The distance of the ship from the shore is 1020m
Explanation:
Given:
Speed, s = 340 m/s
Time, t = 10s
Distance, x = ?
The sound is going to have to go to shore, then come back.
The total round-trip distance is
D = speed X time
D = (340 m/s) * (6.0 s)
D = 2,040 m
But as previously stated, the sound had to get there, then come back. So the actual ship-to-shore distance is only half that.
x = D/2
[tex]x = \frac{2040}{2} \\\\x = 1020m[/tex]
Therefore, the distance of the ship from the shore is 1020m
A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.260 T/s .
(a) What is the magnitude of the electric field induced in the ring?
(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
The magnitude of the induced electric field can be found using Faraday's law of electromagnetic induction. The current flows clockwise as viewed by someone on the south pole of the magnet.
Explanation:To find the magnitude of the electric field induced in the ring, we can use Faraday's law of electromagnetic induction. According to the law, the magnitude of the induced electric field is equal to the rate of change of magnetic flux through the ring. The magnetic flux can be calculated by multiplying the magnetic field strength by the area of the ring. In this case, the magnetic field is decreasing at a rate of 0.260 T/s, and the area of the ring is π(2.50 cm)2. Therefore, the magnitude of the induced electric field is 2π(2.50 cm)2(0.260 T/s).
As for the direction of the current flow, it can be determined using the right-hand rule for induced currents. If you point your right thumb in the direction of the magnetic field (from north to south), the induced current will flow in the direction of your curled fingers, which in this case will be clockwise as viewed by someone on the south pole of the magnet.
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A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 37.5 ∘ with the normal to the top surface of the glass.
θ₃ is 27.12°
Explanation:
Using Snell's law:
n₁ sin θ₁ = n₂ sin θ₂
where,
n₁ = refractive index of material with incident light
n₂ = refractive index of material with refracted light
(a)
1 → air
2 → glas
3 → water
n(water) = n₃ = 1.333
From air to glass,
n₁ sin θ₁ = n₂sin θ₂
The angle of refraction 2 is the angle of incidence when the light comes from glass to water.
From glass to water:
n₁ sin θ₁ = n₃ sin θ₃
1 X sin 37.5° = 1.333 sin θ₃
sin θ₃ = sin (37.5)/ 1.333
sin θ₃ = 0.456
θ₃ = 27.12°
Therefore, θ₃ is 27.12°
20 POINTS! TRUE OR FALSE:
To increase the acceleration of an object, you would reduce its mass or increase the applied force.
Answer: TRUE
Explanation:
A comet has a period of 324 years; in other words, it orbits the Sun in 324 years. Most likely, this comet came from...(Hint: 324 years define a long period comet)
A. Somewhere outside the solar system
B. Oort's cloud
C. The ecliptic plane
D. The asteroid belt
E. The Kuiper belt
Comets are divided into two types. Long-period comets are the comets that take more than two hundred years to finish an orbit throughout the Sun originate from the Oort Cloud.
Explanation:
The Oort Cloud has sufficient distance apart of the Sun than the Kuiper Belt, it seems that the Oort Cloud objects were made closer to the Sun than the Kuiper Belt things.Long-period comets have highly eccentric orbits.The Oort cloud is considered to own a large range commencing from among 2,000 and 5,000 AU to 50,000 AU.The chemical makeup of long-period and short-period comets is quite alike.
Final answer:
A comet with a 324-year orbit is likely from the Oort Cloud, the distant spherical region of icy bodies surrounding our solar system, known as the source of long-period comets. The Kuiper Belt, on the other hand, sources short-period comets.
Explanation:
A comet with a period of 324 years is considered a long-period comet. These comets originate from a region far beyond the inner solar system. The Oort Cloud is a vast, spherical shell of icy bodies that extends up to about 50,000 astronomical units (AU) from the Sun and contains trillions of potential comets. This makes it the most likely source of long-period comets like the one mentioned. In contrast, the Kuiper Belt is a disk-shaped region beyond Neptune, extending to about 50 AU, which is known to source short-period comets, also known as Jupiter-family comets.
Long-period comets like the one with a 324-year orbit are thought to be influenced by gravitational perturbations from nearby passing stars or galactic tides, which can send them towards the inner solar system. Once these comets enter the inner solar system, they can have dramatic interactions with planets or the Sun, sometimes leading to their disintegration or alteration in orbit. An example of such an interaction was when Comet Shoemaker-Levy 9 broke apart and collided with Jupiter in 1994.
A hospital needs 0.100 g of 133 54Xe for a lung-imaging test. If it takes 10 days to receive the shipment, what is the minimal amount mXe of xenon that the hospital should order?
Answer:
the hospital needs to order for a minimum amount of 0.4 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]
Explanation:
Given that:
A hospital needs 0.100 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex]
and it takes 10 days for the shipment to arrive:
the half life [tex]t_{1/2}[/tex] = 5 days
So, since the half life = 5 days ;
decay constant [tex]\lambda = \frac{In_2}{t_{1/2}}[/tex]
where:
[tex]N_o= ??? \\ \\ N = 0.1 00 \\ \\ t = 10 days \\ \\ N(t)= N_oe^{-\lambda t}\\ \\0.1 = N_oe^{\frac{-In_2}{5} *10} \\ \\0.1 = N_oe^{ - In \ 4} \\ \\ 0.1 = N_oe^{ In \frac{1}{ 4}} \\ \\ 0.1 = \frac{N_o}{4} \\ \\ N_o = 0.1*4 \\ \\ N_o = 0.4 \ g[/tex]
Therefore in order to get 0.100 g of [tex]\left \ {{133} \atop {54} \right. Xe[/tex], the hospital needs to order for a minimum amount of 0.4 g of
Three different experiments are conducted that pertain to the oscillatory motion of a pendulum. For each experiment, the length of the pendulum and the mass of the pendulum are indicated in all experiments, the pendulum is released from the same angle with respect to the vertical. If the students collect data bout the kinetic energy of the pendulum as a function of time for each experiment, which of the following claims is true? a. The da ta collected from Experiment 1 will be the same as the data collected from Experiment 2. b. The data collected from Experiment 1 will be the same as the data collected from Experiment 3. C. The data collected from Experiment 2 will be the same as the data collected from Experiment 3. d. The data collected from each experiment will be different.
The data collected from all experiments by the students will be different. Option D is correct.
Kinetic energy:
The kinetic energy of the pendulum depends upon the length and mass and angle of the pendulum.
Since the angle is the same in all three experiments. Mass and length will decide the kinetic energy of the pendulum.
As students measure the kinetic energy against the function of time, the period of the pendulum will also vary.
The period of the pendulum can be calculated by the formula,
[tex]T = 2\pi \sqrt{ \dfrac lg}[/tex]
Where,
[tex]l[/tex] - length
[tex]g[/tex] - gravitational acceleration,
Therefore, the data collected from all experiments by the students will be different. Option D is correct.
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The motion of a simple pendulum is not affected by the mass of the bob; it is only influenced by the length of the pendulum and the acceleration due to gravity. If the pendulum lengths in the experiments are the same, then the kinetic energy data collected will be identical, even if the masses are different.
Explanation:If three different experiments are conducted that pertain to the oscillatory motion of a pendulum with varying lengths and masses, and all are released from the same angle, the kinetic energy as a function of time for each experiment will differ only if the lengths of the pendulums are different. The mass of the pendulum bob does not affect the motion of a simple pendulum; pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity. Therefore, if the lengths of the pendulums in the different experiments are the same, then claim a. The data collected from Experiment 1 will be the same as the data collected from Experiment 2 is true irrespective of the different masses of the pendulum bobs.
The movement of the pendula will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendula are only affected by the period (which is related to the pendulum's length) and by the acceleration due to gravity.
A 3.0 kg object is moving along the x-axis in a region where its potential energy as a function of x is given as U(x) = 4.0x2 , where U is in joules and x is in meters. When the object passes the point x = -0.50 m, its velocity is +2.0 m/s. All forces acting on the object are conservative. Calculate the total mechanical energy of the object Calculate the x-coordinate of any points at which the object has zero kinetic energy. Calculate the magnitude of the momentum of the object at x = 0.60 m. Calculate the magnitude of the acceleration of the object as it passes x = 0.60 m. On the axes below, sketch graphs of the object’s position x versus time t and kinetic energy K versus time t. Assume that x = 0 at time t = 0 . The two graphs should cover the same time interval and use the same scale on the horizontal axes.
Answer:
a) [tex]E_{tot} = 7\,J[/tex], b) [tex]x = \pm\sqrt{\frac{7}{4} }[/tex]c) [tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
Explanation:
a) The total energy of the object is:
[tex]E_{tot} = U + K[/tex]
[tex]E_{tot} = 4\cdot x^{2} + \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]E_{tot} = 4\cdot (-0.50\,m)^{2} + \frac{1}{2}\cdot (3\,kg)\cdot (2\,\frac{m}{s} )^{2}[/tex]
[tex]E_{tot} = 7\,J[/tex]
b) The total energy of the object is:
[tex]E_{tot} = U[/tex]
[tex]7\,J = 4\cdot x^{2}[/tex]
[tex]x = \pm\sqrt{\frac{7}{4} }[/tex]
c) The speed of the object is clear in the total energy expression:
[tex]E_{total} = U + K[/tex]
[tex]K = E_{total}-U[/tex]
[tex]\frac{1}{2}\cdot m \cdot v^{2} = E_{total} - 4\cdot x^{2}[/tex]
[tex]v^{2} = \frac{2\cdot (E_{total}-4\cdot x^{2})}{m}[/tex]
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0.6\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 1.925\,\frac{m}{s}[/tex]
The magnitude of the momentum is:
[tex]p = (3\,kg)\cdot (1.925\,\frac{m}{s} )[/tex]
[tex]p = 5.775\,\frac{kg\cdot m}{s}[/tex]
d) Before calculating the acceleration experimented by the object, it is required to determine the net force exerted on it. There is a relationship between potential energy and net force:
[tex]F = -\frac{dU}{dx}[/tex]
[tex]F = -8\cdot x[/tex]
Acceleration experimented by the object is:
[tex]a = -\frac{8\cdot x}{m}[/tex]
[tex]a = -\frac{8\cdot (0.6\,m)}{3\,kg}[/tex]
[tex]a = -1.6\,\frac{m}{s^{2}}[/tex]
e) The position of the object versus time is found by solving the following differential equation:
[tex]\frac{d^{2}x}{dt} +\frac{8\cdot x}{m} = 0[/tex]
[tex]s^{2}\cdot X(s)- s\cdot v(0) - x(0) + \frac{8}{m}\cdot X(s) = 0[/tex]
[tex](s^{2} + \frac{8}{m})\cdot X(s) = s\cdot v(0)+x(0)[/tex]
[tex]X(s) = \frac{s\cdot v(0)+x(0)}{(s^{2}+\frac{8}{m} )}[/tex]
[tex]X(s) = v(0)\cdot \frac{s}{s^{2}+\frac{8}{m} } +\frac{m\cdot x(0)}{8} \cdot \frac{\frac{8}{m}}{s^{2}+\frac{8}{m}}[/tex]
[tex]x(t) = v(0) \cdot \cos \left(\frac{8}{m}\cdot t \right)+\frac{m\cdot x(0)}{8}\cdot \sin \left(\frac{8}{m}\cdot t \right)[/tex]
The velocity is obtained by deriving the previous expression:
[tex]v(t) = -\frac{8\cdot v(0)}{m}\cdot \sin \left(\frac{8}{m}\cdot t \right)+x(0)\cdot \cos \left(\frac{8}{m}\cdot t \right)[/tex]
Speed of the object at [tex]x = 0[/tex] is:
[tex]v = \sqrt{\frac{2\cdot (E_{total}-4\cdot x^{2})}{m} }[/tex]
[tex]v = \sqrt{\frac{2\cdot [7\,J- 4\cdot (0\,m)^{2}]}{3\,kg} }[/tex]
[tex]v \approx 2.160\,\frac{m}{s}[/tex]
The equation of the motion are:
[tex]x(t) = 2.160\cdot \cos \left(2.667\cdot t \right)[/tex]
[tex]v(t) = -5.76\cdot \sin (2.667\cdot t)[/tex]
The expression for the kinetic energy of the object is:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]K = \frac{1}{2}\cdot (3\,kg)\cdot [33.178\cdot \sin^{2}(2.667\cdot t)][/tex]
[tex]K = 49.767\cdot \sin^{2}(2.667\cdot t)[/tex]
Graphics are included below as attachments. (Position versus time, kinetic energy vs time).
Following are the responses to the given points:
Given:
object mass [tex](m)= 3.0\ kg \\\\[/tex]
Potential energy [tex]U(x)=4.0\ x^2\\\\[/tex]
[tex]\to x=-0.5 \ m\\\\ \to velocity \ (v)=2.0 \ \frac{m}{s}\\\\[/tex]
Solution:
For point (a)
Total Energy [tex](TE) = PE + KE \\\\[/tex]
[tex]\to PE = 4.0 (0.5 m)^2 = 1\ J\\\\ \to KE = \frac{1}{2} mv^2 = 0.5 \times 3.0\ kg (2.0 \frac{m}{s})^2 = 6\ J \\\\\to TE = 7\ J\\\\[/tex]
For point (b)
If an object's potential energy is 7 J, it has 0 kinetic energy.
[tex]\to U(x) = 4x^2 = 7\ J \\\\\to x=+1.2 \ m, -1.2 \ m[/tex]
For point (c)
[tex]\to x=0.60\ m \\\\ \to TE=4x^2 + \frac{1}{2}mv^2 = 7\ J\\\\ \to 4(0.60)^2 +0.5 \times 3.0 \ kg \times v^2 = 7\ J \\\\ \to v=1.92 \ \frac{m}{s}\\\\ \to Momentum (p) = mv \\\\\to p= 3.0\ kg \times 1.92 \ \frac{m}{s}\\\\ \to p= 5.76 kg \ \frac{m}{s}\\\\[/tex]
For point (d)
[tex]\to x_1= 0.6 \ m\\\\ \to v_1 = 1.92 \frac{m}{s}\\\\ \to x_2 = 1.2\ m[/tex] the velocity is found by
[tex]\to 4x^2 + \frac{1}{2} mv^2= 7\ J \\\\ \to 4(1.2)^2 +0.5 \times 3.0\ kg \times v^2 = 7\ J \\\\ \to v_2 = 0.9 \ \frac{m}{s}\\\\[/tex]
Calculating the acceleration:
[tex]\to V^2_{2}-v^2_{1}= 2a(x_2-x_1) \\\\\to (0.9 \ \frac{m}{s})^2 - (1.92\ \frac{m}{s})^2 = 2 a(1.2\ m - 0.6\ m) \\\\\to a= -2.4 \frac{m}{s^2}\\\\[/tex]
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A child's toy consists of a small wedge that has an acute angle θ. The sloping side of the wedge is frictionless, and an object of mass m on it remains at constant height if the wedge is spun at a certain constant speed. The wedge is spun by rotating, as an axis, a vertical rod that is firmly attached to the wedge at the bottom end. Show that, when the object sits at rest at a point at distance L up along the wedge, the speed of the object must be v = (gL sin(θ))1/2.
Answer:
see explanation
Explanation:
Net force along x axis is
[tex]\sum F_x = F \sin \theta= \frac{v^2}{R} ----(1)[/tex]
Net force along y axis is
[tex]\sum F_y = F \cos \theta= mg ----(2)[/tex]
The object can along accelerate down the stamp.
Thus F(net) is at the angle down the stamp
[tex]F_{net}=F_c\\\\F_{net}= \sin \theta mg\\\\F_c = \frac{mv^2}{r} \\[/tex]
where r = L in the direction of acceleration
[tex]\sin \theta mg = \frac{mv^2}{L}[/tex]
[tex]v^2 = gL \sin \theta[/tex]
[tex]v = \sqrt{gL \sin \theta}[/tex]
[tex]v = (gL \sin \theta )^{1/2}[/tex]
The relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].
The given parameters:
inclination of the wedge, = θlet the constant speed = vThe relationship between the distance of the object, speed of the object can be determined by the net force on the toy is calculated follows;
[tex]Fsin(\theta) = \frac{mv^2}{L} \\\\mgsin(\theta) = \frac{mv^2}{L} \\\\gsin(\theta) = \frac{v^2}{L} \\\\v^2 = gL sin(\theta)\\\\v = \sqrt{gL sin(\theta)}[/tex]
Thus, the relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].
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Below is a sketch of the initial state of the situation described in this problem. Draw the most suitable set of coordinate axes for this problem. Note that even though you can choose the y=0y=0 level to be wherever you like, in most situations it is best to set the zero height to coincide with either the initial or final position, so that the calculations for the gravitational potential energy become easier. For this reason, in this particular problem place the origin of your coordinate axes on the black dot marking the performer's initial position. Draw only the positive portion of the coordinate axes. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Answer:
This problem is incomplete, it says like this: The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You dont find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 N/m that will compress with a force of 4400 N. The inside of the gun barrel is coated with teflon, so the average friction force will be only 40 N during the 4 m his moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 m above his initial rest position?
The speed is 15.5 m/s, and the image shows the vector drawing.
Explanation:
For the calculation of the force that is due to the spring is equal to:
F = kx
Where
F = 4400 N
k = 1100 N/m
x = F/k = 4400/1100 = 4 m
For the calculation of the total energy is equal to:
[tex]\frac{1}{2} kx^{2} =\frac{1}{2}mv^{2} +mgh+Fx\\v=\sqrt{\frac{2}{m} (\frac{1}{2}kx^{2} -mgh-Fx) }[/tex]
Where
m = 60 kg
h = 2.5 m
Replacing:
[tex]v=\sqrt{\frac{2}{60}(\frac{1}{2}*1100*4^{2}-60*9.8*2.5-(40*4)) } =15.5m/s[/tex]
The problem is about setting coordinate axes and vectors for a physics problem, with the origin placed at the performer's initial position. The vectors represent various physical quantities depending on their orientation and direction, always starting from the origin.
Explanation:This is a typical problem in physics dealing with vectors, coordinate axes, and natural phenomena like gravitational potential energy. Setting up the appropriate coordinate system is crucial. We set the origin, or the (0,0) point, at the performer's initial position as mentioned in the problem.
The y-axis should be vertical, aligned with the direction of gravity. This makes computations involving gravity much easier. It's important to note that increasing height (direction opposite to that of gravity) is typically represented as positive y.
The vectors originating from the black dot (our origin) represent quantities such as displacement, velocity, force etc. The direction of the vector indicates the direction of the quantity.
For example, if there's a vector pointing directly upwards from the origin, it could represent an upward force or movement. Keep in mind, the orientation and location of the vectors are determined by the physical reality the problem is describing, and they should always start from the origin in the problem's context.
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An electron is placed in a magnetic field that is directed along a z axis. The energy difference between parallel and antiparallel alignments of the z component of the electron's spin magnetic moment with is 7.50 × 10-25 J. What is the magnitude of ?
Answer:
B = 0.0404 т
Explanation:
Given
ΔE = 7.50 x 10⁻²⁵ J
μb = 927.4 x 10⁻²⁶ J/т
ΔE = 2 * μb * B
B = ΔE / 2 μb
B = 7.50 x 10⁻²⁵ / 2 * 927.4 x 10⁻²⁶
B = 7.50 / 1854.8 x 10⁻²⁵⁻⁽⁻²⁶⁾
B = 0.00404 x 10¹
B = 0.0404 т
Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.
What is the force applied between q1 and q2?
Answer:
F = 10.788 N
Explanation:
Given that,
Charge 1, [tex]q_1=6\ \mu C=6\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=2\ \mu C=2\times 10^{-6}\ C[/tex]
Distance between charges, d = 0.1 m
We know that there is a force between charges. It is called electrostatic force. It is given by :
[tex]F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N[/tex]
So, the force applied between charges is 10.788 N.
Answer:
10.8 N
Explanation:
A spring with a constant of 16 N/m has 98 J of energy stored in it when it is extended. How far is the spring extended?
The spring has been extended for 3.5 m
Explanation:
We have the formula,
PE =1/2 K X²
Rewrite the equation as
PE=1/2 K d²
multiply both the sides by 2/K to simplify the equation
2/k . PE= 1/2 K d² . 2/K
√d²=√2PE/K
Cancelling the root value and now we have,
d=√2PE/k
d=√2×98 J / 16N/m
d=√12.25
d=3.5 m
The spring has been extended for 3.5 m
Final answer:
To find the extension of a spring with a constant of 16 N/m and 98 J of energy stored, we use the formula for potential energy. The spring is extended by 3.5 meters.
Explanation:
The question involves calculating the extension of a spring based on the energy stored in it and the spring's constant. The formula for the potential energy stored in a spring is given by
U = 1/2 kx², where U is the potential energy, k is the spring constant, and x is the extension of the spring from its equilibrium position. In this case, the energy (U) is 98 J and the spring constant (k) is 16 N/m.
The spring in question has an energy of 98 J stored in it.
Using the formula for potential energy stored in a spring, we have:
[tex]PE = 1/2 k x^2[/tex]
By substituting the known values, we find that the spring is extended by 0.7 m (70 cm
We rearrange the formula to solve for x:
x = √(2U/k).
Substituting the given values,
x = √(2*98/16)
= √(196/16)
= √(12.25)
= 3.5 meters. Therefore, the spring is extended by 3.5 meters.
A spherical gas-storage tank with an inside diameter of 8.0 m is being constructed to store gas under an internal pressure of 1.62 MPa. The tank will be constructed from steel that has a yield strength of 295 MPa. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness tmin required for the spherical tank.
Answer:
The minimum wall thickness Tmin required for the spherical tank is 65.90mm
Explanation:
Solution
Recall that,
Tmin = The minimum wall thickness =PD/2бp
where D = diameter of 8.0 m
Internal pressure = 1.62 MPa
Then
The yield strength = 295MPa/3.0 = 98.33
thus,
PD/2бp = 1.62 * 8000/ 2 *98.33
= 12960/196.66 = 65.90
Therefore the wall thickness Tmin required for the spherical tank is 65.90mm
Sally places a jar with some pennies into a pool of water and exactly half of the jar is submerged. The volume of the jar is 200 cm3 and the density of the water is 1 g/cm3.
a. What is the buoyant force acting on the jar?
b. What is the weight of the jar and pennies?
c. What is the density of the combined jar and pennies?
Answer:
Bouyant Force = 0.9N
The density of the jar and Penny was not given so we can't determine it's weigh and combined density
Explanation:
Bouyant force Fb = Volume of solid*density of liquid*acceleration
Fb= (200/2)*1*980.665
Fb= 98066.5/100000
Fb= 0.9N
Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T. (a) What is the energy difference between the magnetic moment orientations of the silver atoms in the two subbeams? (b) What is the frequency of the radiation that would induce a transition between these two states? (c) What is the wavelength of this radiation, and (d) to what part of the electromagnetic spectrum does it belong?
Answer:
Explanation
Assume that in the Stern-Gerlach experiment for neutral silver atoms, the magnetic field has a magnitude of B = 0.21 T
A. To calculate the energy difference in the magnetic moment orientation
∆E = 2μB
For example, any electron's magnetic moment is measured to be 9.284764×10^−24 J/T
Then
μ = 9.284764 × 10^-24 J/T
∆E = 2μB
∆E = 2 × 9.284764 × 10^-24 × 0.21
∆E = 3.8996 × 10^-24 J
Then, to eV
1eV = 1.602 × 10^-19J
∆E = 3.8996 × 10^-24 J × 1eV / 1.602 × 10^-19J
∆E = 2.43 × 10^-5 eV
B. Frequency?
To determine the frequency of radiation hitch would induce the transition between the two states is,
∆E = hf
Where h is plank constant
h = 6.626 × 10-34 Js
Then, f = ∆E / h
f = 3.8996 × 10^-24 / 6.626 × 10^-34
f = 5.885 × 10^9 Hz
f ≈ 5.89 GHz
C. The wavelength of the radiation
From wave equation
v = fλ
In electromagnetic, we deal with speed of light, v = c
And the speed of light in vacuum is
c = 3 × 10^8 m/s
c = fλ
λ = c / f
λ = 3 × 10^8 / 5.885 × 10^9
λ = 0.051 m
λ = 5.1 cm
λ = 51 mm
D. It belongs to the microwave
From table
Micro waves ranges from
•Wavelength 10 to 0.01cm
Then we got λ = 5.1 cm, which is in the range.
•Frequency 3GHz to 3 Thz
Then, we got f ≈ 5.89 GHz, which is in the range
•Energy 10^-5 to 0.01 eV
We got ∆E = 2.43 × 10^-5 eV, which is in the range of the microwave
The value above is in microwave range
Describe how the phase angle changes as you move from below resonance to above resonance Based on your results here and other textbook resources, what will happen if your frequency is much lower than resonance? What about much larger?
Answer:
Answer
Explanation:
In a LCR Circuit, the phase difference between voltage and current is usually summarized as;
tan∅ = XL- XC
R
=WL - 1/WC
R
When w=0 ( i.e when it is very low) tan∅ = ∞
or ∅ = -90
When w=0 ( i.e when it is very large) tan∅ =+ ∞
or ∅ = +90
A block of mass 500 g is attached to a spring of spring constant 80 N/m. The other end of the spring is attached to a support while the mass rests on a rough surface with a coefficient of friction of 0.20 that is inclined at angle of 30°. The block is pushed along the surface till the spring compresses by 10 cm and is then released from rest. (a) How much potential energy was stored in the block-spring-support system when the block was just released? (b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched. (c) Determine the position of the block where it just comes to rest on its way up the incline.
Answer:
a) Vki = 0.4 J
b) v = 0.98 m/s
c) final position = 0.11745 m from initial position
Explanation:
Given:-
- The mass of block, m = 500 g
- The spring constant, k = 80 N/m
- The coefficient of the surface, u = 0.20
- The inclination of the block, θ = 30°
- The block is initially compressed, xi = 10 cm
- The block is initially at rest, vi = 0 m/s
Find:-
(a) How much potential energy was stored in the block-spring-support system when the block was just released?
(b) Determine the speed of the block when it crosses the point when the spring is neither compressed nor stretched.
(c) Determine the position of the block where it just comes to rest on its way up the incline.
Solution:-
- The potential energy initially stored in a spring "Vki" with constant "k" which has undergone a displacement of "x" is given by:
Vki = 0.5*k*xi^2
Vki = 0.5*80*0.1^2
Vki = 0.4 J
- The block is released from rest when the energy stored in the spring is dispensed in three forms of energy.
- There would be an increase in the potential energy " Ep " of the block as it moves up.
- There would be an increase in kinetic energy for some time " Ek "
- There would be loss of total energy due to fictitious force ( Ff) the work done against friction will dissipate energy.
- The increase in potential energy "Ep" at displacement xo = 0 from mean position is given by:
ΔEp = m*g*Δh ... change in vertical height h
ΔEp = m*g*( xi - xo )*sin ( θ )
ΔEp = 0.5*9.81*(0.1)*sin ( 30 )
ΔEp = 0.24525 J
- The increase in kinetic energy "Ek" at displacement xi = 10 cm, vi = 0 m/s from initial position to mean position at xo = 0 ,its velocity is "vf" is given by:
ΔEk = 0.5*m*( vf^2 - vi^2 ) ... change in velocity
ΔEk = 0.5*m*( vf )^2
- There would be loss of total energy due to fictitious force ( Ff ) the work done against friction will dissipate energy. First apply the equilibrium conditions on the block normal to slope and determine the contact force ( Nc ):
Nc - m*g*cos ( θ ) = 0
Nc = m*g*cos ( θ )
- The friction force ( Ff ) is given by:
Ff = Nc*u
Ff = u*m*g*cos ( θ )
- The work done by block against friction is given by:
Wf = -Ff*( xi - xo )
Wf = -u*m*g*xi*cos ( θ )
Wf = -0.2*0.5*9.81*0.1*cos ( 30 )
Wf = -0.08495 J
- We can now express the work done principle for the block:
Vki = ΔEp + ΔEk + Wf
ΔEk = - ΔEp - Wf + Vki
0.5*m*( vf )^2 = -0.24525 + 0.08495 +0.4
vf^2 = 4*(-0.24525 + 0.08495 +0.4 )
vf = √0.9588
vf = 0.98 m/s
- We will denote the extension of spring at top most position from mean position as "x".
- From mean position xo = 0 m. The block will further move up the slope and expense all its kinetic energy "Ek" in the form of gain in potential energy and gain in elastic potential energy "Vk" and work is done against friction.
Vki = Ep2 + Wf + Vk2
0.4 = mg*x*sin ( 30 ) + 0.24525 + 0.08495 + u*m*g*x*cos ( θ ) + 0.5*k*x^2
0.5*80*x^2 + x*(0.5*9.81*sin(30) + 0.2*0.5*9.81*cos ( 30 ) ) + 0.3302-0.4 =0
40x^2 + 3.30207x - 0.0698 = 0
Solve the quadratic equation:
x = -0.1 m (10 cm) - initial compression at rest ;
x = 0.01745 m
- So the extension in spring at the rest position is x = 0.01745 m. The position of the next resting point is:
final position = xi + x
final position = 0.1 + 0.01745
final position = 0.11745 m from initial position.
x =
(a) The stored PE will be "0.40 J".
(b) The block's speed will be "0.48 m/s".
(c) The block's position will be at "12 cm".
Potential energy (P.E)According to the question,
Mass, m = 500 h
Spring constant = 80 N/m
Coefficient of friction = 0.20
Inclined angle = 30°
(a) We know the formula,
Spring P.E = [tex]\frac{1}{2}[/tex] kx²
By substituting the values,
= [tex]\frac{1}{2}[/tex] × 80 × (0.1)²
= 0.40 J
(b) We know that,
→ [tex]P.E_s[/tex] = ΔKE + Work done by friction + Δ[tex]PE_g[/tex]
or,
= [tex]\frac{1}{2}[/tex] mv² + [tex]\mu_R[/tex] mgx Cosθ + mgx Sinθ
By substituting the values,
0.40 = [tex]\frac{1}{2}[/tex] × 0.5 × v² + 0.2 × 0.5 × 9.8 × Cos30° × 0.1 + 0.5 × 9.8 × 0.1 × Sin30°
= 0.48 m/s
(c) By using the above formula,
→ 0.40 = [tex]\mu_R[/tex] mgl Cosθ + mgl Sinθ
= 0.2 × 0.5 × 9.8 × Cos30° × l + 0.5 × 9.8 × l × Sin30°
l = 0.12 m or,
= 12 cm
Thus the above responses are correct.
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The angular velocity of a 755-g wheel 15.0 cm in diameter is given by the equation ω(t) = (2.00 rad/s2)t + (1.00 rad/s4)t 3. (a) Through how many radians does the wheel turn during the first 2.00 s of its motion? (b) What is the angular acceleration (in rad/s2) of the wheel at the end of the first 2.00 s of its motion?
Answer:
a
The number of radians turned by the wheel in 2s is [tex]\theta= 8\ radians[/tex]
b
The angular acceleration is [tex]\alpha =14 rad/s^2[/tex]
Explanation:
The angular velocity is given as
[tex]w(t) = (2.00 \ rda/s^2)t + (1.00 rad /s^4)t^3[/tex]
Now generally the integral of angular velocity gives angular displacement
So integrating the equation of angular velocity through the limit 0 to 2 will gives us the angular displacement for 2 sec
This is mathematically evaluated as
[tex]\theta(t ) = \int\limits^2_0 {2t + t^3} \, dt[/tex]
[tex]= [\frac{2t^2}{2} + \frac{t^4}{4}] \left\{ 2} \atop {0}} \right.[/tex]
[tex]= [\frac{2(2^2)}{2} + \frac{2^4}{4}] - 0[/tex]
[tex]= 4 +4[/tex]
[tex]\theta= 8\ radians[/tex]
Now generally the derivative of angular velocity gives angular acceleration
So the value of the derivative of angular velocity equation at t= 2 gives us the angular acceleration
This is mathematically evaluated as
[tex]\frac{dw}{dt} = \alpha (t) = 2 + 3t^2[/tex]
so at t=2
[tex]\alpha (2) = 2 +3(2)^2[/tex]
[tex]\alpha =14 rad/s^2[/tex]
The wheel turns through 8 radians in the first 2 seconds of its motion. The angular acceleration of the wheel at the end of the first 2 seconds is 14.00 rad/s².
Explanation:The questions are related to the concepts of angular velocity and acceleration in the realm of Physics. To solve parts (a) and (b) we would use the principles of rotational motion.
(a) The number of radians the wheel turns in first 2 seconds is given by the integral of the angular velocity function from 0 to 2. The equation becomes ∫ₒ² ω(t) dt = ∫ₒ² ((2.00 rad/s²)t + (1.00 rad/s⁴)t ³ dt) = [t² + 0.25t⁴]ₒ² = 4 rad + 4 rad = 8 rad.
(b) The angular acceleration is given by the derivative of the angular velocity with respect to time which is ω'(t) = 2.00 rad/s² + 3(1.00 rad/s⁴)t². Evaluating at t = 2s gives an angular acceleration of 2.00 rad/s² + 3(1.00 rad/s⁴)(2s)² = 2.00 rad/s² + 12.00 rad/s² = 14.00 rad/s².
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"An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has an internal resistance of 10.0 Ω and the open-circuit voltage across its terminals is 50.0 V. The leads have no appreciable resistance. At time t = 0, the switch is suddenly closed." "When does the maximum current occur?"
Answer:
1.04x[tex]10^{-3}[/tex] s
Explanation:
->The maximum current through the resistor is
[tex]I_{max}[/tex] = V/R = V/[tex]Re^{-t/RC}[/tex]= V/R×[tex]e^{0}[/tex] = V/R
Voltage 'V'=50V
Effective resistance 'R'= 25.0-Ω+ 10.0 Ω= 35.0 Ω
Therefore, [tex]I_{max}[/tex]=50/35=> 1.43 A
->The maximum charge can be determined by
Q = CV
where,
Capacitance of the capacitor 'C' = 30.0µF = 30×10-⁶F
Therefore,
Q=30×10-⁶ x 50=>1.5 x [tex]10^{-3}[/tex]
In order to find that when does the maximum current occur, the time taken given the quantity of charge and the electric current is:
t= Q / I=> 1.5 x [tex]10^{-3}[/tex]/ 1.43
t= 1.04x[tex]10^{-3}[/tex] s
Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant hor-izontal force on a charged object. A 15 g charged projectile is launched with a speed of 6.0 m/s at an angle 35° above the hori-zontal. It lands 2.9 m in front of the launcher. What is the magni-tude of the electric force on the projectile?
Answer:
the magnitude of the electric force on the projectile is 0.0335N
Explanation:
time of flight t = 2·V·sinθ/g
= (2 * 6.0m/s * sin35º) / 9.8m/s²
= 0.702 s
The body travels for this much time and cover horizontal displacement x from the point of lunch
So, use kinematic equation for horizontal motion
horizontal displacement
x = Vcosθ*t + ½at²
2.9 m = 6.0m/s * cos35º * 0.702s + ½a * (0.702s)²
a = -2.23 m/s²
This is the horizontal acceleration of the object.
Since the object is subject to only electric force in horizontal direction, this acceleration is due to electric force only
Therefore,the magnitude of the electric force on the projectile will be
F = m*|a|
= 0.015kg * 2.23m/s²
= 0.0335 N
Thus, the magnitude of the electric force on the projectile is 0.0335N
Answer:
Magnitude of electric force = 0.03345 N
Explanation:
We are given;
Mass; m = 15g = 0.015kg
Angle above horizontal; θ = 35°
Speed; v = 6 m/s
Horizontal displacement; d = 2.9m
Now formula for time of flight is given as;
time of flight; t = (2Vsinθ)/g
Thus, plugging in values, we have
t = (2 x 6.0 x sin35)/9.8
t = (12 x 0.5736)/9.8
t = 0.7024 s
Now, let's find the acceleration
The formula for horizontal displacement is given by;
d = (Vcosθ)t + (1/2)at²
Plugging in the relevant values ;
2.9 = [6(cos35) x 0.7024] + (1/2)a(0.7024)²
2.9 = (4.2144 x 0.8192) + (0.2467)a
2.9 = 3.45 + (0.2467)a
(0.2467)a = 2.9 - 3.45
a = -0.55/0.2467
a = -2.23 m/s²
Since we are looking for the magnitude of the electric force, we will take the absolute value of a. Thus, a = 2.23 m/s²
We know that F = ma
Thus,Force = 0.015kg x 2.23m/s² =
= 0.03345 N
The frequency of a sound wave is 300 Hz and the room temperature is 30 Celsius. What is the wavelength of this sound wave. Two children stretch a jump rope between them and send wave pulses back and forth. The rope is 4 meters long and its mass is .4 kg and the force exerted on it by the children is 50 N. What is the speed of the waves on the rope?
Answer:
Explanation:
Velocity of sound wave at 30 degree = 350 m /s
frequency of sound = 300 Hz .
wavelength of sound in air. = velocity / frequency
= 350 / 300
= 1.167 m
for wave formed in the rope :
velocity of wave in the rope
= [tex]\sqrt{\frac{T}{m} }[/tex]
T is tension in the rope and m is mass per unit length .
m = .4 / 4
= .1
Putting the given values in the equation above
v = [tex]\sqrt{\frac{50}{.1} }[/tex]
v = 22.36 m /s .
velocity of wave in the rope.
= 22.36 m /s .
The hair breaks under a 1.2 N tension. What is the cross-sectional area of a hair?
I need an explanation or the calculation too, so I will not give points for only the answer.
Answer:
bow
Explanation:
A uniform meter stick is suspended from the ceiling of an elevator on one end and is swinging back and forth. The elevator is accelerating upwards with acceleration a.
The period of this meter stick, in terms of its length L, g and a is ______.
Answer:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
Explanation:
It is known that stick is experimenting a Simple Harmonic Movement and, to be exactly, can be modelled as a simple pendulum. The period of oscilation of the stick is:
[tex]T = \frac{2\pi}{\omega}[/tex]
The pendulum is modelled by the Newton's Laws. The Free Body Diagram is presented below:
[tex]\Sigma F_{r} = T - m\cdot g \cdot \cos \theta = m\cdot (\omega^{2}\cdot l + a\cdot \cos \theta)[/tex]
[tex]\Sigma F_{t} = m\cdot g \cdot \sin \theta = m\cdot (l\cdot \alpha - a \cdot \sin \theta)[/tex]
Let assume that pendulum is just experimenting small oscillations, so that:
[tex]\theta \approx \sin \theta[/tex]
Then:
[tex]m\cdot g \cdot \theta = m\cdot (l\cdot \alpha - a\cdot \theta)[/tex]
[tex]g\cdot \theta = l\cdot \alpha - a\cdot \theta[/tex]
[tex](g + a)\cdot \theta = l\cdot \alpha[/tex]
[tex]\alpha = \frac{g+a}{l}\cdot \theta[/tex]
Where [tex]\omega =\sqrt{\frac{g + a}{l} }[/tex].
Finally, the period is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g + a} }[/tex]
A toroidal coil has a mean radius of 16 cm and a cross-sectional area of 0.25 cm2; it is wound uniformly with 1000 turns. A second toroidal coil of 750 turns is wound uniformly over the first coil. Ignoring the variation of the magnetic field within a toroid, determine the mutual inductance of the two coils.
Answer:
Explanation:
Mutual inductance is equal to magnetic flux induced in the secondary coli due to unit current in the primary coil .
magnetic field in a torroid B = μ₀ n I , n is number of turns per unit length and I is current .
B = 4π x 10⁻⁷ x (1000 / 2π x .16 )x 1 ( current = 1 A)
flux in the secondary coil
= B x area of face of coil x no of turns of secondary
= 4π x 10⁻⁷ x (1000 /2π x .16 ) .25 x 10⁻⁴ x 750
= 2 x 1000 x .25 x( 750 /.16) x 10⁻¹¹
2343.75 x 10⁻⁸
= 23.43 x 0⁻⁶ H.
.
Answer:
2.5 x 10^-5 henry
Explanation:
The mutual inductance between the toroids is same.
mean radius of the toroid, r = 16 cm = 0.16 m
Area of crossection, A = 0.25 cm²
Number of turns in the first toroid, N1 = 1000
Number of turns in the second toroid, N2 = 750
The formula for the mutual inductance is given by
[tex]M =\frac{\mu_{0}N_{1}N_{2}A}{l}[/tex]
Where, l is the length
l = 2 x 3.14 x r = 2 x 3.14 x 0.16 = 1.0048 m
[tex]M =\frac{4\pi\times 10^{-7}\times 1000\times 750\times 0.25\times 10^{-4}}{1.0048}[/tex]
M = 2.5 x 10^-5 henry
Thus, the mutual inductance between the two toroid is 2.5 x 10^-5 henry.
An electron in an electron gun is accelerated from rest by a potential of 25 kV applied over a distance of 1 cm.
The final velocity of the electrons is _____.
The mass of the electron is 9.1x10^(-31) kg and its charge is 1.6x10^(-19) C.
Answer:
[tex]9.38\times 10^7 m/s[/tex]
Explanation:
We are given that
Potential ,V=25 kV=[tex]25\times 10^3 V[/tex]
Distance,r =1 cm=[tex]\frac{1}{100}=0.01 m[/tex]
1 m=100 cm
Mass of electron, m=[tex]9.1\times 10^{-31} kg[/tex]
Charge, q=[tex]1.6\times 10^{-19} C[/tex]
We have to find the final velocity of the electron.
Speed of electron,[tex]v=\sqrt{\frac{2qV}{m}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 25\times 10^3}{9.1\times 10^{-31}}[/tex]
v=[tex]9.38\times 10^7 m/s[/tex]
Hence, the final velocity of the electron=[tex]9.38\times 10^7 m/s[/tex]
A wire carrying a 29.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's field, and there is a 2.25 N force on the 3.00 cm of wire in the field. What is the average field strength (in T) between the poles of the magnet
Answer:
2.59 T
Explanation:
Parameters given:
Current flowing through the wire, I = 29 A
Angle between the magnetic field and wire, θ = 90°
Magnetic force, F = 2.25 N
Length of wire, L = 3 cm = 0.03 m
The magnetic force, F, is related to the magnetic field, B, by the equation below:
F = I * L * B * sinθ
Inputting the given parameters:
2.25 = 29 * 0.03 * B * sin90
2.25 = 0.87 * B
=> B = 2.25/0.87
B = 2.59 T
The magnetic field strength between the poles is 2.59 T
A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the bicycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 = [tex]\frac{460\times 32}{34\times 2}[/tex]
ω2 = 216.47
A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet through the pipe and explain the physical processes that cause this motion.
Answer and Explanation:
This experiment is known as Lenz's tube.
The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:
[tex]\varepsilon =-\frac{d\phi_B}{dt}[/tex]
This emf induced on the surface of the tube generates a current within it according to Ohm's Law:
[tex]V=IR[/tex]
This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.
Answer:
Check below for the answer and explanation
Explanation:
According to Faraday's law of electromagnetic induction, if a conductor is exposed to changing magnetic flux, an emf and hence a current is induced in the conductor. The strength of the induced emf is directly proportional to the rate of change of the magnetic flux.
Induced emf, [tex]e =-N\frac{d \phi}{dt}[/tex]
Induced current, I = e/R
In this example, as magnet is dropped down the aluminium pipe, the magnetic flux changes, and current is induced in the pipe.
According to Lenz's law, the direction of the induced current in the conductor opposes the direction of the magnetic flux that produces it.
Based on these stated laws, current is induced in this aluminium pipe and the direction of this induced current opposes the magnetic flux change. The magnetic field is repelled and falls slowly.
The limit to the eye's acuity is actually related to diffraction by the pupil. What is the angle between two just‑resolvable points of light for a 3.75 mm 3.75 mm diameter pupil, assuming the average wavelength of 554 nm 554 nm ?
Answer: [tex]1.8(10)^{-4} rad[/tex]
Explanation:
This problem is related to the Rayleigh Criterion, which provides the following formula to find the acuity or limit of resolution of an optic system with circular aperture (the eye in this case):
[tex]\theta=1.22\frac{\lambda}{D}[/tex]
Where:
[tex]\theta[/tex] is the angle of resolution (related to the acuity)
[tex]\lambda=554 nm=554(10)^{-9} m[/tex] is the wavelength of the light
[tex]D=3.75 mm=3.75(10)^{-3} m[/tex] is the diameter of the pupil
Solving:
[tex]\theta=1.22 \frac{554(10)^{-9} m}{3.75(10)^{-3} m}[/tex]
[tex]\theta=1.8(10)^{-4} rad[/tex]
The human eye's acuity is limited by light's diffraction through the pupil, which determines the minimum angle, Δθ, between two resolvable points. This is calculated using the formula Δθ = 1.22 (λ / D), where λ is the light's wavelength and D the diameter of the pupil.
Explanation:The resolution of human visual acuity is related to the diffraction of light by the eye's pupil. This diffraction effect can be worked out using a formula that quantifies resolving power (Δθ), when the diameter of the aperture (in this case, the pupil size) and the wavelength of incident light are known.
In your example, a 3.75 mm diameter pupil encountering light of 554 nm wavelength can identify two points of light provided they subtend an angle (Δθ) at the pupil calculated by using the formula: Δθ = 1.22 (λ / D). For all light's range of visible wavelengths, the angle is extremely tiny - on the order of tens of thousands of a degree. This shows that the eye’s ability to distinguish two separate points of light is limited by the diffraction of light by the eye’s pupil.
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A change in velocity means a change in?