Answer:
Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m
Explanation:
From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month
Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds
Now, the square of the period of any planet is given by the formula;
T² = (4π²/GM)α³
Where;
G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²
M is mass of sun which has a value of 1.99 x 10^(30) kg
α is the semi major axis of the comets orbit.
Thus, making α the subject, we have;
α = ∛(GMT²/4π²)
So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)
α = 1.89 x 10^(13) m
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, O3O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 HzHz . What is the wavelength λλlambda of the radiation absorbed by ozone? Express your answer in nanometers.
λ = ___.
Answer:
λ = 3.2 x 10⁻⁷ m = 320 nm
Explanation:
The relationship between the velocity of electromagnetic waves (UV rays) and the their frequency is:
v = fλ
where,
v = c = speed of the electromagnetic waves (UV rays) = speed of light
c = 3 x 10⁸ m/s
f = frequency of the electromagnetic waves (UV rays) = 9.38 x 10¹⁴ Hz
λ = wavelength of the electromagnetic waves (UV rays) = ?
Therefore, substituting the values in the relation, we get:
3 x 10⁸ m/s = (9.38 x 10¹⁴ Hz)(λ)
λ = (3 x 10⁸ m/s)/(9.38 x 10¹⁴ Hz)
λ = 3.2 x 10⁻⁷ m = 320 nm
So, the radiation of 320 nm wavelength is absorbed by Ozone.
Answer:
0.03052 nm
Explanation:
From c = f¥
Where c = speed of electromagnetic wave in vacuum = 3x10^8 m/s
f = frequency of the wave
¥ = wavelenght of the wave.
¥ = c/f = (3x10^8)/(9.38×10^14)
= 3.052x10^-7 m
= 0.03052 nm
Young's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.2 mm from the center of the central bright fringe. What is the separation of the two slits?
Answer:
Explanation:
wave length of light λ = 502 nm
screen distance D = 1.2 m
width of one fringe = 10.2 mm / 20
= .51 mm
fringe width = λ D / a , a is separation of slits
Puting the values given
.51 x 10⁻³ = 502 x 10⁻⁹ x 1.2 / a
a = 502 x 10⁻⁹ x 1.2 / .51 x 10⁻³
= 1181.17 x 10⁻⁶ m
1.18 x 10⁻³ m
= 1.18 mm .
In this experiment we will observe the magnetic fields produced by a current carrying wire. A long wire is suspended vertically, passing through a horizontal platform. The wire is connected to a power supply, allowing a current of 5.0 Amps to flow. A compass is placed on the platform stand near the vertical wire. When no current is present in the wire, the compass needle points in the same direction any where around the wire due to the Earth’s magnetic field. When a current flows through the wire, the compass needle deflects in a direction tangent to a circle, which is the direction of the magnetic field created by the current carrying wire.
Answer:
See explanation
Explanation:
Solution:-
Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.
Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.
( See attachments )
- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:
[tex]B = \frac{uo*I}{2\pi *r}[/tex]
Where,
I : The current,
r : The shortest distance to the wire,
uo : The permeability of free space. = 4π * 10^-7 T. m/A
- Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.
- The magnetic field of a long straight wire has more implications than one might first suspect. Each segment of current produces a magnetic field like that of a long straight wire, and the total field of any shape current is the vector sum of the fields due to each segment. The formal statement of the direction and magnitude of the field due to each segment is called the Biot-Savart law. Integral calculus is needed to sum the field for an arbitrary shape current. The Biot-Savart law is written in its complete form as:
[tex]B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}[/tex]
Where the integral sums over,
1) The wire length where vector dl = direction of current (in or out of plane)
2) r is the distance between the location of dl and the location at which the magnetic field is being calculated
3) r^ is a unit vector in the direction of r.
Answer:
B
Explanation:
n 38 g rifle bullet traveling at 410 m/s buries itself in a 4.2 kg pendulum hanging on a 2.8 m long string, which makes the pendulum swing upward in an arc. Determine the maximum vertical displacement of the pendulum (height).(g = 9.8 m/s2)
Answer:
68cm
Explanation:
You can solve this problem by using the momentum conservation and energy conservation. By using the conservation of the momentum you get
[tex]p_f=p_i\\mv_1+Mv_2=(m+M)v[/tex]
m: mass of the bullet
M: mass of the pendulum
v1: velocity of the bullet = 410m/s
v2: velocity of the pendulum =0m/s
v: velocity of both bullet ad pendulum joint
By replacing you can find v:
[tex](0.038kg)(410m/s)+0=(0.038kg+4.2kg)v\\\\v=3.67\frac{m}{s}[/tex]
this value of v is used as the velocity of the total kinetic energy of the block of pendulum and bullet. This energy equals the potential energy for the maximum height reached by the block:
[tex]E_{fp}=E_{ki}\\\\(m+M)gh=\frac{1}{2}mv^2[/tex]
g: 9.8/s^2
h: height
By doing h the subject of the equation and replacing you obtain:
[tex](0.038kg+4.2kg)(9.8m/s^2)h=\frac{1}{2}(0.038kg+4.2kg)(3.67m/s)^2\\\\h=0.68m[/tex]
hence, the heigth is 68cm
Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the collision is half its speed before the collision. Block B was at rest before the collision. The mass of block A is 7 kg and the mass of block B is 2 kg. What is the minimum initial speed (in m/s) that block A must have for block B to swing through a complete vertical circle?
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The minimum velocity of A is [tex]v_A= 4m/s[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]L = 1m[/tex]
The initial speed of block A is [tex]u_A[/tex]
The final speed of block A is [tex]v_A = \frac{1}{2}u_A[/tex]
The initial speed of block B is [tex]u_B = 0[/tex]
The mass of block A is [tex]m_A = 7kg[/tex] gh
The mass of block B is [tex]m_B = 2 kg[/tex]
According to the principle of conservation of momentum
[tex]m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}[/tex]
Since block B at initial is at rest
[tex]m_A u_A = m_Bv_B + m_A \frac{u_A}{2}[/tex]
[tex]m_A u_A - m_A \frac{u_A}{2} = m_Bv_B[/tex]
[tex]m_A \frac{u_A}{2} = m_Bv_B[/tex]
making [tex]v_B[/tex] the subject of the formula
[tex]v_B =m_A \frac{u_A}{2 m_B}[/tex]
Substituting values
[tex]v_B =\frac{7 u_A}{4}[/tex]
This [tex]v__B[/tex] is the velocity at bottom of the vertical circle just at the collision with mass A
Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of [tex]v__B'[/tex] at the top of the vertical circle
The angular centripetal acceleration would be mathematically represented
[tex]a= \frac{v^2_{B}'}{L}[/tex]
Note that this acceleration would be toward the center of the circle
Now the forces acting at the top of the circle can be represented mathematically as
[tex]T + mg = m \frac{v^2_{B}'}{L}[/tex]
Where T is the tension on the string
According to the law of energy conservation
The energy at bottom of the vertical circle = The energy at the top of
the vertical circle
This can be mathematically represented as
[tex]\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L[/tex]
From above
[tex](T + mg) L = m v^2_{B}'[/tex]
Substitute this into above equation
[tex]\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L[/tex]
[tex]\frac{49 mv_A^2}{16} = T + 5mgL[/tex]
The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero
This is mathematically represented as
[tex]\frac{49 mv_A^2}{16} = 5mgL[/tex]
making [tex]v_A[/tex] the subject
[tex]v_A = \sqrt{\frac{80mgL}{49m} }[/tex]
substituting values
[tex]v_A = \sqrt{\frac{80* 9.8 *1}{49} }[/tex]
[tex]v_A= 4m/s[/tex]
In your research lab, a very thin, flat piece of glass with refractive index 1.30 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. Whenλ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.
1) Use these measurements to calculate the thickness of the glass.
2) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
Answer:
A) The thickness of the glass is 868 nm
B) The wavelength is 3472 nm
Explanation:
We are given;
Refractive index = 1.30
Wavelength = 496 nm
Next wavelength = 386 nm
Now, we need to calculate the thickness of the glass
Using formula for constructive interference which is given as;
2nt = (m + ½)λ
Where;
m is the order of the interference
λ is wavelength
t is thickness
Now, for the first wavelength, we have;
2nt = (m + ½)496 - - - - eq1
for the second wavelength, we have;
2nt = (m + 1 + ½)386
2nt = (m + 3/2)386 - - - - eq2
Thus, combining eq1 and eq2, we have;
(m + ½)496 = (m + 3/2)386
496m + 248 = 386m + 579
496m - 386m = 579 - 248
110m = 331
m = 331/110
m = 3
Put 3 for m in eq 1;
2nt = (3 + ½)496
2nt = 1736
t = 1736/(2 x 1)
t = 868 nm
B) now we need to calculate the longest wavelength.
From earlier, we saw that ;
2nt = (m + ½)λ
Making wavelength λ the subject, we have;
λ = 2nt/(m + ½)
The longest wavelength will be at m = 0
Thus,
λ = (2 x 1 x 868)/(0 + ½)
λ = 3472 nm
A glass lens, nglass=1.52, has a 127 nm thick antireflective film coating one side, nfilm=1.39. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating? Assume that nair=1.00.
Answer:
The wavelength is [tex]\lambda = 706nm[/tex]
Explanation:
From the question we are told that
The refractive index of the glass is [tex]n__{glass}} = 1.52[/tex]
The thickness of film is [tex]D = 127nm = 127*10^{-9}m[/tex]
The refractive index of film [tex]n__{film}} = 1.39[/tex]
The refractive index of air is [tex]n__{air}} = 1.00[/tex]
Generally the thickness of the film can be obtained mathematically from this expression
[tex]D = \frac{\lambda}{4 * n__{film}} }[/tex]
Where [tex]\lambda[/tex] is the wavelength
Making the wavelength the subject of the formula
[tex]\lambda = 4 * n__{film}} * D[/tex]
Substituting values
[tex]\lambda = 4 *1.39 * 127 *10^{-9}[/tex]
[tex]\lambda =7.06 *10^{-7}m = 706nm[/tex]
Answer:
[tex]\lambda = 706.12 nm[/tex]
Explanation:
The optical path length for the reflection of light = [tex]2 n_{film} t[/tex]
For destructive interference, [tex]2 n_{film} t = \frac{\lambda}{2}[/tex]
The thickness of the anti-reflective film = 127 nm
The largest wavelength of the reflected light, [tex]\lambda = 4n_{film} t[/tex]
[tex]\lambda = 4 * 1.39 *127 * 10^{-9}[/tex]
[tex]\lambda = 706.12 * 10^{-9} m\\\lambda = 706.12 nm[/tex]
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initially, the fluid level is such that the block is at a depth d and the tension in the thread is T. Then, the fluid level is decreased such that the depth is 0.5d. What is the tension in the thread when the block is at the new depth?
Answer:
(C) T
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence the tension.
Note: The figure of question is added in the attachment
The tension in the thread when the block is at the new depth is :
(C) T
Buoyant forceThe definition of buoyancy refers to whether something can drift in water or discuss, or the control of water or other fluids to keep water above water, or an idealistic disposition.
The tension T at equilibrium will be equal to the Buoyant force.
The Buoyant force is given by:
Fb = density x acceleration due to gravity x volume displaced
The change in height doesn't affect the Buoyant force and hence, the tension.
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Imagine that an incident light wave encounters a thin film. Some of that light wave will be reflected off of the top surface of the thin film (call this Wave 1), while the rest will be transmitted through the film. Some of this transmitted light (Wave 2) will reflect off the back surface of the film and move back through the thin film and out into the original medium of the incident light wave. If the thin film has a thickness t, then what is the path length difference between Wave 1 and Wave 2?
a) t/4
b) t/2
c) t
d) 2t
Answer
D, 2t
Explanation:
See attached file
"2t" will be the path length difference between Wave 1 and Wave 2. A complete solution is below.
According to the question,
The path difference between Wave 1 and 2 will be:
When [tex]\Theta = 0^{\circ}[/tex],
→ [tex]P.d= \Delta x[/tex]
[tex]= AB+BC[/tex]
hence,
Path difference,
= [tex]t+t[/tex]
= [tex]2t[/tex]
Thus the above answer i.e., "option d" is appropriate.
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A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 Joules. What is the maximum height h reached by the ball? Consider air friction to be negligible.
A. h ≈ 30 m
B. h ≈ 40 m
C. h ≈ 50 m
D. h ≈ 10 m
E. h ≈ 20 m
Given that,
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.
We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,
[tex]mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N[/tex] ........(1)
Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,
[tex]mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m[/tex]
So, the correct option is (E) "h = 20 m".
The maximum height reached by the ball is about 20 meters, as determined by using the conservation of mechanical energy principle, considering that the total mechanical energy at the height of 10 meters was 100 Joules.
Explanation:To solve for the maximum height reached by the ball, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of the ball remains constant in the absence of air friction.
At 10 meters above the ground, the ball has a potential energy (PE) of 50 Joules and a kinetic energy (KE) of 50 Joules. Therefore, the total mechanical energy at that height is:
PE + KE = 50 J + 50 J = 100 J
As the ball rises, its kinetic energy is converted into potential energy until the kinetic energy becomes zero at the maximum height. The total mechanical energy at maximum height will be equal to the potential energy:
PE at maximum height = total mechanical energy = 100 J
Using the formula for gravitational potential energy, PE = mgh (where m is mass, g is the acceleration due to gravity (9.81 m/s²), and h is the height), and knowing that the PE at 10 meters is 50 J, we can find the mass of the ball:
50 J = m * 9.81 m/s² * 10 m
m = 50 J / (9.81 m/s² * 10 m) = 0.51 kg
With the mass of the ball, we can now calculate the maximum height using the total mechanical energy:
100 J = 0.51 kg * 9.81 m/s² * h
h = 100 J / (0.51 kg * 9.81 m/s²) ≈ 20 meters
Therefore, the maximum height h reached by the ball is about 20 meters.
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 14.5 m/s , shatters the glass as it passes through, and leaves the window at 11.5 m/s with no change of direction.
What is the direction of the impulse that the glass imparts to the baseball? Calculate the magnitude of this impulse (a positive number)?
Explanation:
Mass of baseball, m = 0.148 kg
Initial speed of the ball, u = 14.5 m/s
Final speed of the ball, v = 11.5 m/s
After crashing through the pane of a second-floor window, the ball shatters the glass as it passes through, and leaves the window at 11.5 m/s with no change of direction. So, the direction of the impulse that the glass imparts to the baseball is in opposite direction to the direction of the balls path.
The change in momentum of the ball is called impulse. It is given by :
[tex]J=m(v-u)\\\\J=0.148\times (11.5-14.5) \\\\J=-0.444\ kg-m/s\\\\|J|=0.444\ kg-m/s[/tex]
Hence, this is the required solution.
"One possible explanation for a galaxy's type invokes the angular momentum of the protogalactic cloud from which it formed. Suppose a galaxy forms from a protogalactic cloud with a lot of angular momentum. Assuming its type has not changed as a result of other interactions, we'd expect this galaxy to be ______."
Answer:
Spiral galaxy
Explanation:
If a galaxy forms from a protogalactic cloud with a lot of angular momentum, we would expect this galaxy to be a spiral galaxy. This is due to the conservation of angular momentum, where greater angular momentum usually results in a flatter, disk-like shape seen in spiral galaxies. Other factors also affect galaxy formation.
Explanation:In the subject of astrophysics, the angular momentum of the protogalactic cloud has significant implications for the type of galaxy formed. Specifically, if a galaxy forms from a protogalactic cloud with a high angular momentum, we'd expect this galaxy to be a spiral galaxy.
Angular momentum is a property of rotating bodies, determined by the mass, shape, and spin rate of the object. In the context of galaxies, a protogalactic cloud with higher angular momentum is more likely to flatten out into a spinning disk shape, leading to the formation of a spiral galaxy. This is due to the conservation of angular momentum. In contrast, lower angular momentum could lead to the formation of elliptical galaxies, which are less flat and more randomly arranged.
In the realm of galaxy formation and evolution, these are widely accepted ideas, though it's important to note that other factors, like the gas content of the protogalactic cloud and interactions with other galaxies, can also play a major role.
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Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have the same radius R. It is determined that the planets have angular momenta of the same magnitude L about their suns, and that the orbital periods are in the ratio of three to one; i.e., T1 = 3T2. The ratio m1/m2 of the masses of the two planets is
(A) 1
(C) (3)^1/2
(C) 2
(D) 3
(E) 9
Answer:
(D) 3
Explanation:
The angular momentum is given by:
[tex]\vec{L}=\vec{r}\ X \ \vec{p}[/tex]
Thus, the magnitude of the angular momenta of both solar systems are given by:
[tex]L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}[/tex]
where we have taken that both systems has the same radius.
By taking into account that T1=3T2, we have
[tex]L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1[/tex]
but L1=L2=L:
[tex]L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3[/tex]
Hence, the answer is (D) 3
HOPE THIS HELPS!!
If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine the principal stresses in the pipe at point A, which is located on the outer surface of the pipe. Express your answers, separated by a comma, to three significant figures.
The principal stress in the pipe at point A, which is located on the outer surface of the pipe is 132.773 psi
Finding the principal stress:It is given that l = 12 in
The force F = 60 lb
The radiuses are as follows:
outer radius, R = 3.90 in/2
R = 1.95 in
Inner radius, r = 3.65 in/2
r = 1.825 in
The angle between the force applied and the distanced from the axis is 30°
So we get the torque:
T = l×Fsin 30°
T = 12 × 60 × (0.5)
T = 360 lb-in
Now, the angle of rotation or angular displacement ωt is given as
ωt = π(R⁴ - r⁴)/(2R)
ωt = π((1.95 in)⁴ - (1.825 in)⁴)/(2×1.95 in)
ωt = 2.7114 in³
Then the principal stress in the pipe at point A is:
principal stress = T/ωt
principal stress = (360 lb-in)/(2.7114 in³)
principal stress = 132.773 lb/in²
principal stress = 132.773 psi
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Start with light that non polarized of intensity I (initial) It will travel through one polarizer that is set to 90 degrees to the x axis. It will then travel through 6 additional polarizers that decrease in angle size by 15 degrees. Polarizer 1 at 75 degrees Polarizer 2 at 60 degrees Polarizer 3 at 45 degrees Polarizer 4 at 30 degrees Polarizer 5 at 15 degrees Polarizer 6 at 0 degrees. Find the final intensity of light that travels through all polarizers as a percentage of the initial intensity
Answer:
35.35 is the final intensity
Explanation:
See attached file for calculation
Suppose your telescope has perfect mirrors and is limited only by diffraction. If you are viewing with light of 400 nm wavelength, what diameter mirror in a space telescope would have 1.0-km resolution of objects on Mars when viewed from above the earth at a time when the Earth-Mars distance of 80 million kilometers
Answer:
39 m
Explanation:
We are given that
Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
y=1 km=[tex]1000 m[/tex]
1 km=1000 m
Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]
1million km=[tex]10^9 m[/tex]
[tex]sin\theta=\frac{1.22\lambda}{d}[/tex]
[tex]sin\theta=\frac{y}{x}[/tex]
[tex]\frac{y}{x}=\frac{1.22\lambda}{d}[/tex]
[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]
[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]
[tex]d=39.04 m\approx 39 m[/tex]
Diameter of mirror =39 m
To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.
Explanation:This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.
Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.
Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.
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Suppose you had a collection of a large number of hypothetical quantum objects, each of whose individual energy levels were -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. If nearly all of these identical objects were in the ground state, what would be the energies of dark spectral lines in an absorption spectrum if visible white light (1.8 to 3.1 eV) passes through the material? Enter the energies in order of increasing energy, followed by entering 0 in any later boxes for which there is no dark line within the visible spectrum. (That is, if your answers were 1, 2, and 3 eV, you would enter 1 in the first box, 2 in the second box, and 3 in the third box. If your answers were 1 and 2 eV, you would enter 1 in the first box, 2 in the second box, and 0 in the third box. If your answer is just 1 eV, you would enter 1 in the first box, 0 in the second box, and 0 in the third box.) Smallest energy of a dark line: eV Next larger energy of a dark line (or 0): eV Next larger energy of a dark line (or 0): eV Additional Materials
The energy differences between the levels of the quantum objects and the photons in the visible spectrum determine the dark spectral lines in an absorption spectrum.
Explanation:In order to determine the energies of the dark spectral lines in an absorption spectrum, we need to consider the energy differences between the energy levels of the quantum objects and the energy levels of the photons in the visible spectrum. The energy levels of the quantum objects are -4.6 eV, -3.0 eV, -2.1 eV, and -1.6 eV. The range of photon energies for visible light is 1.63 to 3.26 eV. Only photons with energies matching the energy differences between the levels of the quantum objects will be absorbed, resulting in dark lines in the spectrum.
Based on the given information, the dark spectral lines within the visible spectrum would be:
Smallest energy of a dark line: -4.6 eVNext larger energy of a dark line (or 0): -3.0 eVNext larger energy of a dark line (or 0): -2.1 eVLearn more about Energy differences and absorption spectrum here:https://brainly.com/question/31230618
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Having straight ears (E) is dominant over droopy ears
(e). A parent with straight ears is crossed with a parent
with droopy ears. The Punnett square will show the
possible genotypes of the offspring.
What is the genotype of each square?
A:
B:
C:
D:
that is... A. B
................c. D
The Punnett square for the cross between a parent with straight ears and a parent with droopy ears would result in offspring with the genotypes EE, EE, Ee, and Ee.
Explanation:The Punnett square is a tool used in genetics to predict the possible genotypes of offspring based on the genotypes of the parents. In this case, the Punnett square for the cross between a parent with straight ears (genotype EE) and a parent with droopy ears (genotype ee) would look like:
EEeEEEEeEeEe
The genotypes of each square are as follows:
A: EE
B: EE
C: Ee
D: Ee
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The 1.18-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in the horizontal position θ = 0 where the spring is unstretched. If the bar is observed to momentarily stop in the position θ = 46°, determine the spring constant k. For your computed value of k, what is magnitude of the angular velocity of the bar when θ = 30°.
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 = [tex]W_{y}[/tex] / W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
[tex]Em_{f}[/tex] = K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the earth with a speed of 0.99543c relative to the earth. A scientist at rest on the earth's surface measures that the particle is created at an altitude of 43.0 km .
(a) As measured by the scientist, how much time does it take the particle to travel the 45.0 km to the surface of the earth?
(b) Use the length-contraction formula to calculate the distance from where the particle is created to the surface of the earth as measured in the particle’s frame.
(c) In the particle’s frame, how much time does it take the particle to travel from where it is created to the surface of the earth? Calculate this time both by the time dilation formula and from the distance calculated in part (b). Do the two results agree?
Answer:
Check attachment
The question have two distance
I decided to use the one in the question "a" in attachment and I will use the other one here
Explanation:
Given that,
Speed of particle relative to the earth is
V = 0.99543c
Where c is speed of light
c = 3 × 10^8 m/s
Particle height as detected by scientist is 43km
The initial length is 43km
Lo = 43km
Lo = 43,000m
A. Time taken for the particle to reach the earth surface?
Speed = distance / Time
Time = distance / speed
t = L / V
t = 43,000 / 0.99543c
t = 43,000 / (0.99543 × 3 × 10^8)
t = 1.4399 × 10^-4 seconds
b. Initial Lenght is given as Lo = 43km
Using length contraction formula
L = Lo√(1 — u² / c²)
L = 43√[1 — (0.99543c)² / c²]
L = 43√[1 — 0.990881c² / c²]
L = 43√[1 — 0.990881]
L = 43 × √(9.1191 × 10^-3)
L = 43 × 0.095494
L = 4.1062 km
c. Using time dilation formula
∆to = ∆t√(1 — u² / c²)
∆t is gotten from question a
∆t = 1.4399 × 10^-4 seconds
∆to = ∆t√[1 — (0.99543c)² / c²]
∆to = ∆t√[1 — 0.990881c² / c²]
∆to = ∆t√[1 — 0.990881]
∆to = ∆t × √(9.1191 × 10^-3)
∆to = 1.4399 × 10^-4 × 0.095494
∆to = 1.375 × 10^-5 seconds
To check if the time dilation agree
t = L / V
t = 4.1062 × 1000 / 0.99543c
t = 4.1062 × 1000 / 0.99543 × 3 × 10^8
t = 1.375 × 10^-5 seconds
The time dilation agreed
A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.
(a) How much time does it take a pulse to travel the full length of the string?
(b) What is the weight W?
(c) How many wavelengths are on the string at any instant of time?
(d) What is the equation for waves traveling ?down? the string?
The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
Answer:
A) 0.0534 seconds
B) 0.67N
C) 41
D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
Explanation:
we are given weight of string = 0.0125N
Thus, since weight = mg
Then, mass of string = 0.0125/9.8
Mass of string = 1.275 x 10⁻³ kg
Length of string; L= 1.5 m .
mass per unit length; μ = (1.275 x 10⁻³)/1.5
μ = 0.85 x 10⁻³ kg/m
We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
Now if we compare it to the general equation of motion of standing wave on a string which is:
y(x,t) = Acos(Kx − ω t)
We can deduce that
angular velocity;ω = 4830 rad/s
Wave number;k = 172 rad/m
A) Velocity is given by the formula;
V = ω/k
Thus, V = 4830/172 m/s
V = 28.08 m /s
Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds
B) We know that in strings,
V² = F/μ
Where μ is mass per unit length and V is velocity.
Thus, F = V²*μ =28.08² x 0.85 x 10⁻³
F = 0.67N
C) Formula for wave length is given as; wave length;λ = 2π /k
λ = 2 x π/ 172
λ = 0.0365 m
Thus, number of wave lengths over whole length of string
= 1.5/0.0365 = 41
D) The equation for waves traveling down the string
= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)
The required time is: 0.395 s. Weight and wavelengths are 0.013 and 1.5. The equation for waves traveling down the string is: y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
a) To find the time it takes for a pulse to travel the full length of the string, we need to determine the velocity of the wave. The velocity of a wave on a string is given by the equation [tex]v = \sqrt{(T/\mu)[/tex], where T is the tension and μ is the linear mass density of the string.
Given that the weight is 0.0125 N and the length is 1.50 m.
Here, mass = w/g = 0.0125/9.8 = 1.275 x 10⁻³ kg
we can calculate μ as follows: μ = m/L = (1.275 x 10⁻³)/(1.50 m) = 0.00085 kg/m.
Substituting the values into the equation:
we get v = [tex]\sqrt{W/0.00085}[/tex]. As it is given that the tension of the string is constant and equal to W.
Thus, v = [tex]\sqrt{14.7}[/tex] = 3.8 m/s
The time it takes for a pulse to travel the full length of the string can be calculated using the formula time = distance/velocity = 1.50/3.8. So, time = 0.395 s.
b) In case of string, V² = F/μ
μ x V² = F
This implies:
F = 0.00085 x [tex](3.8)^2[/tex] = 0.0123
C) We know, f (frequency) = 1/t = 1/0.395 = 2.53
Also, v = λf. This implies:
λ = v/f = 3.8/2.53 = 1.5
D) Finally, to find the equation for waves traveling down the string, we need to adjust the sign inside the cosine function to account for the wave traveling in the opposite direction. It is given that:
y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)
So, the equation becomes y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
Complete Question:
Question
A 1.50-m string of weight 0.0125 N is tied to the ceil-ing at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation. Assume that the tension of the string is constant and equal to W.
y(x, t) = (8.50 mm)cos(172 rad/m)x+4830 rad/s)t).
(a) How much time does it take a pulse to travel the full length of the string?
(b) What is the weight W?
(c) How many wavelengths are on the string at any instant of time?
(d) What is the equation for waves traveling ?down? the string?
Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two pumping stations are located at the same elevation 13 km apart. The pressure drop between the stations is 1.4 MPa. The diameter of the pipe is 0.6 m and has a roughness of ε = 0.12 mm. Determine the volume flow rate.
Answer:
1.0416 m∧3/sec
Explanation:
check the pictures below for the solution
The coil in a 60-Hz ac generator has 125 turns, each having an area of 3.0 × 10-2 m2 and is rotated in a uniform 0.12-T magnetic field. What is the peak output voltage of this generator?
Answer:
Explanation:
Given that, .
Frequency
f = 60Hz
Number of turns
N = 125turns
Surface area of coil
A = 3 × 10^-2 m²
Magnetic field
B = 0.12T
Voltage peak to peak? I.e the EMF
EMF is given as
ε = —dΦ/dt
Where Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Where N is number of turns
B is magnetic field
A is the cross sectional area
And θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, your magnetic flux becomes
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
I.e Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
The peak to peak voltage is 169.65 V
The required value of peak output voltage is 169.65 Volts.
Given data:
The frequency of ac generator is, f = 60 Hz.
The number of turns of coil is, n = 125 turns.
The area of each coil is, [tex]A =3.0 \times 10^{-2} \;\rm m^{2}[/tex].
The strength of magnetic field is, B = 0.12 T.
To start with this problem, we need to find the peak emf first. The peak output voltage is nothing but the value of this peak emf only. The expression for the peak EMF is given as,
ε = —dΦ/dt
here,
Φ is magnetic flux and it is given as
Φ = NBA Cosθ
Here,
θ is the resulting angle from the dot product of area and magnetic field
Where θ =ωt and ω = 2πf
Then, θ = 2πft
So, the expression for the magnetic flux becomes,
Φ = NBA Cos(2πft)
Now, dΦ / dt = —NBA•2πf Sin(2πft)
dΦ / dt = —2πf • NBA Sin(2πft)
So, ε = —dΦ/dt
Then,
ε = 2πf • NBA Sin(2πft)
So, the maximum peak to peak emf will occur when the sine function is 1
Sin(2πft) = 1
So, the required peak to peak emf is
ε = 2πf • NBA
Substituting all the given parameters
ε = 2π × 60 × 125 × 0.12 × 3 × 10^-2
ε = 169.65 Volts
Thus, we can conclude that the required value of peak output voltage is 169.65 Volts.
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The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The strength of the magnetic field that seperates the ions is Bo=0.750 T. A stream of single charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?
Answer:
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Explanation:
Given,
Magnetic field, B = 0.5 T
Electric field, E = 1.2 x 10⁵ V/m
strength of the magnetic field that separates the ions, Bo=0.750 T
Radius, r = 2.32 cm
Relation of charge to mass ratio is given by
[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]
[tex]m=\dfrac{qBB_0R}{E}[/tex]
Substituting all the values
[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
The mass of the Li ions is approximately [tex]\( 1.64 \times 10^{-26} \)[/tex] kg.
To find the mass of the Li ions, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:
[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]
Where:
- r is the radius of the circular path
- m is the mass of the ion
- v is the velocity of the ion
- q is the charge of the ion
- B is the magnetic field strength
We can also use the formula for the force experienced by a charged particle moving in both electric and magnetic fields:
[tex]\[ F = q \cdot (E + v \times B) \][/tex]
Since the particle moves in a circular path, the electric force F must be equal to the magnetic force [tex]\( q \cdot v \cdot B \)[/tex], where v is the speed of the particle.
We can rearrange the formulas to solve for m:
[tex]\[ m = \frac{q \cdot r \cdot B}{v} \][/tex]
Now, we need to find \( v \). Since the particle passes through both the electric and magnetic fields without deviation, the forces acting on it must be balanced. Therefore:
[tex]\[ F_{electric} = F_{magnetic} \]\[ q \cdot E = q \cdot v \cdot B \][/tex]
Solving for v:
[tex]\[ v = \frac{E}{B} \][/tex]
Now, substituting v into the equation for m :
[tex]\[ m = \frac{q \cdot r \cdot B}{\frac{E}{B}} \]\[ m = \frac{q \cdot r \cdot B^2}{E} \][/tex]
Now we can calculate \( m \) using the given values:
[tex]\[ m = \frac{(1.6 \times 10^{-19} C) \times (0.0232 m) \times (0.75 T)^2}{1.2 \times 10^5 V/m} \]\[ m= 1.64 \times 10^{-26} kg \][/tex]
A particular gun (10kg) is able to fire 20 gram bullets at a speed of 350 m/s. From this information, calculate roughly how much energy contained in the powder in the bullet (Hint: you will need to calculate the recoil speed of the gun.) A. 2455 J B. 620 J C. Cannot be determined. D. 1230 J
Answer:
The answer is
D. 1230j
Explanation:
When a bullet is shot out of a gun the person firing experiences a backward impact, which is the recoil force, while the force propelling the bullet out of the gun is the propulsive force
given data
Mass of gun M=10kg
Mass of bullet m=20g----kg=20/1000 =0.02kg
Propulsive speed of bullet = 350m/s
Hence the moment of the bullet will be equal and opposite to that of the gun
mv=MV
where V is the recoil velocity which we are solving for
V=mv/M
V=0.02*350/10
V=7/10
V=0.7m/s
The energy contained in the bullet can be gotten using
KE=1/2m(v-V)²
KE=1/2*0.02(350-0.7)²
KE=1/2*0.02(349.3)²
KE=1/2*0.02*122010.49
KE=1/2*2440.20
KE=1220.1J
roughly the energy is 1230J
The volume of a gas varies inversely as the pressure and directly as the temperature (in degrees Kelvin). If acertain gas occupies a volume of 2.2 liters at a temperature of 340 K and a pressure of 16 newtons per squarecentimeter, find the volume when the temperature is 408 K and the pressure is 24 newtons per squarecentimeter.
Final answer:
The volume of the gas at 408 K and 24 N/cm² will be 1.76 liters, calculated using the combined gas law that integrates Boyle's law and Charles's law for the given changes in pressure and temperature.
Explanation:
The volume of a gas is dependent on both its pressure and temperature, following Boyle's law and Charles's law, respectively. The relationship is such that volume varies inversely with pressure and directly with temperature (in kelvins) when the amount of gas is constant. To find the new volume when both the pressure and temperature change, we use the combined gas law, which integrates both Boyle's and Charles's laws:
V₁/T₁ * P₁ = V₂/T₂ * P₂
Given that at P₁ = 16 N/cm², T₁ = 340 K, and V₁= 2.2 liters, we can find V₂ when P₂ = 24 N/cm² and T₂ = 408 K using the formula:
V₂ = (V₁ * T₂/T₁) * (P₁/P₂)
Plugging in the values:
V₂ = (2.2 liters * 408 K / 340 K) * (16 N/cm² / 24 N/cm²)
V₂ = (2.2 liters * 1.2) * (2 / 3)
V₂ = 2.64 liters * (2 / 3)
V₂ = 1.76 liters
So, the volume of the gas when the temperature is 408 K and the pressure is 24 N/cm² would be 1.76 liters.
A proton is released from rest in a uniform electric field that has a magnitude of 8.0 x 104 V/m. The proton undergoes a displacement of 0.50 m in the direction of E Find the change in electric potential, the voltage, between point A and B. Find the change in potential energy of the proton-field system for this displacement. Hint: U
Answer:
Explanation:
The magnitude of electric field = 8 x 10⁴ V /m
there is a potential difference of 8 x 10⁴ V on a separation of 1 m
so on a separation of .5 m , potential drop or change in potential will be equal to
.5 x 8 x 10⁴
= 4 x 10⁴ V .
The increase in kinetic energy of proton = V X Q
V is potential drop x Q is charge on proton
= 4 x 10⁴ x 1.6 x 10⁻¹⁹
= 6.4 x 10⁻¹⁵ J
potential energy of the proton-field system will be correspondingly decreased by the same amount or by an amount of
- 6.4 x 10⁻¹⁵ J .
The change in electric potential is 4.0 x 10^4 volts. The change in potential energy of the proton-field system, given the charge of a proton, is calculated to be 6.4 x 10^-15 joules.
Explanation:The change in electric potential (or voltage) is calculated by multiplying the electric field strength by the displacement, so in this case it is (8.0 x 10^4 V/m) * (0.50 m) = 4.0 x 10^4 volts.
The change in potential energy of the proton-field system can be found by multiplying the change in electric potential by the charge of a proton. The charge of a proton is 1.6 x 10^-19 C, so the change in potential energy is (4.0 x 10^4 volts) * (1.6 x 10^-19 C), which equals to 6.4 x 10^-15 joules.
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A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at 1200C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 6 10-11 m2/s, and the diffusion flux is found to be 1.2 10-7 kg/m2 -s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4 kg/m3. How far into the sheet from this high-pressure side will the concentration be 2.0 kg/m3? Assume a linear concentration profile.
Answer:
The answer is 0.001 m
Explanation:
Solution
Recall that,
A sheet of steel 1.5 mm thick has nitrogen atmospheres on both sides at = 1200°C
The diffusion coefficient for nitrogen in steel at this temperature is =6 * 10-11 m2/s
The diffusion flux is = 1.2 *10^-7 kg/m2s
Concentration of nitrogen in the steel at the high-pressure surface is= 4 kg/m3.
The high-pressure side will the concentration is estimated to be = 2.0 kg/m3
Now,
The flux = -D dC/dx
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
∫ˣ₀ dx = -5x^10-4 ∫²₄ dC
so,
x = (2-4) kg/m3 (-5x10-4 m4/kg)
where x = .001 m
Therefore x = 0.001 m
The distance from this high-pressure side will the concentration is 0.001 m
Calculation of the distance:Since the diffusion coefficient for nitrogen in steel at this temperature is 6 x 10-11 m2/s, the diffusion flux is found to be 1.2x 10-7 kg/m2 -s.
We know that
The flux =[tex]-D\ dC\div dx[/tex]
So,
1.2 x 10-7 kg/m2s = - 6 x 10-11 m2/s dC/dx
dC/dx = -2000
Now
-2000 = 4 - 2/0-x_B
x_B = 2/2000
x_B = 1*10^-3m
x_B = 0.001m
Hence, The distance from this high-pressure side will the concentration is 0.001 m
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Suppose that a pendulum clock keeps time exactly correctly on Earth. Now suppose that you take this clock to planet X and find that the clock’s hour hand makes α revolutions every hour. Find a formula for the magnitude of the acceleration due to gravity near the surface of planet X. What would α be on the moon?
Answer:
The gravity at planet X is [tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
The value of [tex]\alpha[/tex] on the moon is [tex]\alpha = 730.38 \ revolutions[/tex]
Explanation:
From the question we are told that
The clocks hour hand makes [tex]\alpha[/tex] revolution every 1 hour which is 3600 sec
this implies that the time peroid for 1 revolution would be [tex]= \frac{3600}{\alpha }sec[/tex]
The peroid for a pendulum is mathematically represented as
[tex]T = 2 \pi\sqrt{\frac{L}{g} }[/tex]
Where L is the pendulum length
g is the acceleration due to gravity
Let assume that we have a pendulum that counts in second on earth
This implies that its peroid would be = 2 second
i.e one second to swimg forward and one second to swing back to its original position
Now the length of this pendulum on earth is
[tex]L = \frac{gT^2}{4 \pi^2}[/tex] [Making L the subject in above equation]
Substituting values
[tex]L = \frac{9.8 * (2)^2}{4 * (3.142)^2}[/tex]
[tex]= 1[/tex]
When the same pendulum is taken to planet X the peroid would be
[tex]T = \frac{3600}{\alpha }[/tex]
Recall this value was obtained above for 1 revolution (from start point to end point back to start point)
So the acceleration due to gravity on this planet would be mathematically represented as
[tex]g = \frac{4 \pi L }{T^2}[/tex] [making g the subject in the above equation]
substituting values
[tex]g = \frac{4 * 3.142^2 * 1}{[ \frac{3600}{\alpha } ]^2}[/tex]
[tex]g= 3.05*10^{-6} \alpha^2 m/s^2[/tex]
On moon the acceleration due to gravity has a constant value of
[tex]g = 1.625 m/s^2[/tex]
The period of this pendulum on the moon can be mathematically evaluated as
[tex]T = 2\pi \sqrt{\frac{L}{g} }[/tex]
substituting value
[tex]T = 2 *3.142 \sqrt{\frac{1}{1.625} }[/tex]
[tex]= 4.929s[/tex]
given that
[tex]1 \ revolution ----> 4.929s\\ \\ \alpha \ revolution -------> 3600 \ s[/tex] {Note 1 revolution takes a peroid }
Making [tex]\alpha[/tex] the subject of the formula
[tex]\alpha =\frac{3600}{4.929}[/tex]
[tex]\alpha = 730.38 \ revolutions[/tex]
A 910-kg object is released from rest at an altitude of 1200 km above the north pole of the earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg, the polar radius of the earth is 6357 km) 2.7 km/s 3.2 km/s 4.8 km/s 4.5 km/s 2.2 km/s
The object strikes the surface of the Earth with a speed of approximately 11.2 km/s.
Ignoring atmospheric friction, the object's total mechanical energy (the sum of its gravitational potential energy and kinetic energy) is conserved throughout its fall. We can use this principle to calculate the object's speed at the Earth's surface.
Here's how:
Calculate the gravitational potential energy (PE) at the initial altitude:
PE_i = -G * M_earth * m / (R_earth + h)
where:
G is the gravitational constant (6.67 × 10^-11 N m²/kg²)
M_earth is the mass of the Earth (5.97 × 10^24 kg)
m is the mass of the object (910 kg)
R_earth is the Earth's polar radius (6357 km)
h is the initial altitude (1200 km)
Set the final PE (PE_f) to 0:
Since all the PE will be converted to kinetic energy when the object reaches the surface, PE_f = 0.
Apply the conservation of mechanical energy:
E_i = PE_i + KE_i = PE_f + KE_f = KE_f (as PE_i = PE_f = 0)
where:
KE_i is the initial kinetic energy (0 J, as the object is at rest)
KE_f is the final kinetic energy
Solve for the final velocity (v_f):
KE_f = 1/2 * m * v_f^2
v_f = sqrt(2 * E_i / m)
Plug in the values and calculate v_f:
v_f = sqrt(2 * (-G * M_earth * m / (R_earth + h)) / m)
v_f ≈ 11,180 m/s
Convert m/s to km/s:
v_f ≈ 11.2 km/s