Answer: 3.78 m
Explanation:
Volume = 1.9 * 1.9 * 1.9
Volume = 6.859 m³
Density of iron assumed to be 7870 kg/m³
Recall,
Density = mass / volume, so
Mass = density * volume
Mass = (7870 * 6.859)
Mass = 53980.33 kg
This mass we calculated, is the mass of water the new cube would have to displace, absolute minimum, so that it can float.
Now density of fresh water is assumed to be (1000 kg/m³).
Thus, the volume of water that will be displaced has to be
Volume = mass / density
Volume = (53980.33 / 1,000)
Volume = 53.98 m³.
Cube root of 53.88 = 3.78 m
Therefore, the minimum side length is 3.78 m
Answer:
3.781488032m or greater
Explanation:
General Rule of thumb is that any thing that is less dense than water will float in it.
So
density of water is 997Kg/m^3 and density of iron on the other hand is 7860kg/m^3.
so to make iron less dense it has to be in larger volume.
mass of the given cube is 53911.74Kg and we have contain it in volume such that density is less than water, mathematically this translates to .
[tex]\frac{53911.74Kg}{V}\leq \frac{997Kg}{m^3}[/tex] left side is density of iron and right side is density of water, solving v in this inequality gives us.
[tex]v \geq 54.0739m^3[/tex] which means each side is [tex]s \geq 3.7814880m[/tex]. so the minum is just greater than that.
A vertical wire carries current in the upward direction. An electron is traveling parallel to the wire. What is the angle ααalpha between the velocity of the electron and the magnetic field of the wire? Express your answer in degrees. View Available Hint(s) ααalpha = nothing degreesdegrees
Answer:
First of all note that The magnetic field produced by the vertical wire will be into on the right hand side and it will be out of the page on the left hand side
Assuming that the electron beam is coming from the right hand side of the page parallel to the wire, The direction of the velocity vector(V) is left and the direction of magnetic field due the wire(B) is into the page. If u use right hand rule, you will get the direction downwards but as the formula also depends on q , the charge on electron is negative .Therefore the direction will be inverted i.e Upwards.
If you assume the electron beam coming from left hand side.Then also u will get the same answer.
So, the angle α between the velocity of the electron and the magnetic field of the wire is 90°.
Explanation:
For an electron traveling parallel to a wire carrying an upward direction current, the angle between its velocity and the magnetic field of the wire, according to the right-hand rule, is 90 degrees.
Explanation:The setting of this problem involves a vertical wire carrying an upward direction current and an electron traveling parallel to it. According to the right-hand rule in magnetism, which states that if your thumb points in the direction of the current, then your fingers will curl in the direction of the magnetic field, the magnetic field of the wire would form concentric circles around the wire.
For an electron traveling parallel to the wire, according to this rule, it would be always at a right angle or 90 degrees to the magnetic field as it moves along the circumference of these imaginary concentric circles. Therefore, the angle alpha (ααalpha) between the velocity of the electron and the magnetic field of the wire is 90 degrees.
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An ice skater is spinning on frictionless ice with her arms extended outward. She then pulls her arms in toward her body, reducing her moment of inertia. Her angular momentum is conserved, so as she reduces her moment of inertia, her angular velocity increases and she spins faster. Compared to her initial rotational kinetic energy, her final rotational kinetic energy is _________
Answer:
larger, because her angular speed is larger.
Explanation:
The rotational kinetic energy is proportional to the square of the angular velocity while it is linearly proportional to the moment of inertia. So the increase of angular speed will have a larger effect of the kinetic energy than the decrease of the moment of inertia.
Answer:
Rotational kinetic energy will Increase
Explanation:
Rotational kinetic energy KE is
KE = 1/2 x I x w^2
Where I is moment of inertia,
w is angular velocity.
It can be seen that increasing angular velocity increases rotational kinetic energy.
A 3.0-Ω resistor is connected in parallel with a 6.0-Ω resistor. This combination is then connected in series with a 4.0-Ω resistor. The resistors are connected across an ideal 12-volt battery. How much power is dissipated in the 3.0-Ω resistor? Group of answer choices
a. 12 w
b. 2.7 w
c. 6.0 w
d. 5.3 w
To solve this problem we must find the values of the equivalent resistances in both section 1 and section 2. Later we will calculate the total current and the total voltage. With the established values we can find the values of the currents in the 3 Ohms resistance and the power there.
The equivalent resistance in section 1 would be
[tex]R_{eq1} = \frac{(3\Omega)(6\Omega)}{(3+6)\Omega}[/tex]
[tex]R_{eq1} = 2\Omega[/tex]
The equivalent resistance in section 2 would be
[tex]R_{eq2} = R_{eq1} +4\Omega[/tex]
[tex]R_{eq2} = 6\Omega[/tex]
Now the total current will be,
[tex]I_t = \frac{V_t}{R_{eq2}}[/tex]
[tex]I_t = \frac{12V}{6\Omega}[/tex]
[tex]I_t = 2.0A[/tex]
Finally the total Voltage will be,
[tex]V = IR_{eq1}[/tex]
[tex]V = (2.0A)(2.0\Omega)[/tex]
[tex]V = 4V[/tex]
Since the voltage across the 3 and 6 Ohms resistor is the same, because they are in parallel, the current in section 3 would be
[tex]I_{3.0\Omega} = \frac{V}{R}[/tex]
[tex]I_{3.0\Omega} = \frac{4.0V}{3.0\Omega}[/tex]
[tex]I_{3.0\Omega} = 1.3A[/tex]
Finally the power ratio is the product between the current and the voltage then,
[tex]P_{3.0\Omega} = I_{3.0\Omega} V[/tex]
[tex]P_{3.0\Omega} = (1.3A)(4.0V)[/tex]
[tex]P_{3.0\Omega} = 5.3W[/tex]
Therefore the correct answer is D.
Final answer:
To find the power dissipated in the 3.0-Ω resistor, calculate the total resistance and current, then apply Ohm's law to find the voltage across and current through the resistor. Finally, use the power formula P = V²/R, resulting in a power dissipation of (a) 12 W for the 3.0-Ω resistor.
Explanation:
The question deals with finding the power dissipated in the 3.0-Ω resistor. To answer this, we first need to find the total resistance of the circuit and the current through the circuit. We then apply this current to the parallel resistors to find the voltage across and the current through the 3.0-Ω resistor before calculating its power dissipation.
First, calculate the resistance of the resistors in parallel. Using the formula for resistors in parallel:
1/Rparallel = 1/R₁ + 1/R₂
1/Rparallel = 1/3.0 + 1/6.0
1/Rparallel = 1/3 + 1/6 = 2/6 + 1/6 = 3/6
1/Rparallel = 1/2
Rparallel = 2.0 Ω
Now, add the series resistor to find the total resistance:
Rtotal = Rparallel + Rseries
Rtotal = 2.0 + 4.0 = 6.0 Ω
Then, using Ohm's law, calculate the total current from the battery:
I = V/Rtotal
I = 12V/6.0Ω
I = 2.0 A
The current through the 3.0-Ω resistor in parallel is the same as the total current, so we use Ohm's law V = IR to find the voltage across the 3.0-Ω resistor:
V3.0-Ω = 2.0 A × 3.0 Ω = 6.0 V
Finally, calculate the power dissipated by the 3.0-Ω resistor:
P = V2/R
P = (6.0 V)2/3.0 Ω
P = 36 W/3.0 Ω
P = 12 W
Therefore, the power dissipated in the 3.0-Ω resistor is 12 W, which corresponds to choice (a).
A 0.62112 slug uniform disc A is pinned to a 0.12422 slug uniform rod CB which is pinned to a 0.03106 slug collar C. C slides on a smooth vertical rod and the disc rolls without slipping. Determine: a) the angular acceleration ofA and the acceleration of C when released from rest and b)The velocity of B and C when the rod is horizontal. Ans:b) vc
Answer:
Explanation: check the attached document for step by step solution.
1. If only the vertical magnet were present, the field would point _____ at the point P.2. If only the horizontal magnet were present, the field would point _______.3. The vector combination of two magnetic fields from the two magnets points __________.Options areUpward and to the leftDownward and to the rightto the leftto the rightdownward and to the leftupwardupward and to the rightdownward
Answer:
1. downward
2. to the left
3. downward and to the left
Explanation:
This is gotten by using vector law of triangle Addition which states that If 2 vectors acting simultaneously on a body are represented both in magnitude and direction by 2 sides of a triangle taken in an order then the resultant(both magnitude and direction) of these vectors is given by 3rd side of that triangle taken in opposite order.
Domestic cats have vertical pupils. Imagine a cat is observing two small birds sitting side-by-side on a telephone wire. If the slit width of the cat's pupils is a = 0.550 mm and the average wavelength of the ambient light is λ = 519 nm, what is the angular resolution (in rad) for the two birds?
To solve this problem we will apply the concepts related to angular resolution based on wavelength and the slit width at this case of the cat's pupils. This relationship is given as,
[tex]\theta = \frac{\lambda}{a}[/tex]
Here,
[tex]\lambda[/tex] = Wavelength
a = Slit width
[tex]\theta = \frac{519*10^{-9}m}{0.55*10^{-3}m}[/tex]
[tex]\theta = 9.43*10^{-4} Rad[/tex]
Therefore the angular resolution is [tex]9.43*10^{-4}rad[/tex]
Final answer:
The angular resolution for a cat observing birds can be estimated using the formula θ = 1.22 λ / D, with a pupil width of 0.550 mm and a wavelength of 519 nm, yields an angular resolution of approximately 1.152 × 10⁻⁶ radians.
Explanation:
The question refers to angular resolution in optics, specifically related to the diffraction limit of a cat's eyes observing birds. In physics, the angular resolution for a circular aperture, like the pupil of a cat's eye, can be estimated using the formula θ = 1.22 λ / D, where θ is the angular resolution in radians, λ is the wavelength of the light, and D is the diameter of the aperture (the pupil in this case). Given that the cat's pupil has a slit width of a = 0.550 mm (which we will use in place of the diameter for this rough estimate) and the light has an average wavelength of λ = 519 nm, the angular resolution θ can be calculated as follows:
θ = 1.22 × 519 x 10⁻⁹ m / 0.550 x 10^-3 m
θ = 1.22 × 519 / 550 × 10⁻⁶
θ = 1.22 × 0.944 × 10⁻⁶
θ = 1.152 × 10⁻⁶ radians
Given the specified conditions, this calculation estimates the angular resolution for the cat's eyes. Note, however, that this is a simplification and doesn't take into account the complexities of a vertical slit pupil versus a circular aperture.
A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released from rest 2.0 mm from a very wide plane of positive charge, after which it accelerates toward the plane and collides with a speed of 4.5 m/s.
What is the surface charge density of the plane?
Answer:
[tex]6.75\mu C/m^2[/tex]
Explanation:
We are given that
Diameter,d=[tex]1\mu m=1\time 10^{-6} m[/tex]
[tex]1\mu m=10^{-6} m[/tex]
Radius,r=[tex]\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m[/tex]
Density,[tex]\rho=900kg/m^3[/tex]
Total number of electrons,n=39
Charge on electron =[tex]1.6\times 10^{-19} C[/tex]
Total charge=[tex]q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C[/tex]
Distance,s=2mm=[tex]2\times 10^{-3} m[/tex]
Mass =[tex]density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg[/tex]
Initial velocity,u=0
Final speed,v=4.5 m/s
[tex]v^2-u^2=2as[/tex]
[tex](4.5)^2-0=2a(2\times 10^{-3})[/tex]
[tex]20.25=4a\times 10^{-3}[/tex]
[tex]a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2[/tex]
Force,F=ma
[tex]qE=ma[/tex]
[tex]q(\frac{\sigma}{2\epsilon_0})=ma[/tex]
[tex]\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}[/tex]
[tex]\epsilon_0=8.85\times 10^{-12}[/tex]
[tex]\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2[/tex]
A particle enters a uniform magnetic field initially traveling perpendicular to the field lines and is bent with a circular arc of radius R. If this particle was traveling two times as fast, the radius of its circular arc would be? EXPLAIN
Answer:
Explanation:
In a scenario where a particle of charge overrightarrow{B} enters a magnetic field with a velocity overrightarrow{V}, it experiences a force overrightarrow{F} given by: overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B}).
implies F=BqVSin\theta.
Where theta is the angle between the velocity vector of the particle and the magnetic field vector.
When a particle enters the magnetic field at an angle 90, it moves in a circular path as it experiences a centripetal force, given by F=\frac{mV^2}{R}.
Where R is the radius of the circle, V is its velocity and m is its mass
Thus, magnetic force becomes F=BqVSin90^o=\frac{mV^2{R}\implies R=\frac{mV}{Bq}.
The equation changes as below, when velocity is doubled, let us assume that the radius is given by R_1.
R_1=\frac{2mV}{Bq}=2R.
Therefore, it is obvious that the velocity of a charged particle in a circular arc is directly proportional to the radius of the arc. The radius of the circular arc doubles when the velocity of the charged particle in the circular orbit doubles only if the mass, charge and magnetic field of the particle remains constant
Hence when velocity is doubled radius of the circle also gets doubled.
When the speed of the particle doubles, the radius of the arc also doubles.
The magnetic force on the particle is calculated as follows;
[tex]F = qvB[/tex]
The centripetal force on the particle is calculated as follows;
[tex]F_c = \frac{mv^2}{R}[/tex]
The speed of the particle is calculated as follows;
[tex]\frac{mv^2}{R} = qvB\\\\mv = qBR\\\\v = \frac{qBR}{m} \\\\\frac{v_1}{R_1} = \frac{v_2}{R_2}[/tex]
when the speed of the particle doubles;
[tex]\frac{v_1}{R_1} = \frac{2v_1}{R_2} \\\\R_2v_1 = 2R_1v_1\\\\R_2 = 2R_1[/tex]
Thus, we can conclude that when the speed of the particle doubles, the radius of the arc also doubles.
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A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.45 m high and 12.4 m long incline. The total mass is 60.0 kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.70 m/s. Determine the work done (in J) by the girl as the bike travels down the incline.
Answer:
Explanation:
Given that,
Initial speed of the girl is
u = 1.4m/s
Height she is going is
H = 2.45m
Incline plane she will pass to that height
L = 12.4m
Mass of girl and bicycle is
M=60kg
Frictional force that oppose motion is
Fr = 41N
Speed at lower end of inclined plane
V2 = 6.7m/s
Work done by the girl when the car travel downward
Using conservation of energy
K.E(top) + P.E(top) + work = K.E(bottom) + P.E(bottom) + Wfr
Where Wfr is work done by friction
Wfr = Fr × d
P.E(bottom) is zero, sicne the height is zero at the ground
K.E is given as ½mv²
Then,
½M•u² + MgH + W = ½M•V2² + 0 + Fr×d
½ × 60 × 1.4² + 60×9.8 × 2.45 + W = ½ × 60 × 6.7² + 41 × 12.4
58.8 + 1440.5 + W = 1855.1
W = 1885.1 —58.8 —1440.5
W = 355.8 J
You build a grandfather clock, whose timing is based on a pendulum. You measure its period to be 2s on Earth. You then travel with the clock to the distant planet CornTeen and measure the period of the clock to be 4s. By what factor is the gravitational acceleration constant g different on planet CornTeen compared to g on Earth
Answer:
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
Explanation:
The period of the simple pendulum is:
[tex]T = 2\pi\cdot \sqrt{\frac{l}{g} }[/tex]
Where:
[tex]l[/tex] - Cord length, in m.
[tex]g[/tex] - Gravity constant, in [tex]\frac{m}{s^{2}}[/tex].
Given that the same pendulum is test on each planet, the following relation is formed:
[tex]T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}[/tex]
The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{T_{1}}{T_{2}} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \left(\frac{2\,s}{4\,s} \right)^{2}[/tex]
[tex]\frac{g_{2}}{g_{1}} = \frac{1}{4}[/tex]
The rho− meson has a charge of −e, a spin quantum number of 1, and a mass 1 507 times that of the electron. The possible values for its spin magnetic quantum number are −1, 0, and 1. Imagine that the electrons in atoms are replaced by rho− mesons. Select all of the following which are possible sets of quantum numbers (n, ℓ, mℓ, s, ms) for rho− mesons in the 3d subshell.
A. (2, 2, 1, 1, 0)
B. (3, 2, -1, 1, 1)
C. (3, 2. -1, 1, 1/2)
D. (3, 2, 0, 1, 1)
E. (3, 2, 0, 1, -1)
F. (3, 2, -1, 1, 0)
Answer:
Look up attached file
Explanation:
One end of a string is attached to a rigid wall on a tabletop. The string is run over a frictionless pulley and the other end of the string is attached to a stationary hanging mass. The distance between the wall and the pulley is 0.405 meters, When the mass on the hook is 25.4 kg, the horizontal portion of the string oscillates with a fundamental frequency of 261.6 Hz (the same frequency as the middle C note on a piano). Calculate the linear mass density of the string.
Answer:
The linear mass density is of the string [tex]\mu= 5.51*10^{-3} kg / m[/tex]
Explanation:
From the question we are told that
The distance between wall and pulley is [tex]d = 0.405m[/tex]
The mass on the hook is [tex]m = 25.4\ kg[/tex]
The frequency of oscillation is [tex]f = 261.6 Hz[/tex]
Generally, the frequency of oscillation is mathematically represented as
[tex]f = \frac{1}{2d} \sqrt{\frac{T}{\mu} }[/tex]
Where T is the tension mathematically represented as
T = mg
Substituting values
[tex]T = 25.4 *9.8[/tex]
[tex]=248.92N[/tex]
[tex]\mu[/tex] is the mass linear density
Making [tex]\mu[/tex] the subject of the formula above
[tex]\mu = \frac{T}{(2df)^2}[/tex]
Substituting values
[tex]\mu = \frac{248.92}{(2 * 0.405 * 261.6)^2}[/tex]
[tex]\mu= 5.51*10^{-3} kg / m[/tex]
Answer:
0.005550 Kg/m
Explanation:
The picture attached below shows the full explanation
Although the transmission of light, and electromagnetic radiation generally, is correctly described by wave (physical) optics, there are situations for which ray (geometric) optics gives a sufficiently good approximation. For each of the situations described in the following, determine whether ray optics may be used or wave optics must be used. Part (a) Green laser light of wavelength 530 nm is incident on a 26-cm diameter mirror. Ray None of these. WavePart (b) Red laser light of wavelength 722 nm is incident on a molecule of size 114 nm. Wave Ray None of these
Answer:
a) the geometric optics is adequate (Ray)
b) wave optics should be used for the second case
Explanation:
In general, the approximation of the geometric optics is adequate when the dimensions of the system are much greater than the wavelengths and the wave optics should be used for cases in which the size of the system is from the length of the wave.
Let's apply this to our case
a) in this case the size of the system is d = 26 cm=0.26 m and the wavelength is alm = 530 10⁺⁹ m
in this case
d >>> λ
therefore the geometric optics is adequate (Ray)
b) in this case the system has a size of d = 114 10⁻⁹ m with a wavelength of λ= 722 nm
for this case d of the order of lam
therefore wave optics should be used for the second case
The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the wire represent the direction of current). For the following wires, which all also carry current , indicate the magnitude (in mT) and direction of the magnetic field at the center (red point) of each configuration.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The magnetic field is [tex]B_{net} = \frac{1}{4} * mT[/tex]
And the direction is [tex]-\r k[/tex]
Explanation:
From the question we are told that
The magnetic field at the center is [tex]B = 1mT[/tex]
Generally magnetic field is mathematically represented as
[tex]B = \frac{\mu_o I}{2R}[/tex]
We are told that it is equal to 1mT
So
[tex]B = \frac{\mu_o I}{2R} = 1mT[/tex]
From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)
[tex]\frac{\mu_o I}{2R} = 1mT[/tex]
This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)
[tex]B_1 = \frac{1}{2} \frac{\mu_o I}{2R}[/tex]
and for the larger semi-circular loop is
[tex]B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction
So the net magnetic field would be
[tex]B_{net} = B_1 - B_2[/tex]
[tex]= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}[/tex]
[tex]=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}[/tex]
[tex]=\frac{\mu_o I}{8R}[/tex]
[tex]= \frac{1}{4} \frac{\mu_o I}{2R}[/tex]
Recall [tex]\frac{\mu_o I}{2R} = 1mT[/tex]
So
[tex]B_{net} = \frac{1}{4} * mT[/tex]
Using the Right-hand rule we see that the direction is into the page which is [tex]-k[/tex]
Two in-phase loudspeakers, which emit sound in all directions, are sitting side by side. One of them is moved sideways by 3.0 m, then forward by 7.0 m. Afterward, constructive interference is observed 14, 12, and 34 the distance between the speakers along the line that joins them, and at no other positions along this line.What is the maximum possible wavelength of the sound waves?Express your answer with the appropriate units.
The maximum possible wavelength for constructive interference from two shifted loudspeakers at given interference maxima positions is calculated to be 0.5738 m.
Explanation:The question is about the phenomenon of constructive interference of sound waves emitted from two loudspeakers that are shifted apart. The problem requires identifying the maximum possible wavelength of the sound waves produced by the speakers.
Constructive interference occurs when two waves meet and their crests (high points) and troughs (low points) align. This occurs effectively when the path difference is either zero or a multiple of the wavelength. When the path difference is specifically an integral multiple of the wavelength, the interference is constructive and has a maximum value.
Given that the loudspeakers are moved to a distance of 3.0 m sideways and 7.0 m forward, the distance separating the two is calculated using Pythagoras's theorem as √((7)^2+(3)^2) = √58 m. The maxima of the constructive interference are at positions 14, 12, and 34 the distance between the speakers, thus the maximum possible wavelength is 2 * √58 / 34 = 0.5738 m.
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An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E. Part A What is the system's potential energy when its kinetic energy is equal to 34E
The correct question is;
An object of mass m attached to a spring of force constant K oscillates with simple harmonic motion. The maximum displacement from equilibrium is A and the total mechanical energy of the system is E.
What is the system's potential energy when its kinetic energy is equal to ¾E?
Answer:
P.E = ⅛KA²
Explanation:
From conservation of energy, the total energy in the system is given as the sum of potential and kinetic energy.
Thus,
Total Energy; E = K.E.+P.E.
In simple harmonic motion, the total energy is given by;
E = ½KA²
We are told that kinetic energy is ¾E.
Thus, ½KA² = ¾(½KA²) + P.E
P.E = ½KA² - ⅜KA²
P.E = ⅛KA²
If enough heat was REMOVED from B, it would change into ____________.
A transparent oil with index of refraction 1.28 spills on the surface of water (index of refraction 1.33), producing a maximum of reflection with normally incident orange light (wavelength 600 nm in air). Assuming the maximum occurs in the first order, determine the thickness of the oil slick. nm
To solve this problem we will apply the concepts related to the principle of destructive and constructive interference. Mathematically this expression can be given as
[tex]2nt = m\lambda[/tex]
Here,
n = Index of refraction
t = Thickness
m = Order of the reflection
[tex]\lambda[/tex] = Wavelength
We have all of this values, therefore replacing,
[tex]2(1.28)t = (1)(600nm)[/tex]
[tex]t = 233nm[/tex]
Therefore the thickness of the oil slick is 233nm
g You're a safety engineer reviewing plans for a university's new high-rise dorm. The elevator motors draw 20 A and behave electrically like 2.4-H inductors. You're concerned about dangerous voltages developing across the switch when a motor is turned off, and you recommend that a resistor be wired in parallel with each motor. Part A What should be the resistance in order to limit the emf to 100 V
To solve the problem it is necessary to apply Ohm's law. From this it is established that the voltage is the equivalent to the product between the current and the resistance, therefore we have to,
[tex]V = IR[/tex]
Here,
V = Voltage
I = Current
R = Resistance
Rearranging to find the resistance,
[tex]R = \frac{V}{ I}[/tex]
Replacing,
[tex]R = \frac{100V}{20A}[/tex]
[tex]R = 5\Omega[/tex]
Therefore the resistance should be [tex]5\Omega[/tex]
The resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.
Ohm's law:
It states that the voltage is the equival to the product of the current and the resistance.
[tex]\bold{V =I\times R}[/tex]
Where,
V = Voltage = 100 Volts
I = Current = 20 Ampere
R = Resistance
Put the values and solve it for R
[tex]\bold {R =\dfrac{100}{20}}\\\\\bold {R = 5\Omega}[/tex]
Therefore, the resistance should be [tex]\bold { 5\Omega}[/tex] in order to limit the emf to 100 V.
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A string of 18 identical Christmas tree lights are connected in series to a 130 V source. The string dissipates 61 W.
What is the equivalent resistance of the light string?
Answer in units of Ω.
What is the resistance of a single light? Answer in units of Ω.
How much power is dissipated in a single light?
Answer in units of W.
One of the bulbs quits burning. The string has a wire that shorts out the bulb filament when it quits burning, dropping the resistance of that bulb to zero. All the rest of the bulbs remain burning.
What is the resistance of the light string now?
Answer in units of Ω.
How much power is dissipated by the string now?
Answer in units of W.
Answer:
(a) 277.05 Ω
(b) 15.39 Ω
(c) 3.76 W
Explanation:
(a)
Applying,
P = V²/R.......................... Equation 1
Where P = Power dissipated by the string. V = Voltage source, R = equivalent resistance of the light string
Make R the subject of the equation
R = V²/P................... Equation 2
Given: V = 130, P = 61 W
Substitute into equation 2
R = 130²/61
R = 277.05 Ω
(b) The resistance of a single light is given as
R' = R/18 (since the light are connected in series and the are identical)
Where R' = Resistance of the single light.
R' = 277.05/18
R' = 15.39 Ω
(c)
Heat dissipated in a single light is given as
P' = I²R'..................... Equation 3
Where P' = heat dissipated in a single light, I = current flowing through each light.
We can calculate for I using
P = VI
make I the subject of the equation
I = P/V
I = 61/130
I = 0.469 A.
Also given: R' = 15.39 Ω
Substitute into equation 3
P' = 0.496²(15.39)
P' = 3.76 W
(a)The equivalent resistance of the light string is 277.05 Ω
(b)The power is dissipated in a single ligh15.39 Ω
(c)The resistance of the light string now3.76 W
Calculation of Power is dissipates
(a) P is = V²/R.......................... Equation 1
Where P is = Power dissipated by the string.
Then V = Voltage source,
After that R is = the equivalent resistance of the light string
Now Make R the subject of the equation
R is = V²/P................... Equation 2
Then Given: V = 130,
P = 61 W
After that Substitute into equation 2
Then R = 130²/61
Therefore, R = 277.05 Ω
(b) When The resistance of a single light is given as
R' is = R/18 (since the light are connected in series and are identical)
Now Where R' is = Resistance of the single light.
R' is = 277.05/18
Therefore, R' is = 15.39 Ω
(c) When Heat dissipated in a single light is given as
P' is = I²R'..................... Equation 3
Where P' is = heat dissipated in a single light,
Then I = current flowing through each light.
Now We can calculate for I using
P is = VI
Now we make I the subject of the equation
After that I = P/V
Then I = 61/130
I is = 0.469 A.
Also given: R' is = 15.39 Ω
Then Substitute into equation 3
P' is = 0.496²(15.39)
Therefore, P' is = 3.76 W
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The projectile partially fills the end of the 0.3 m pipe. Calculate the force required to hold the projectile in position when the mean velocity in the pipe is 6 m
Final answer:
To hold the projectile in position, the force required is 20 Newtons.
Explanation:
To calculate the force required to hold the projectile in position, we need to consider the acceleration of the projectile in the pipe.
From the given information, we can calculate the acceleration using the formula:
Acceleration = Change in velocity / Time taken
Since the mean velocity is given as 6 m/s, we can use this as the final velocity and assume the initial velocity is 0 m/s.
Substituting the values in the formula, we have:
Acceleration = (6 m/s - 0 m/s) / 0.3 m = 20 m/s^2
Now we can calculate the force using Newton's second law:
Force = Mass * Acceleration
Assuming the mass of the projectile is 1 kg, we have:
Force = 1 kg * 20 m/s^2 = 20 N
Light from an LED with a wavelength of 4.90 ✕ 102 nm is incident on (and perpendicular to) a pair of slits separated by 0.310 mm. An interference pattern is formed on a screen 2.20 m from the slits. Find the distance (in mm) between the first and second dark fringes of the interference pattern.
Answer:
Δx = 3.477 x 10⁻³ m = 3.477 mm
Explanation:
The distance between two consecutive dark fringes is given by the following formula, in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes = ?
λ = wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L = Distance between slits and screen = 2.2 m
d = slit separation = 0.31 mm = 0.31 x 10⁻³ m
Therefore,
Δx = (4.9 x 10⁻⁷ m)(2.2 m)/(0.31 x 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
The distance (in mm) between the first and second dark fringes of the
interference pattern is 3.477 mm
This is calculated by using the formula in Young's Double Slit experiment:
Δx = λL/d
where,
Δx = distance between two consecutive dark fringes which is unknown
λ which is wavelength of light = 4.9 x 10² nm = 4.9 x 10⁻⁷ m
L which is distance between slits and screen = 2.2 m
d which is slit separation = 0.31 mm = 0.31 x 10⁻³ m
We then substitute them into the equation
Δx = (4.9 x 10⁻⁷ m ×2.2 m)/(0.31 × 10⁻³ m)
Δx = 3.477 x 10⁻³ m = 3.477 mm
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A two-stage rocket is traveling at 1210m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 40m/s relative to the second stage after the explosion. The first stage is three times as massive as the second stage.
What is the speed of the second stage after the separation?
Answer:
R
Explanation:
Given that,
Two stage rocket traveling at
V = 1210 m/s with respect to earth
First stage
When fuel Is run out , explosive bolts releases and push rocket backward at speed is
V1 = 40m/s relative to second stage
Therefore
V1 = 40 - V2
The first stage is 3 times as massive as the second stage
I.e Mass of first stage is 3 times the second stage
Let Mass of second stage be
M2 = M
Then, M1 = 3M
Velocity of second stage V2?
Applying conservation of linear momentum
Momentum before explosion = momentum after explosion
Momentum p=mv
Then,
(M1+M2)V = —M1•V1 + M2•V2
(3M+M)•1210 = —3M•(40-V2) +M•V2
4M × 1210 = —120M + 3M•V2 + MV2
4840M = —120M + 4M•V2
4840M + 120M = 4M•V2
4960M = 4M•V2
Then, V2 = 4960M / 4M
V2 = 1240 m/s
Why might these "Mental Maps" be inaccurate or differ between different people?
Answer:
Catastrophic events of weather related outcomes.
Explanation:
The mental map of a person from a certain place may change due to long periods of time outside, since this gives the mind time to forget certain important details, they may also change according to people's experience and perception, places, regions and environments since these places are also changing and the perception we had is no longer the same. A mental map is a first person perspective of an area that an individual possesses. This type of subconscious map shows a person how a place looks and how to interact with it.
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A proton and a deuteron are moving with equal velocities perpendicular to a uniform magnetic field. A deuteron has the same charge as the proton but has twice its mass. The ratio of the magnetic force on the proton to that on the deuteron is:
a. 0.5.
b. 1.
c. 2.
d. There is no magnetic force in this case.
Answer:
option (b)
Explanation:
mass of proton, mp = m
mass of deuteron, md = 2m
charge on proton, qp = q
charge on deuteron, qd = q
The magnetic force on the charged particle when it is moving is given by
F = q v B Sinθ
where, θ is the angle between the velocity and magnetic field.
Here, θ = 90°
Let v is the velocity of both the particle when they enters in the magnetic field.
The force on proton is given by
Fp = q x v x B ...... (1)
The force on deuteron is
Fd = q x v x B .... (2)
Divide equation (1) by equation (2)
Fp / Fd = 1
Thus, the ratio of force on proton to the force on deuteron is 1 : 1.
Thus, option (b) is correct.
Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very long, straight wire. If you were unfortunate enough to be 4.9 m away from such a lightning bolt, how large a magnetic field would you experience
Answer:
how large a magnetic field would you experience = 8.16 x 10∧-4T
Explanation:
I = 20KA = 20,000A
r = 4.9 m
how large a magnetic field would you experience = u.I/2πr
how large a magnetic field would you experience = (4π x10∧-7) × 20000/2π × 4.9
how large a magnetic field would you experience = 8.16 x 10∧-4T
Answer: 8.16*10^-4 T
Explanation:
Given
Current of the lightening bolt, I = 20 kA
Distance from the strike of the lightening bolt, r = 4.9 m
To solve, we use the formula
B = [μ(0) * I] / 2πr, where
B = magnetic field of the lightening
μ = permeability constant = 4π*10^-7 N/A²
I = current of the lightening
r = distance from the lightening strike
B = [(4 * 3.142*10^-7) * 20*10^3] / (2 * 3.142 * 4.9)
B = (12.568*10^-7 * 20*10^3) / 6.284 * 4.9
B = 0.025 / 30.79
B = 8.16*10^-4 T
The magnetic field to be experienced would be 8.16*10^-4 T large
A 150kg person stands on a compression spring with spring constant 10000n/m and nominal length of 0.50.what is the total length of the loaded spring
Answer:
The total length of the spring would be 0.65 m
Explanation:
The Concept
Hooke's law evaluates the increment of spring in relation to the force acting on the body. Hooke's law states that for a spring undergoing deformation, the force applied is directly proportional to the deformation experienced by the spring. Hooke's law is represented thus;
F = k x ..................1
where F is the force applied to the spring
k is the spring constant
x is the spring stretch or extension
Step by Step Calculations
We have to obtain x before adding it to the nominal length, We make x the subject formula in equation 1;
x = F/k
but F = m x g
so, x = (m x g)/k
given that, the mass of the person m =150 kg
g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex]
k is the spring constant = 10000 N/m
then x = (9.81 m/[tex]s^{2}[/tex] x 150 kg)/10000 N/m
x = 0.14715 m
the extension experienced by the spring after the compression is 0.14715 m
The total length of the spring would be;
L = 0.14715 m + 0.5 m = 0.64715
L ≈ 0.65 m
Therefore the total length of the spring would be 0.65 m
g If 6.35 moles of a monatomic ideal gas at a temperature of 320 K are expanded isothermally from a volume of 1.45 L to a volume of 3.95 L, calculate: a) the change in the internal energy of the gas. b) the work done by the gas. c) the heat flow into or out of the gas.
Answer:
(a) change in the internal energy of the gas is zero
(b) the work done by the gas is 16.93 kJ
(c) the heat flow is 16.93 kJ, which is into the gas
Explanation:
Given;
number of moles of gas, n = 6.35 moles
temperature of the gas, T = 320 K
initial volume of the gas, V₁ = 1.45 L
final volume of the gas, V₂ = 3.95 L
Part (a)
For isothermal expansion, temperature is constant and internal energy will also be constant.
Therefore, change in the internal energy of the gas is zero since the gas expanded isothermally (constant temperature).
ΔU = Q - W
where;
ΔU is change in internal energy
Q is heat transferred to the system
W is the work done by the system
Thus, Q = W
ΔU = 0
Part (b)
the work done by the gas
[tex]W = nRTln{[\frac{V_2}{V_1}][/tex]
where;
R is gas constant = 8.314 J/mol.K
[tex]W = (6.35)(8.314)(320)ln{[\frac{3.95}{1.45}]}\\\\W =16930.4\ J\\\\W = 16.93\ kJ[/tex]
Part (c)
the heat flow into or out of the gas
Q = ΔU + W
Q = 0 + 16.93 kJ
Q = 16.93 kJ
Since the heat flow is positive, then it is heat flow into the gas.
what is the electric potential at point A in the electric field created by a point charge of 5.5 • 10^-12 C? estimate k as 9.00 • 10^9
The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
Explanation:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
The electric potential at a point in the electric field created by a point charge is calculated by the formula V = kQ/r. The provided constants are [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex] and [tex]k = 9.00 \times 10^9[/tex] The distance r from the point charge is necessary to calculate the electric potential.
Explanation:The electric potential at point A in the electric field created by a point charge can be found using the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the value of the point charge, and r is the distance from the point charge.
Given, the charge [tex]Q = 5.5 \times 10^{-12}\, \text{C}[/tex], and Coulomb's constant [tex]k = 9.00 \times 10^9[/tex]
The distance r is not given in the question. Therefore, we can't calculate the electric potential at point A. However, if distance r from the point charge were provided (in meters), we could substitute k, Q, and r into the formula and solve for V.
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Because all blocks weigh the same, their masses must also be equal. Density is defined as mass per unit volume, so we know the densities of the red and blue blocks are different because their volumes are not the same. Since the volume of a blue block is one-half the volume of a red block, the density of a blue block is ________ the density of a red block.
a.half
b.equal to
Answer:
the density of a blue block is ___twice _____ the density of a red block.
Explanation:
If both blocks have the same mass, then the block with the lesser volume will be more densely packed when compared to the block with the larger volume. This is because the molecules are more closely packed.
Check the image below for detailed calculations for proof
If the blue block and red block have the same mass but different volumes - specifically, the blue block's volume is half that of the red block - then the density of the blue block is twice that of the red block.
Explanation:Given the blocks weigh the same, their masses are equal. Density is defined as mass per unit volume (mass/volume), meaning the density is influenced by both the mass and the volume of an object. As the red and blue blocks have different volumes, their densities must be different if their masses are the same.
In the case of the blue block and the red block, the blue block has half the volume of the red block but the same mass. Thus, because the blue block's volume is half that of the red block but the mass is the same, the density of the blue block is twice the density of a red block, not half. This is because the proportion of mass to volume in the blue block is larger than in the red block. In other words: Density of blue block = 2 * Density of red block.
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