constant with a value [tex]k = 1.38 *10^{-23} \ J/K[/tex]
=> [tex]N = \frac{PV}{ k T}[/tex]
=> [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]
=> [tex]N = 3.708*10^{7} \ \ atoms[/tex]
A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.
Required:
Calculate the electric energy dissipated in this process.
Answer:
Explanation:
change in magnetic flux = .16 x .16 x .510 - 0
= .013056 weber .
rate of change of flux = change in flux / time
= .013056 / 40.5 x 10⁻³
= .32237
voltage induced = .32237 V
electrical energy dissipated = v² / R where v is voltage , R is resistance
= .32237² / 6.35
= 16.36 x 10⁻³ J .
A plano-convex glass lens of radius of curvature 1.4 m rests on an optically flat glass plate. The arrangement is illuminated from above with monochromatic light of 520-nm wavelength. The indexes of refraction of the lens and plate are 1.6. Determine the radii of the first and second bright fringes in the reflected light.
Given that,
Radius of curvature = 1.4 m
Wavelength = 520 nm
Refraction indexes = 1.6
We know tha,
The condition for constructive interference as,
[tex]t=(m+\dfrac{1}{2})\dfrac{\lambda}{2}[/tex]
Where, [tex]\lambda=wavelength[/tex]
We need to calculate the radius of first bright fringes
Using formula of radius
[tex]r_{1}=\sqrt{2tR}[/tex]
Put the value of t
[tex]r_{1}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(0+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=0.603\ mm[/tex]
We need to calculate the radius of second bright fringes
Using formula of radius
[tex]r_{2}=\sqrt{2\times(m+\dfrac{1}{2})\dfrac{\lambda}{2}\times R}[/tex]
Put the value into the formula
[tex]r_{1}=\sqrt{2\times(1+\dfrac{1}{2})\dfrac{520\times10^{-9}}{2}\times1.4}[/tex]
[tex]r_{1}=1.04\ mm[/tex]
Hence, The radius of first bright fringe is 0.603 mm
The radius of second bright fringe is 1.04 mm.
Earth’s Moon has a diameter of 3,474 km and orbits at an average distance of 384,000 km. At that distance it subtends and angle just slightly larger than half a degree in Earth’s sky. Pluto’s moon Charon has a diameter of 1,186 km and orbits at a distance of 19,600 km from the dwarf planet. Compare the appearance of Charon in Pluto’s skies with the Moon in Earth’s skies. Describe where in the sky Charon would appear as seen from various locations on Pluto.
The reason for arriving at the above solutions is as follows:
The given dimensions and distance from the Earth of the Moon are;
The diameter of the Moon, d = 3,474 km
The average distance of the Moon from the Earth, R = 384,000 km
Required:
The comparison between Charon's appearance in Pluto and the Moon's appearance on Earth Earth
Solution:
The distance of the Moon's travels in an orbit, C = 2·π·R
∴ C = 2 × π × 384,000 km
The angle subtended by the Moon, θ = d/C × 360°
∴ θ = 3,474/(2 × π × 384,000) × 360° ≈ 0.518°
Pluto's moon Charon, has the following parameters;
The diameter of the Charon, d₂ = 1,186 km
The average distance of the Charon from Pluto, R₂ = 19,600 km
Therefore, the distance of the Moon's travels in an orbit, C₂ = 2·π·R₂
∴ C₂ = 2 × π × 19,600 km
The angle subtended by the Moon, θ₂ = d₂/C₂ × 360°
∴ θ₂ = 1,186/(2 × π × 19,900) × 360° ≈ 3.415°
The angle subtended by Charon in Pluto's sky ≈ 3.415°
Charon therefore, appears 7 times larger in Pluto's skies than the Moon's appearance in Earth's skies
Required:
The appearance of Charon as seen from different locations on Pluto
Solution:
Charon is gravitationally locked to Pluto, therefore, the same side of Pluto is faced with the same side of Charon
Therefore;
Charon appears constantly overhead from the side of Pluto locked to CharonCharon appears constantly at the horizon from the poles on either side of the axis of rotation of Pluto and CharonLearn more about Pluto's moon Charon here:
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If a convex lens were made out of very thin clear plastic filled with air, and were then placed underwater where n = 1.33 and where the lens would have an effective index of refraction n = 1, the lens would act in the same way
a. as a flat refracting surface between water and air as seen from the water side.
b. as a concave mirror in air.
c. as a concave lens in air.
d. as the glasses worn by a farsighted person.
e. as a convex lens in air.
Answer:
D. A convex lens in air
Explanation:
This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 29.0and 58.0, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 110 after it passes through the stack.
If the incident intensity is kept constant, what is the intensity of the light after it has passed through the stack if the second polarizer is removed?
Answer:
I₂ = 143.79
Explanation:
To solve this problem, work them in two parts. A first one where we look for the intensity of the incident light in the set and a second one where we silence the light transmuted by the other set,
Let's start with the set of three curling irons
Beautiful light falls on the first polarized is not polarized, therefore only half the radiation passes
I₁ = I₀ / 2
this light reaches the second polarized and must comply with the Mule law
I₂ = I₁ cos² tea
The angle between the first polarized and the second is Tea = 29.0º
I₂ = I / 2 cos² 29
The light that comes out of the third polarized is
I₃ = I₂ cos² tea
the angle between the third - second polarizer is
tea = 58-29
tea = 29th
I3 = (I₀ / 2 cos² 29) cos² 29
indicate the output intensity
I3 = 110
we clear
I₀ = 2I3 / cos4 29
I₀ = 2 110 / cos4 29
I₀ = 375.96 W / cm²
Now we have the incident intensity in the new set of three polarizers
back to the for the first polarizer
I₁ = I₀ / 2
when passing the second polarizer
I₂ = I1 cos² 29
I2 = IO /2 cos²29
let's calculate
I₂ = 375.96 / 2 cos² 29
I₂ = 143.79