Answer:
The length is [tex]l = 8.6 \ m[/tex]
Explanation:
From the question we are told that
The frequencies of the two successive harmonics are [tex]f_1 = 220 \ Hz[/tex] , [tex]f_2 = 240 \ Hz[/tex]
The speed of sound in the air is [tex]v_s = 343 \ m/s[/tex]
Generally the frequency of a given harmonic is mathematically represented as
[tex]f_n = \frac{n v }{2l}[/tex]
Here n defines the position of the harmonics
Now since the position of both harmonic is not know but we know that they successive then we can represented them mathematically as
[tex]220 = \frac{n v}{2l}[/tex]
and
[tex]240 = \frac{(n+1) v}{2l}[/tex]
So
[tex]\frac{(n + 1 ) v}{2l} - \frac{n v}{2l} = 240-220[/tex]
=> [tex]\frac{v}{2l} = 20[/tex]
=> [tex]l = 8.6 \ m[/tex]
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low- mass sail and the energy and momentum of sunlight for propulsion.
Should the sail be absorbing or reflective? Why?
a. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is smaller than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
b. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is larger than for reflective sail, therefore the radiation pressure is larger for the absorbing sail
c. The sail should be absorbing because in this case the momentum transferred to the sail per unit area per unit time is smaller than for reflective sail, therefore the radiation pressure is larger for the absorbing sail.
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail
Answer:
d. The sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
Explanation:
Let us take the momentum of a photon unit as u
we know that the rate of change of momentum is proportional to the force exerted.
For a absorbing surface, the photon is absorbed, therefore the final momentum is zero. From this we can say that
F = (u - 0)/t = u/t
for a unit time, the force is proportional to the momentum of the wave due to its energy density. Therefore,
F = u
For a reflecting surface, the momentum of the wave strikes the sail and changes direction. Since we know that the speed of light does not change, then the force is proportional to
F = (u - (-u))/t = 2u/t
just as the we did above, it becomes
F = 2u.
From this we can see that the force for a reflective sail is twice of that for an absorbing sail, and we know that the pressure is proportional to the force for a given area. From these, we conclude that the sail should be reflective because in this case the momentum transferred to the sail per unit area per unit time is larger than for absorbing sail, therefore the radiation pressure is larger for the reflective sail.
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 998.0 rad/s while it is being read, and then is allowed to come to rest over 0.502 seconds , what is the magnitude of the average angular acceleration of the disk?
Answer:
1988.05 rad/s^2
Explanation:
The angular speed of the optical disk ω = 998.0 rad/s
the time taken to come to rest t = 0.502 s
The magnitude of the average angular acceleration ∝ = ω/t
∝ = 998.0/0.502 = 1988.05 rad/s^2
The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3
Answer:
1) M = 11.3V2) 195.49 gramsExplanation:
1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;
M ∝ V
M = kV
For every proportionality sign, there will always be a proportionality constant 'k'
Since the proportionality constant is the density (D) of the metal, the equation will become;
M = DV
Given the density to be 11.3 g/cm3, the equation will become;
M = 11.3V
Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V
2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;
M = 11.3V
M = 11.3 * 17.3
M = 195.49 grams
Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.
550 J of heat is added to the gas in an isothermal process. As the gas expands, pushing against the piston, how much work does it do
Answer:
The work done by the system is 550 J
Explanation:
Given;
heat added to the system, Q = 550 J
Apply the first law of thermodynamics;
ΔU = Q - W
Where;
ΔU is change in internal energy
Q is the heat added to the system
W is the work done by the system
During an isothermal process, the temperature of the system is constant for the entire process. During this process, the change in the internal energy is zero.
0 = Q - W
W = Q
W = 550 J
Therefore, the work done by the system is 550 J
A person is being pulled by gravity with a force of 500 N. What is the force with which the person pulls Earth?
1,000 N
O100 N
500 N
0 250 N
Answer:
The correct answer is 500 N
Explanation:
This is an exercise in Newton's third law or law of action and reaction
The Earth exerts a force on the person, which we call a weight of 500 N directed downwards, we can call this action and the person exerts a force on the Earth of equal magnitude 500N and in the opposite direction, that is directed upwards.
Which force we call action does not matter, the analysis and conclusions are the same
The correct answer is 500N
A circular conducting loop of radius 31.0 cm is located in a region of homogeneous magnetic field of magnitude 0.700 T pointing perpendicular to the plane of the loop. the loop is connected in series with a resistor of 265 ohms. The magnetic field is now increased at a constant rate by a factor of 2.30 in 29.0 s.
Calculate the magnitude of induced emf in the loop while the magnetic field is increasing.
With the magnetic field held constant a ts its new value of 1.61 T, calculate the magnitude of its induced voltage in the loop while it is pulled horizontally out of the magnetic field region during a time interval of 3.90s.
Answer:
(a) The magnitude of induced emf in the loop while the magnetic field is increasing is 9.5 mV
(b) The magnitude of the induced voltage at a constant magnetic field is 124.7 mV
Explanation:
Given;
radius of the circular loop, r = 31.0 cm = 0.31 m
initial magnetic field, B₁ = 0.7 T
final magnetic field, B₂ = 2.3B₁ = 2.3 X 0.7 T = 1.61 T
duration of change in the field, t = 29
(a) The magnitude of induced emf in the loop while the magnetic field is increasing.
[tex]E = A*\frac{\delta B}{\delta t} \\\\[/tex]
[tex]E = A*\frac{B_2 -B_1}{\delta t}[/tex]
Where;
A is the area of the circular loop
A = πr²
A = π(0.31)² = 0.302 m²
[tex]E = A*\frac{B_2 -B_1}{\delta t} \\\\E = 0.302*\frac{1.61-0.7}{29} \\\\E = 0.0095 \ V\\\\E = 9.5 \ mV[/tex]
(b) the magnitude of the induced voltage at a constant magnetic field
E = A x B/t
E = (0.302 x 1.61) / 3.9
E = 0.1247 V
E = 124.7 mV
Therefore, the magnitude of the induced voltage at a constant magnetic field is 124.7 mV
A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
How could a country benefit from making it into space?
Answer:
space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held?a. The tube should be held horizontally, parallel to the ground b. The potential difference between the ends of the tube does not depend on the tube's orientation. c. The tube should be held vertically perpendicular to the ground
Answer:
b) True. potencial diferencie does not depend on orientation
Explanation:
In this exercise we are asked to show which statements are true.
The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.
It does not refer to the height of the system.
We can now review the claims
a) False. Potential not to be refers to height
b) True. Does not depend on orientation
c) False The potential does not refer to the altitude but to the Earth's charge
An electric heater draws 13 amperes of current when connected to 120 volts. If the price of electricity is $0.10/kWh, what would be the approximate cost of running the heater for 8 hours?
(A) $0.19
(B) $0.29
(C) $0.75
(D) $1.25
(E) $1.55
Answer:
C $0.75 my friend I wish it is right answer
It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?
Answer:
Explanation:
Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/
α = 26.56°
Laser angle with Normal = 90 - 26.56 = 63.44 °
Assuming a red laser, refractive index in water is 1.331.
Angle of refraction in water is given by:
Ref Ind = Sin i / Sin r
1.331 = Sin 63.44 / Sin r
Sin r = 0.8945 / 1.331 = 0.6721
Angle r = 42.22°
For the path in water:
Tan 42.22 = x / 3.2
x = 2.9m where x is the lateral displacement of the laser ince it hits the water
So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool
A mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 6 feet below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to 1 2 the instantaneous velocity. Find the equation of motion x(t) if the mass is driven by an external force equal to f(t)
Answer:I don’t know
Explanation:
A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?
Answer:
A. 4.82 cm
B. 24.66 cm
Explanation:
The depth of water = 19.6 cm
Distance of fish = 6.40 cm
Index of refraction of water = 1.33
(A). Now use the below formula to compute the apparent depth.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]
(B). the depth of the fish in the mirror.
[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]
Now find the depth of reflection of the fish in the bottom of the tank.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]
A radiation worker is subject to a dose of 200 mrad/h of maximum QF neutrons for one 40 h work week. How many times the yearly allowable effective dose did she receive?
Answer:
16 times.
Explanation:
The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr
Time consumed, t = 40 hr
The magnitude of Q.F for the neutrons, Q.F = 2
Thus the effective radiation dose is:
[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]
Thus, the effective dose allowable yearly = 16 times
Complete each of the statements
A. Lines of force are lines used to represent ________ an ________ electric field
B. The intensity of an electric field is the coefficient between the _________ that in the field exerts on a test ___________ located at that point and the value of said charge
C. The electric field is uniform if at any point in the field its _________ and ________ is the same
D. The van der graff generator is a _________ machine which has two __________ that are driven by a _________ that generates a rotation
Answer:
A: magnitude and direction
B: Force that the field exerts on a test charge
C: its magnitude and direction is the same.
D: electrostatic machine
two rollers that are driven by a motor that generates a rotation
Explanation:
NASA is doing research on the concept of solar sailing. A solar sailing craft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion.
A) Should the sail be absorptive or reflective? Why?
B)The total power output of the sun is 3.90 × 1026 W . How large a sail is necessary to propel a 1.06 × 104 kg spacecraft against the gravitational force of the sun?
Answer:
A = 6.8 km²
Explanation:
A) The sail should be reflective. This is so that, it can produce the maximum radiation pressure.
B) let's begin with the formula used to calculate the average solar sail in orbit around the sun. Thus;
F_rad = 2IA/c
I is given by the formula;
I = P/(4πr²)
Thus;
F_rad = (2A/c) × (P/(4πr²)) = PA/2cπr²
Where;
A is the area of the sail
r is the distance of the sail from the sun
c is the speed of light = 3 × 10^(8) m/s
P is total power output of the sun = 3.90 × 10^(26) W
Now,F_rad = F_g
Where F_g is gravitational force.
Thus;
PA/2cπr² = G•m•M_sun/r²
r² will cancel out to givw;
PA/2cπ = G•m•M_sun
Making A the subject, we have;
A = (2•c•π•G•m•M_sun)/P
Now, m = 1.06 × 10⁴ kg and M_sun has a standard value of 1.99 × 10^(30) kg
G is gravitational constant and has a value of 6.67 × 10^(-11) Nm²/kg²
Thus;
A = (2 × 3 × 10^(8) × π × 6.67 × 10^(-11) × 1.06 × 10^(4) × 1.99 × 10^(30))/(3.90 × 10^(26))
A = 6.8 × 10^(6) m² = 6.8 km²
"A satellite requires 88.5 min to orbit Earth once. Assume a circular orbit. 1) What is the circumference of the satellites orbit
Answer:
circumference of the satellite orbit = 4.13 × 10⁷ m
Explanation:
Given that:
the time period T = 88.5 min = 88.5 × 60 = 5310 sec
The mass of the earth [tex]M_e[/tex] = 5.98 × 10²⁴ kg
if the radius of orbit is r,
Then,
[tex]\dfrac{V^2}{r} = \dfrac{GM_e}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e r}{r^2}[/tex]
[tex]{V^2} = \dfrac{GM_e }{r}[/tex]
[tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Similarly :
[tex]T = \sqrt{\dfrac{ 2 \pi r} {V} }[/tex]
where; [tex]{V} =\sqrt{ \dfrac{GM_e }{r}}[/tex]
Then:
[tex]T = {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {GM_e }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ {6.674\times 10^{-11} \times 5.98 \times 10^{24} }} }[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {\sqrt{ 3.991052 \times 10^{14} }}[/tex]
[tex]5310= {\dfrac{ 2 \pi r^{3/2}} {19977617.48}[/tex]
[tex]5310 \times 19977617.48= 2 \pi r^{3/2}}[/tex]
[tex]1.06081149 \times 10^{11}= 2 \pi r^{3/2}}[/tex]
[tex]\dfrac{1.06081149 \times 10^{11}}{2 \pi}= r^{3/2}}[/tex]
[tex]r^{3/2}} = \dfrac{1.06081149 \times 10^{11}}{2 \pi}[/tex]
[tex]r^{3/2}} = 1.68833392 \times 10^{10}[/tex]
[tex]r= (1.68833392 \times 10^{10})^{2/3}}[/tex]
[tex]r= 2565.38^2[/tex]
r = 6579225 m
The circumference of the satellites orbit can now be determined by using the formula:
circumference = 2π r
circumference = 2π × 6579225 m
circumference = 41338489.85 m
circumference of the satellite orbit = 4.13 × 10⁷ m
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work. This heat engine violates the second law but not the first law of thermodynamics. Why is this true?
Answer:
It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J
But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law
Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?
a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100
Answer:
f. 80 and 90
Explanation:
1 x 10⁻¹² W/m² sound intensity falls within 0 sound level
1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level
1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level
1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level
1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level
1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level
1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level
1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level
1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level
1 x 10⁻³ W/m² sound intensity falls within 90 sound level
Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.
f. 80 and 90
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses. But he loses them while travelling. Fortunately he has his old pair as a spare. (a) If the lenses of the old pair have a power of 2.25 diopters, what is his near point (measured from the eye) when wearing the old glasses, if they rest 2.0 cm in front of the eye
Answer:
30.93 cm
Explanation:
Given that:
A person with a near point of 85 cm, but excellent distance vision normally wears corrective glasses
The power of the old pair of lens p = 2.25 diopters
The focal point length = 1/p
The focal point length = 1/2.25
The focal point length = 0.444 m
The focal point length = 44.4 cm
The near point of the person from the glass = (85 -2)cm , This is because the glasses are usually 2 cm from the lens
The near point of the person from the glass = 83 cm
Let consider s' to be the image on the same sides of the lens,
∴ s' = -83 cm
We known that:
the focal length of a mirror image 1/f =1/u +1/v
Assume the near point is at an excellent distance s from the glass where the person wears the corrective glasses.
Then:
1/f = 1/s + 1/s'
1/s = 1/f - 1/s'
1/s = (s' -f)/fs'
s = fs'/(s'-f)
s =( 44.4× -83)/(-83 - 44.4)
s = - 3685.2 / - 127.4
s = 28.93 cm
Thus , the near distance point measured from the eye wearing the old glasses, if they rest 2.0 cm in front of the eye = (28.93 +2.0)cm
= 30.93 cm
A) Hooke's law is described mathematically using the formula Fsp = -ku. Which statement is correct about the spring force, Fsp?
A.It is a vector quantity
B.It is the force doing the push or pull,
C.It is always a positive force.
D.It is larger than the applied force.
1. Which example best describes a restoring force?
B) the force applied to restore a spring to its original length
2. A spring is compressed, resulting in its displacement to the right. What happens to the spring when it is released?
C) The spring exerts a restoring force to the left and returns to its equilibrium position.
3. A 2-N force is applied to a spring, and there is displacement of 0.4 m. How much would the spring be displaced if a 5-N force was applied?
D) 1 m
4. Hooke’s law is described mathematically using the formula Fsp=−kx. Which statement is correct about the spring force, Fsp?
D)It is a vector quantity.
5. What happens to the displacement vector when the spring constant has a higher value and the applied force remains constant?
A) It decreases in magnatude.
Hope this Helps!! Sorry its late
What is the answer?
Answer: i think it is d. none of them.
Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.
A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer
3.340J
Explanation;
Using the relation. Energy stored in capacitor = U = 7.72 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.
Initial separation between the plates =d1= 3.30mm .
Final separation = d2 = 1.45 mm
(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same
Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy *d2/d1
Final energy = Uf =7.72*1.45/3.30
(A) Final energy = Uf = 3.340J
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
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
6. If you wanted to develop a telescope, what kind of lenses would you use for the objective lens (the lens that collects the light) and the eyepiece? Explain your reasoning. Draw a picture with ray tracing of your setup.
Answer:
objetive: a converging lens for large diameter lenses
eyepiece you must select a lens with a small focal length and the diameter is not important
The selected lenses should decrease chromatic aberration.
Explanation:
A telescope is an instrument that collects light from very distant objects, therefore very weak.
Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.
For the eyepiece you must select a lens with a small focal length and the diameter is not important
the telescope magnification is
m = f_objective / F_ocular
The selected lenses should decrease chromatic aberration.
In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.
Explanation:
A 137 kg horizontal platform is a uniform disk of radius 1.53 m and can rotate about the vertical axis through its center. A 68.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 25.9 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
The moment of inertia is [tex]I= 312.09 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The mass of the platform is m = 137 kg
The radius is r = 1.53 m
The mass of the person is [tex]m_p = 68.7 \ kg[/tex]
The distance of the person from the center is [tex]d_c =1.19 \ m[/tex]
The mass of the dog is [tex]m_d = 25.9 \ kg[/tex]
The distance of the dog from the person [tex]d_d = 1.45 \ m[/tex]
Generally the moment of inertia of the system is mathematically represented as
[tex]I = I_1 + I_2 + I_3[/tex]
Where [tex]I_1[/tex] is the moment of inertia of the platform which mathematically represented as
[tex]I_1 = \frac{m * r^2}{2}[/tex]
substituting values
[tex]I_1 = \frac{ 137 * (1.53)^2}{2}[/tex]
[tex]I_1 = 160.35 \ kg\cdot m^2[/tex]
Also [tex]I_2[/tex] is the moment of inertia of the person about the axis which is mathematically represented as
[tex]I_2 = m_p * d_c^2[/tex]
substituting values
[tex]I_2 = 68.7 * 1.19^2[/tex]
[tex]I_2 = 97.29 \ kg \cdot m^2[/tex]
Also [tex]I_3[/tex] is the moment of inertia of the dog about the axis which is mathematically represented as
[tex]I_3 = m_d * d_d^2[/tex]
substituting values
[tex]I_3 = 25.9 * 1.45^2[/tex]
[tex]I_3 = 54.45 \ kg \cdot m^2[/tex]
Thus
[tex]I= 160.35 + 97.29 + 54.45[/tex]
[tex]I= 312.09 \ kg \cdot m^2[/tex]
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, and compare that to your feeling now. How have your feeling changed? Are you more comfortable with peer assessment? Have you learned something new while assessing your peer's work?
Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
