Answer:
2035
Explanation:
The doctor does not travel with the woman, and therefore, he won't experience any relativistic effect on his time. The doctor will judge time by the time here on earth. Technically, the last new year's day the doctor, who is here on earth, would expect the woman to celebrate will be in 2020 + 15 years = 2035
If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed
Answer:
The frequency will change by a factor of 0.4
Explanation:
T = 2(pi)*sqrt(L/g)
Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.
A viewing screen is separated from a double slit by 5.20 m. The distance between the two slits is 0.0300 mm. Monochromatic light is directed toward the double slit and forms an interference pattern on the screen. The first dark fringe is 3.70 cm from the center line on the screen.
Required:
a. Determine the wavelength of light.
b. Calculate the distance between the adjacent bright fringes.
Answer:
The wavelength of this light is approximately [tex]427\; \rm nm[/tex] ([tex]4.27\times 10^{-7}\; \rm m[/tex].)The distance between the first and central maxima is approximately [tex]7.40\; \rm cm[/tex] (about twice the distance between the first dark fringe and the central maximum.)Explanation:
WavelengthConvert all lengths to meters:
Separation of the two slits: [tex]0.0300\; \rm mm = 3.00\times 10^{-5}\; \rm m[/tex].Distance between the first dark fringe and the center of the screen: [tex]3.70\; \rm cm = 3.70\times 10^{-2}\; \rm m[/tex].Refer to the diagram attached (not to scale.) Assuming that the screen is parallel to the line joining the two slits. The following two angles are alternate interior angles and should be equal to each other:
The angle between the filter and the beam of light from the lower slit, andThe angle between the screen and that same beam of light.These two angles are marked with two grey sectors on the attached diagram. Let the value of these two angles be [tex]\theta[/tex].
The path difference between the two beams is approximately equal to the length of the segment highlighted in green. In order to produce the first dark fringe from the center of the screen (the first minimum,) the length of that segment should be [tex]\lambda / 2[/tex] (one-half the wavelength of the light.)
Therefore:
[tex]\displaystyle \cos \theta \approx \frac{\text{Path difference}}{\text{Slit separation}} = \frac{\lambda / 2}{3.00\times 10^{-5}\; \rm m}[/tex].
On the other hand:
[tex]\begin{aligned} \cot \theta &\approx \frac{\text{Distance between central peak and first minimum}}{\text{Distance between the screen and the slits}} \\ &= \frac{3.70\times 10^{-2}\; \rm m}{5.20\; \rm m} \approx 0.00711538\end{aligned}[/tex].
Because the cotangent of [tex]\theta[/tex] is very close to zero,
[tex]\cos \theta \approx \cot \theta \approx 0.00711538[/tex].
[tex]\displaystyle \frac{\lambda /2}{3.00\times 10^{-5}\; \rm m} \approx \cos\theta\approx 0.00711538[/tex].
[tex]\begin{aligned}\lambda &\approx 2\times 0.00711538 \times \left(3.00\times 10^{-5}\; \rm m\right) \\ &\approx 4.26 \times 10^{-7}\; \rm m = 426\; \rm nm\end{aligned}[/tex].
Distance between two adjacent maximaIf the path difference is increased by one wavelength, then the intersection of the two beams would move from one bright fringe to the next one.
The path difference required for the central maximum is [tex]0[/tex].The path difference required for the first maximum is [tex]\lambda[/tex].The path difference required for the second maximum is [tex]2\,\lambda[/tex].On the other hand, if the distance between the maximum and the center of the screen is much smaller than the distance between the screen and the filter, then:
[tex]\begin{aligned}&\frac{\text{Distance between image and center of screen}}{\text{Distance between the screen and the slits}} \\ &\approx \cot \theta \\ &\approx \cos \theta \\ &\approx \frac{\text{Path difference}}{\text{Slit separation}}\end{aligned}[/tex].
Under that assumption, the distance between the maximum and the center of the screen is approximately proportional to the path difference. The distance between the image (the first minimum) and the center of the screen is [tex]3.70\; \rm cm[/tex] when the path difference is [tex]\lambda / 2[/tex]. The path difference required for the first maximum is twice as much as that. Therefore, the distance between the first maximum and the center of the screen would be twice the difference between the first minimum and the center of the screen: [tex]2 \times 3.70\; \rm cm = 7.40\; \rm cm[/tex].
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to
Answer:
Ok, the question is incomplete buy ill try to answer this in a general way.
Suppose that you have no-polarized light.
When that light hits one polaroid, the light becomes polarized along some line, and has an intensity I0.
Now, when polarized light hits a polaroid which axis is at an angle θ with respect to the polarization of the light, the intensity of the resulting beam is given by the Malus's law:
I(θ) = I0*cos^2(θ)
For example, if the axis of the polaroid is exactly the same as the one of the polarized light, then we have θ = 0°
and:
I(0°) = I0*cos^2(0°) = I0
So the intensity does not change.
Now, knowing the initial intensity, you can find the angle needed to get a given intensity.
For example, if the question was:
"At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to A"
We should solve:
I(θ) = A = I0*cos^2(θ)
(A/i0) = cos^2(θ)
√(A/I0) = cos(θ)
Acos(√(A/I0)) = θ
You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?
Answer:
Here,
v (final velocity) = 0
u (initial velocity) = u
a = ?
s = 1m
t = 0.4s
using the first equation of motion,
0 = u + 0.4a
= -0.4a = u
using the second equation of motion:
1 = 0.4u + 0.08a
from the bold equation
1 = 0.4(-0.4a) + 0.08a
1 = -0.16a + 0.08a
1 = -0.08a
a = -1/0.08
a = -100/8
a = -12.5 m/s/s
please make me brainly, i am 1 brainly away from the next rank
Why was Bohr's atomic model replaced by the
modern atomic model?
Answer:
Explanation:
Bohr's atomic model was replaced by the modern atomic model because of its limitations, which included :
(a) Only applicable for Hydrogen and like atoms ( He+1, Li+2 )
(b) Couldn't explain Zeeman Effect (splitting of spectral lines due external magnetic field ) and Stark Effect (splitting of spectral lines due to external electric field).
(c) Inconsistent with De-Broglie's Dual nature of matter and Heisenberg Uncertainty principal, etc.
A circular coil of wire 8.40 cm in diameter has 17.0 turns and carries a current of 3.20 A . The coil is in a region where the magnetic field is 0.610 T.Required:a. What orientation of the coil gives the maximum torque on the coil ?b. What is this maximum torque in part (A) ?c. For what orientation of the coil is the magnitude of the torque 71.0 % of the maximum found in part (B)?
Answer:
a) for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
b) therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
c) Given the torque is 71.0% of its maximum value; Ф = 45.24⁰ ≈ 45⁰
Explanation:
Given that; Diameter is 8.40 cm,
Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m
Number of turns (N) = 17
Current in the loop (I) = 3.20 A
Magnetic field (B) = 0.610 T
Let the angle between the loop's area vector A and the magnetic field B be
Now. the area of the loop is;
A = πR²
A = 3.14 ( 0.042 )²
A = 0.005539 m²
Torque on the loop (t) = NIABsinФ
t = 17 × 3.20 ×0.005539 × 0.610 × sinФ
t = 0.1838sinФ N.m
for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
Given the torque is 71.0% of its maximum value
t = 0.71 × tmax
t = 0.71 × 0.1838
t = 0.1305
Now
0.1305 N.m = 0.1838 sinФ N.m
sinФ = 0.1305 / 0.1838
sinФ = 0.71001
Ф = sin⁻¹ 0.71001
Ф = 45.24⁰ ≈ 45⁰
What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with a wavelength of 580 nm
Answer:
Explanation:
In case of soap film , light gets reflected from denser medium , hence interference takes place between two waves , one reflected from upper and second from lower surface . For destructive interference the condition is
2μt = nλ where μ is refractive index of water , t is thickness , λ is wavelength of light and n is an integer .
2 x 1.34 x t = 1 x 580
t = 216.42 nm .
Thickness must be 216.42 nm .
Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).
Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answer:
a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².
Explanation:
a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.
τ = Frsinθ
= 185 N × 1.73 m × sin90°
= 320.05 Nm
So, the torque τ = 320.05 Nm
b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m
α = angular acceleration of grindstone
τ = Iα
α = τ/I = τ/(MR²/2) = 2τ/MR²
substituting the values of the variables, we have
α = 2τ/MR²
= 2 × 320.05 Nm/[76 kg × (1.73 m)²]
= 640.1 Nm/227.4604 kgm²
= 2.814 rad/s²
So, the angular acceleration α = 2.814 rad/s²
c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.
So τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm
The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm
Since τ'' = Iα
α' = τ''/I where α' = its new angular acceleration
α' = 2τ/MR²
= 2 × 319.75 Nm/[76 kg × (1.73 m)²]
= 639.5 Nm/227.4604 kgm²
= 2.811 rad/s²
So, the angular acceleration α' = 2.811 rad/s²
Somebody please help it’s urgent!!!!
In the tug of war game, none of the teams won. What can you conclude about the forces of the two teams ? Write all the evidence to support your answer.
Answer:
Explanation:
We can conclude that the forces of the two teams are equal and opposite and hence they cancel each other. Therefore none of the teams won as the rope did not move.
hope this helps
plz mark as brainliest!!!!!!!
What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 566 nm light? The index of refraction of the coating material is 1.46 and the index of the glass is 1.71.
Answer:
The thickness is [tex]t = 1.415 *10^{-7 } \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 566 \ nm = 566 *10^{-9} \ m[/tex]
The index of refraction of glass is [tex]n_g = 1.71[/tex]
The index of refraction of the coating is [tex]n= 1.46[/tex]
Generally the condition for destructive interference is
[tex]2 t = (m + \frac{1}{2} ) * \frac{\lambda }{n }[/tex]
Here m is the order of the interference pattern and given from the question that we are considering minimizing reflection m = 0
t = thickness of the coating
substituting values
[tex]2 t = (0 + \frac{1}{2} ) * \frac{ 566 *10^{-9}}{ 1.46 }[/tex]
=> [tex]t = 1.415 *10^{-7 } \ m[/tex]
A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?
Answer:
(a) nearsighted
(b) diverging
(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D
Explanation:
(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.
(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)
(c) the absolute value of the focal length (f) is given by the formula:
[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]
So it would normally be written with a negative signs in front indicating a divergent lens.
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
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
What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops
Answer:
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force
Explanation:
galvanometer is an instrument that can detect and measure small current in an electrical circuit.
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.
The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.
What happens to the magnetic field when you reverse the direction of current by sliding the battery voltage bar past 0 volts
Answer:
The polarity of the magnetic field changes
Explanation:
This because The magnetic field generated is always perpendicular to the direction of the current and parallel to the solonoid. Hence if we reverse the current the direction of magnetism also reverses. In other words the magnetic poles gets reversed (North pole becomes south pole and the south pole becomes the north pole)
A string of holiday lights has 15 bulbs with equal resistances. If one of the bulbs
is removed, the other bulbs still glow. But when the entire string of bulbs is
connected to a 120-V outlet, the current through the bulbs is 5.0 A. What is the
resistance of each bulb?
Answer:
Resistance of each bulb = 360 ohms
Explanation:
Let each bulb have a resistance r .
Since, even after removing one of the bulbs, the circuit is closed and the other bulbs glow. Therfore, the bulbs are connected in Parallel connection.
[tex] \frac{1}{r(equivalent)} = \frac{1}{r1} + \frac{1}{r2} + + + + \frac{1}{r15} [/tex]
[tex] \frac{1}{r(equivalent)} = \frac{15}{r} [/tex]
R(equivalent) = r/15
Now, As per Ohms Law :
V = I * R(equivalent)
120 V = 5 A * r/15
r = 360 ohms
Adjust the mass of the refrigerator by stacking different objects on top of it. If the mass of the refrigerator is increased (with the Applied Force held constant), what happens to the acceleration
Answer:
The acceleration of the refrigerator together with the objects decreases.
Explanation:
If the mass of the refrigerator is increased by stacking more masses (objects) on it,
and the force applied remains constant, then we know from
F = ma
where
F is the applied force
m is the total mass of the refrigerator and the objects
a is the acceleration of the masses.
If F is constant, and m is increased, the acceleration will decrease
Answer:
The acceleration decreases.
Explanation:
its right
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
A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 from the speaker.
At what rate does this speaker produce energy?
What is the intensity of this sound 9.50 from the speaker?
What is the total amount of energy received each second by the walls (including windows and doors) of the room in which this speaker is located?
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2, separated by a distance of 1.70 mm. A 25.0-V potential difference is applied to these plates. Calculate: a. the electric field between the plates b. the surface charge density c. the capacitance d. the charge on each plate.
Answer:
(a) 1.47 x 10⁴ V/m
(b) 1.28 x 10⁻⁷C/m²
(c) 3.9 x 10⁻¹²F
(d) 9.75 x 10⁻¹¹C
Explanation:
(a) For a parallel plate capacitor, the electric field E between the plates is given by;
E = V / d -----------(i)
Where;
V = potential difference applied to the plates
d = distance between these plates
From the question;
V = 25.0V
d = 1.70mm = 0.0017m
Substitute these values into equation (i) as follows;
E = 25.0 / 0.0017
E = 1.47 x 10⁴ V/m
(c) The capacitance of the capacitor is given by
C = Aε₀ / d
Where
C = capacitance
A = Area of the plates = 7.60cm² = 0.00076m²
ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m
d = 1.70mm = 0.0017m
C = 0.00076 x 8.85 x 10⁻¹² / 0.0017
C = 3.9 x 10⁻¹²F
(d) The charge, Q, on each plate can be found as follows;
Q = C V
Q = 3.9 x 10⁻¹² x 25.0
Q = 9.75 x 10⁻¹¹C
Now since we have found other quantities, it is way easier to find the surface charge density.
(b) The surface charge density, σ, is the ratio of the charge Q on each plate to the area A of the plates. i.e
σ = Q / A
σ = 9.75 x 10⁻¹¹ / 0.00076
σ = 1.28 x 10⁻⁷C/m²
Which statement accurately describes the inner planets? Uranus is one of the inner planets. The inner planets formed when the solar system cooled. The inner planets are also called terrestrial planets. The inner planets are larger than the outer planets.
The correct answer is C. The inner planets are also called terrestrial planets.
Explanation:
Our solar system includes a total of eight planets. Additionally, planets are classified into broad categories including inner planets and outer planets. The inner planets category applies to planets such as Earth, Mercury, or Mars because these are located within the asteroid belt (region of asteroids between Mars and Jupiter). Moreover, inner planets differ from others due to their composition as they are composed of rocks and metals. Also, due to this composition, these are known as terrestrial planets. According to this, the statement that best describes inner planets is "The inner planets are also called terrestrial planets".
Answer:
The answer is c.) The inner planets are also called terrestrial planets.
Explanation:
A deep-space vehicle moves away from the Earth with a speed of 0.870c. An astronaut on the vehicle measures a time interval of 3.10 s to rotate her body through 1.00 rev as she floats in the vehicle. What time interval is required for this rotation according to an observer on the Earth
Answer:
t₀ = 1.55 s
Explanation:
According to Einstein's Theory of Relativity, when an object moves with a speed comparable to speed of light, the time interval measured for the event, by an observer in motion relative to the event is not the same as measured by an observer at rest.
It is given as:
t = t₀/[√(1 - v²/c²)]
where,
t = time measured by astronaut in motion = 3.1 s
t₀ = time required according to observer on earth = ?
v = relative velocity = 0.87 c
c = speed of light
3.1 s = t₀/[√(1 - 0.87²c²/c²)]
(3.1 s)(0.5) = t₀
t₀ = 1.55 s
Answer:
The time interval required for this rotation according to an observer on the Earth = [tex]6.29sec[/tex]Explanation:
Time interval required for this rotation according to an observer on the Earth is given as [tex]\delta t[/tex]
where,
[tex]t_o = 3.1\\\\v = 0.87[/tex]
[tex]\delta t = \frac{t_o}{\sqrt{1-\frac{v^2}{c^2}}}\\\\\delta t = \frac{3.1}{\sqrt{1-(\frac{0.87c}{c})^2}}\\\\\delta t = 6.29sec[/tex]
For more information visit
Exercise 2.4.5: Suppose we add possible friction to Exercise 2.4.4. Further, suppose you do not know the spring constant, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz, for the 2 kg weight you measured 0.8 Hz. a) Find k (spring constant) and c (damping constant). Find a formula for the mass in terms of the frequency in Hz. Note that there may be more than one possible mass for a given frequency. b) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know that the mass of the unknown object is more than a kilogram.
Answer:
a) k = 95.54 N / m, c = 19.55 , b) m₃ = 0.9078 kg
Explanation:
In a simple harmonic movement with friction, we can assume that this is provided by the speed
fr = -c v
when solving the system the angular value remains
w² = w₀² + (c / 2m)²
They give two conditions
1) m₁ = 1 kg
f₁ = 1.1 Hz
the angular velocity is related to frequency
w = 2π f₁
Let's find the angular velocity without friction is
w₂ = k / m₁
we substitute
(2π f₁)² = k / m₁ + (c / 2m₁)²
2) m₂ = 2 kg
f₂ = 0.8 Hz
(2π f₂)² = k / m₂ + (c / 2m₂)²
we have a system of two equations with two unknowns, so we can solve it
we solve (c / 2m)² is we equalize the expression
(2π f₁)² - k / m₁ = (2π f₂²) 2 - k / m₁
k (1 / m₂ - 1 / m₁) = 4π² (f₂² - f₁²)
k = 4π² (f₂² -f₁²) / (1 / m₂ - 1 / m₁)
a) Let's calculate
k = 4 π² (0.8² -1.1²) / (½ -1/1)
k = 39.4784 (1.21) / (-0.5)
k = 95.54 N / m
now we can find the constant of friction
(2π f₁) 2 = k / m₁ + (c / 2m₁)²
c2 = ((2π f₁)² - k / m₁) 4m₁²
c2 = (4ππ² f₁² - k / m₁) 4 m₁²
let's calculate
c² = (4π² 1,1² - 95,54 / 1) 4 1²
c² = (47.768885 - 95.54) 8
c² = -382.1689
c = 19.55
b) f₃ = 0.2 Hz
m₃ =?
(2πf₃)² = k / m₃ + (c / 2m₃) 2
we substitute the values
(4π² 0.2²) = 95.54 / m₃ + 382.1689 2/4 m₃²
1.579 = 95.54 / m₃ + 95.542225 / m₃²
let's call
x = 1 / m₃
x² = 1 / m₃²
- 1.579 + 95.54 x + 95.542225 x² = 0
60.5080 x² + 60.5080 x -1 = 0
x² + x - 1.65 10⁻² = 0
x = [1 ±√ (1- 4 (-1.65 10⁻²)] / 2
x = [1 ± 1.03] / 2
x₁ = 1.015 kg
x₂ = -0.015 kg
Since the mass must be positive we eliminate the second results
x₁ = 1 / m₃
m₃ = 1 / x₁
m₃ = 1 / 1.1015
A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i v, start subscript, i, end subscript. Later, its speed increases to 4v_i4v i 4, v, start subscript, i, end subscript. How does the magnitude of the car's centripetal acceleration change after the linear speed increases
Answer:
The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Explanation:
The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.
When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.
So the centripetal acceleration, a' = 16v²/R.
To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16
a'/a = 16
a' = 16a.
So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the wavelength (in cm) of the first harmonic?
Answer:
200cm
Explanation:
Answer:
100cm
Explanation:
Using
F= ( N/2L)(√T/u)
F1 will now be (0.5*2)( √600/0.015)
=> L( wavelength)= 200/2cm = 100cm
A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no appreciable resistance, and a switch. If the emf across the inductor is 80% of its maximum value 4 s after the switch is closed, what is the resistance of the resistor
Answer: R = 394.36ohm
Explanation: In a LR circuit, voltage for a resistor in function of time is given by:
[tex]V(t) = \epsilon. e^{-t.\frac{L}{R} }[/tex]
ε is emf
L is indutance of inductor
R is resistance of resistor
After 4s, emf = 0.8*19, so:
[tex]0.8*19 = 19. e^{-4.\frac{22}{R} }[/tex]
[tex]0.8 = e^{-\frac{88}{R} }[/tex]
[tex]ln(0.8) = ln(e^{-\frac{88}{R} })[/tex]
[tex]ln(0.8) = -\frac{88}{R}[/tex]
[tex]R = -\frac{88}{ln(0.8)}[/tex]
R = 394.36
In this LR circuit, the resistance of the resistor is 394.36ohms.
A vertical spring stretches 3.8 cm when a 13-g object is hung from it. The object is replaced with a block of mass 20 g that oscillates in simple harmonic motion. Calculate the period of motion.
Answer:
The period of motion is 0.5 second.
Explanation:
Given;
extension of the spring, x = 3.8 cm = 0.038 m
mass of the object, m = 13 g = 0.013 kg
Determine the force constant of the spring, k;
F = kx
k = F / x
k = mg / x
k = (0.013 x 9.8) / 0.038
k = 3.353 N/m
When the object is replaced with a block of mass 20 g, the period of motion is calculated as;
[tex]T = 2\pi\sqrt{\frac{m}{k} } \\\\T = 2\pi\sqrt{\frac{0.02}{3.353} } \\\\T = 0.5 \ second[/tex]
Therefore, the period of motion is 0.5 second.
