Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have [tex]m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}[/tex] , where:
[tex]l[/tex] is the length
[tex]m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3[/tex]
We divide the first equation by the second equation to get:
[tex]\frac{m}{R} = \frac{d A^2}{\rho}[/tex]
[tex]A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2[/tex]
Using this Area, we find the diameter of the wire:
[tex]D = \sqrt{\frac{4A}{\pi}}[/tex]
[tex]D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}[/tex]
[tex]D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm[/tex]
To find the length, we multiply the two equations stated initially:
[tex]mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}[/tex]
[tex]l^2 = 8.534\\l = 2.92 \ m[/tex]
If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, estimate how many frames are needed to show a two hour long movie.
1 hour = 3600 seconds
2 hrs = 7200 sec
(24 frame/sec) x (72 sec) =
172,800 frames.
If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second, then the number of frames needed to show a two-hour-long movie would be 172800 frames
What is multiplication?Finding the product of two or more numbers in mathematics is done by multiplying the numbers. It is one of the fundamental operations in mathematics that we perform on a daily basis.
As given in the problem If each frame of a motion picture film is 35 mm high, and 24 frames go by in a second,
The number of seconds in 2 hours = 2 ×3600
=7200 seconds
The number of frames goes by in one second = 24 frames
the number of frames in a 2-hour movie = 24×7200
=172800 frames
Thus, the number of frames needed to show a two-hour-long movie would be 172800 frames
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A couple is defined as two parallel forces, separated by a distance, that have equal magnitudes but opposite directions. A couple only produces a rotation in a specified direction. The moment produced by a couple is called a couple moment.
A submarine hatch door is to be opened by applying two oppositely oriented forces of equal magnitude F=265N at points A and B on the hatch door wheel. The radii of the wheel's inner and outer rings are r1 = 0.470 m and r2 = 0.200 m, respectively.
Calculate the moments MA and MB about point D for the forces applied at points A and B. Then, determine the resulting couple moment MR. Assume that a positive moment produces a counterclockwise rotation whereas a negative moment produces a clockwise rotation.
Answer:
MA = 178 Nm
MB = 72 Nm
MR = 249 Nm
For the moments MA the lever arm was taken as r1 + r2 which is the distance of the point of application of the force at point A to point D.
For the moment MB, the distance of the point of application of force is r1 - r2 which is the distance from the outer ring to the inner ring.
The couple moment is given by F × r1. Which is basically a sum of the moments of both forces applied on the wheel.
Explanation:
See the attachment for detail of the calculation.
The moment MA for the force at point A is 124.55 N*m, the moment MB for the force at point B is 53 N*m, and the resulting couple moment MR is 71.55 N*m.
Explanation:To calculate the moments MA and MB about point D for the forces applied at points A and B, we can use the formula:
M = F * r
Where M is the moment, F is the force, and r is the radius. For point A, the radius is r1 = 0.470 m and the force is F = 265 N. So, MA = 265 N * 0.470 m = 124.55 N*m (counterclockwise).
Similarly, for point B, the radius is r2 = 0.200 m and the force is F = 265 N. So, MB = 265 N * 0.200 m = 53 N*m (clockwise).
To determine the resulting couple moment MR, we can subtract the clockwise moment (MB) from the counterclockwise moment (MA). MR = MA - MB = 124.55 N*m - 53 N*m = 71.55 N*m (counterclockwise).
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On the moon, all free-fall distance functions are of the forms(t)=0.81t^2 where t is in seconds and s is in meters. An object is dropped from a height of 200
meters above the moon. After 9 sec, consider parts (a) through(d) below.
a)
How far has the object fallen?
b)
How fast is it traveling?
c)
What is its acceleration?
d)
Explain the meaning of the second derivative of this free-fall function.
After 9 seconds, an object dropped on the moon will have fallen 65.61 meters, be traveling at 14.58 meters per second, with an acceleration of 1.62 meters per second squared. The second derivative of the distance function represents the moon's gravitational acceleration.
Explanation:An object is dropped on the moon from a height of 200 meters, and the distance function of free-fall due to lunar gravity is given by s(t)=0.81t^2. We are asked to determine various properties of the object's motion after 9 seconds.
a) Distance Fallen
To find how far the object has fallen, we plug t=9 into the distance function to get s(9)=0.81(9)^2=65.61 meters. This is the distance the object has fallen, not the distance from its starting height.
b) Velocity
The object's velocity can be determined by finding the derivative of the distance function, which gives us v(t) = 2(0.81)t = 1.62t. Substituting t=9, we find v(9) = 1.62(9) = 14.58 meters per second.
c) Acceleration
The acceleration of the object is constant and equal to twice the coefficient of t^2 in the distance function, which is 1.62 meters per second squared on the moon.
d) Meaning of the Second Derivative
The second derivative of the free-fall distance function represents the acceleration due to gravity on the moon. It's constant at 1.62 meters per second squared, indicating uniform acceleration during free-fall.
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘
Answer:
The conditions are not given in the question. Here is the complete question.
High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from a climber's body under the following conditions. The specific heat of ice is 2100 J/kg⋅C∘, the latent heat of fusion is 333 kJ/kg, the specific heat of water is 4186 J/kg⋅C∘
a) eat 1.0kg of -15 degrees Celsius snow which your body warms to body temperature of 37 degrees Celsius ?
b) melt 1.0kg of -15 degrees Celsius snow using a stove and drink the resulting 1.0kg of water at 2 degrees Celsius, which your body has to warm to 37 degrees Celsius ?
Explanation:
Let's calculate the heat required to convert ice from -15°C to 0°C and then the heat required to convert it into the water.
a)
Heat required to convert -15°C ice to 0°C.
[tex]ms_{ice}[/tex]ΔT = (1.0)(2.100×10³)(15) = 3.150×[tex]10^{4}[/tex]J
Heat required to convert 1.0 kg ice to water.
[tex]mL_{ice} = 1[/tex]×[tex]3.33[/tex]×[tex]10^{5}[/tex] = 3.33×[tex]10^{5}[/tex]J
Heat required to convert 1.0 kg water at 0°C to 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×37 = 1.548×[tex]10^{5}[/tex] J
Total heat required = 3.150×[tex]10^{4}[/tex] + 3.33×[tex]10^{5}[/tex] + 1.548×[tex]10^{5}[/tex]
= 5.19×[tex]10^{5}[/tex] J
b)
Heat required to warm 1.0 kg water at 2°C to water at 37°C.
[tex]ms_{water}[/tex]ΔT = 1×4.186×[tex]10^{3}[/tex]×35
= 1.465×[tex]10^{5}[/tex]J
At the start of meiosis II, the cell exhibits characteristics of a haploid cell preparing for mitosis. It has one set of homologous chromosomes, each with two chromatids, equivalent to a haploid cell in the G₂ phase of interphase.
Explanation:At the beginning of meiosis II, a cell appears much like a haploid cell preparing to undergo mitosis. After completion of meiosis I, the cell does not duplicate its chromosomes, so at the onset of meiosis II, each dividing cell has only one set of homologous chromosomes, each with two chromatids. This results in half the number of sister chromatids to separate out as a diploid cell undergoing mitosis. In terms of chromosomal content, cells at the start of meiosis II are similar to those of haploid cells in the G₂ phase of interphase, where they are preparing for mitosis.
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A bicycle is rolling down a circular portion of a path; this portion of the path has a radius of 9.30 m. As the drawing illustrates, the angular displacement of the bicycle is 1.130 rad. What is the angle (in radians) through which each bicycle wheel (radius = 0.300 m) rotates?
Answer:
The angle through which each bicycle wheel rotates is 35.03 rad.
Explanation:
Given;
the radius of the circular path, R = 9.30 m
the angular displacement of the bicycle, θ = 1.130 rad
the radius of the bicycle wheel, r = 0.3
S = θR
where;
s is the distance of the circular path
S = 1.13 x 9.3 = 10.509 m
The angle (in radians) through which each bicycle wheel of radius 0.300 m rotates is given as;
θr = 10.509 m
θ = 10.509 / 0.3
θ = 35.03 rad.
Therefore, the angle through which each bicycle wheel of radius 0.300 m rotates is 35.03 rad.
Two protons are released from rest when they are 0.750 {\rm nm} apart.
a) What is the maximum speed they will reach?
b) What is the max acceleration they will reach?
Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
By clicking the two-slit barrier and dragging it, you can change its position. Always allow a few seconds for the simulation to catch up with the changes you are making before analyzing the results. When the two-slit barrier is brought closer to the screen (and farther away from the source), what happens to the distance between two adjacent bright (or dark) fringes?
A. It becomes larger.
B. It becomes smaller.
C. It stays the same.
Answer:
the answer the correct one is B
Explanation:
The interference phenomenon is described by
d sin θ = m λ
The spectrum is recorded on a screen, so we can use trigonometry
tan θ = y / L
In this experiment the angles are very small, so
tan θ = sin θ /cos θ = sin θ
We replace
d y / L = m La m
y = m Lam L / d
When the slits approach the screen the value of L decreases,
Therefore the value of the separation between the slits must also decrease
When reviewing the answer the correct one is B
The inner and outer surfaces of a cell membrane carry a negative and positive charge respectively. Because of these charges, a potential difference of about 70 mV exists across the membrane. The thickness of the membrane is 8 nm. If the membrane were empty (filled with air), what would the magnitude of the electric field inside the membrane
Answer:
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
Explanation:
The electric field due to electric potential at a distance Δs is given by
E=ΔV/Δs
We have to find the magnitude electric field in the membrane
Ecell= -ΔV/Δs
[tex]E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m[/tex]
The magnitude of the electric field inside the membrane is 8.8×10⁶V/m
The magnitude of the electric field inside the membrane, if it were empty, would be approximately [tex]\( 8.75 \times 10^6 \) V/m.[/tex]
To find the magnitude of the electric field inside the membrane, we can use the formula for electric field strength ( E ) due to a uniform field between two parallel plates, which is given by:
[tex]\[ E = \frac{V}{d} \][/tex]
where ( V ) is the potential difference across the plates (in volts) and ( d ) is the separation between the plates (in meters). In this case, the potential difference ( V ) is given as 70 mV, which we need to convert to volts:
[tex]\[ V = 70 \text{ mV} = 70 \times 10^{-3} \text{ V} \][/tex]
The thickness of the membrane ( d ) is given as 8 nm, which we need to convert to meters:
[tex]\[ d = 8 \text{ nm} = 8 \times 10^{-9} \text{ m} \][/tex]
Now we can plug these values into the formula for the electric field:
[tex]\[ E = \frac{70 \times 10^{-3} \text{ V}}{8 \times 10^{-9} \text{ m}} \][/tex]
[tex]\[ E = \frac{70}{8} \times 10^{(3 - (-9))} \text{ V/m} \][/tex]
[tex]\[ E = 8.75 \times 10^{6} \text{ V/m} \][/tex]
You are operating an 80kg reciprocating machine. The manufacturer notified you thatthere is an imbalance mass of 3kg on the rotating shaft, which has a 10cm diameter.The system was designed to have negligible damping.P.1.1What is the steady state amplitude of the machine’s displacement if you are operatingat very high frequencies?
Answer:
Explanation:
The system can be modeled as,
Using magnitude relationship for imbalance system.
Check attachment for solution
A 2.0-kg ball is moving with a constant speed of 5.0 m/s in a horizontal circle whose diameter is 1.0 m. What is the magnitude of the net force on the ball
Answer:
F = 100.0 N
Explanation:
in order to the ball keeps moving at a constant speed in a circle, instead of moving along a straight line, there must be an acceleration that accounts for the change in direction of the ball.This acceleration, called centripetal, is directed at any time, towards the center of the circle.As any acceleration, as dictated by Newton's 2nd law, it must be produced by a force, called centripetal force.The magnitude of this force is related with the mass, the speed and the radius of the circle, as follows:[tex]F_{c} = m *\frac{v^{2} }{r} = 2.0 kg *\frac{(5.0m/s)^{2} }{0.5m} =100.0 N[/tex]
The magnitude of the net force on the ball is 100.0 N100.0N
Explanation:As the ball moves round the circle, a centripetal acceleration which is directed towards the center of the circle will keep the ball from falling off. This acceleration produces a force called centripetal force (F).
Since this is the only force acting on the ball, then the net force acting on the ball to keep it moving round the circle is the centripetal force.
F = m a [according to Newton's second law of motion] --------------(i)
Where;
m = mass of the ball.
a = centripetal acceleration. = [tex]\frac{v^2}{r}[/tex]
v = speed of the ball.
r = radius of the circle.
Substitute a = [tex]\frac{v^2}{r}[/tex] into equation (i) as follows;
F = m x [tex]\frac{v^2}{r}[/tex] --------------------(ii)
From the question;
m = 2.0kg
v = 5.0m/s
r = diameter / 2 [diameter = 1.0m]
r = 1.0 / 2
r = 0.5m
Substitute these values into equation (ii) as follows;
F = 2.0 x [tex]\frac{5.0^{2} }{0.5}[/tex]
F = 2.0 x [tex]\frac{25.0}{0.5}[/tex]
F = 2.0 x 50.0
F = 100.0N
Therefore, the magnitude of the net force on the ball is 100.0N
The surface of the dock is 6 feet above the water. If you pull the rope in at a rate of 2 ft/sec, how quickly is the boat approaching the dock at the moment when there is 10 feet of rope still left to pull in
Answer:
The boat is approaching the dock at a rate of 2.5 ft/s.
Explanation:
Let the rope length be 'l' at any time 't', the distance of boat from dock be 'b' at any time 't'.
Given:
The height of dock above water (h) = 6 feet
Rate of pull of rope or rate of change of rope is, [tex]\frac{dl}{dt}=2\ ft/s[/tex]
As clear from the question, the height is fixed and only the length 'l' and distance 'b' varies with time 't'.
Now, the above situation represents a right angled triangle as shown below.
Using Pythagoras Theorem, we have:
[tex]l^2=h^2+b^2\\\\l^2=6^2+b^2\\\\l^2=36+b^2----------(1)[/tex]
Now, differentiating the above equation with time 't', we get:
[tex]2l\frac{dl}{dt}=0+2b\frac{db}{dt}\\\\l\frac{dl}{dt}=b\frac{db}{dt}\\\\\frac{db}{dt}=\frac{l}{b}\frac{dl}{dt}------(2)[/tex]
Now, the distance 'b' can be calculated using 'l=10 ft' in equation (1). This gives,
[tex]b^2=10^2-36\\\\b=\sqrt{64}=8\ ft[/tex]
Now, substituting all the given values in equation (2) and solve for [tex]\frac{db}{dt}[/tex]. This gives,
[tex]\frac{db}{dt}=\frac{10}{8}\times 2\\\\\frac{db}{dt}=2.5\ ft/s[/tex]
Therefore, the boat is approaching the dock at a rate of 2.5 ft/s.
The six metals have the work functions, W.
Part A Rank these metals on the basis of their cutoff frequency. Rank from largest to smallest. To rank items as equivalent, overlap them.
Part B Rank these metals on the basis of the maximum wavelength of light needed to free electrons from their surface. Rank from largest to smallest. To rank items as equivalent, overlap them.
Part C Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted electrons. (If no electrons are emitted from a metal, the maximum kinetic energy is zero, so rank that metal as smallest.) Rank from largest to smallest. To rank items as equivalent, overlap them.
Cesium= w= 2.1 eV Aliminium= w= 4.1 eV Beryllium= 5.0 eV Potassium= 2.3 eV Platinium= w= 6.4 eV Magnisium=w= 3.7 eV
Answer:
Explanation:
W (Ce) = 2.1 eV
W (Al) = 4.1 eV
W (Be) = 5 eV
W (K) = 2.3 eV
W (Pt) = 6.4 eV
W (Mg) = 3.7 eV
Part A:
Work function is directly proportional to the cut off frequency.
let f denotes the cut off frequency.
So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)
Part B:
Maximum wavelength for the emission is inversely proportional to the cut off frequency
So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)
Part C:
E = 3.10 eV
Let K is the maximum kinetic energy
K = E - W
K (Ce) = 3.1 - 2.1 = 1 eV
K (Al) 3.1 - 4.1 = not possible
K (Be) = 3.1 - 5 = not possible
K (K) = 3.1 - 2.3 = 0.8 eV
K (Pt) = .1 - 6.4 = not possible
K (Mg) = 3.1 - 3.7 = not possible
So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)
Metals can be ranked based on their work functions, which determine cutoff frequency, the maximum wavelength of light needed to free electrons, and maximum kinetic energy of emitted electrons under specific light. Lower work functions result in higher cutoff frequencies and longer required wavelengths, and more energetic emitted electrons when under 400 nm light.
Explanation:The ranking of the metals, in terms of cutoff frequency, maximum wavelength of light, and maximum kinetic energy, is based on the work function of each metal, which is the minimum energy required to free an electron from the metal's surface. The cutoff frequency is inversely related to the work function, so the metal with the lowest work function (cesium, 2.1 eV) will have the highest cutoff frequency, and the metal with the highest work function (platinum, 6.4 eV) will have the lowest cutoff frequency.
Conversely, the maximum wavelength of light needed to free electrons is directly related to the work function, so cesium will require the longest wavelength and platinum will require the shortest wavelength. When illuminated with 400 nm (3.10 eV) light, the kinetic energy of the emitted electrons will be the difference between the energy of the light and the work function, so cesium (with the smallest work function) will have the most energetic electrons, and metals with a work function larger than 3.10 eV (like aluminum, beryllium, and platinum) will not emit electrons at all.
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A simply supported wood beam with a span of L = 15 ft supports a uniformly distributed load of w0 = 270 lb/ft. The allowable bending stress of the wood is 1.95 ksi. If the aspect ratio of the solid rectangular wood beam is specified as h/b = 1.75, calculate the minimum width b that can be used for the beam.
The minimum width b that can be used for the beam is 4.32 in
To determine the minimum width (b) of the solid rectangular wood beam, we employ principles of structural engineering.
The maximum bending moment (M) for a simply supported beam with a uniformly distributed load (\(w_0\)) occurs at the center and is given by [tex]\(M = \frac{w_0 L^2}{8}\).[/tex]
The section modulus (S) for a rectangular cross-section is [tex]\(S = \frac{b \times h^2}{6}\).[/tex]
The bending stress (\(σ_b\)) is given by[tex]\(σ_b = \frac{M}{S}\)[/tex].
Setting \(σ_b\) equal to the allowable bending stress [tex](\(1.95 \, \text{ksi}\))[/tex]and using the specified aspect ratio (h/b = 1.75), we can solve for b.
This ensures that the wood beam meets structural safety criteria.
The calculation yields a minimum width b of 4.32 in, ensuring that the beam is structurally sound, withstanding the specified uniformly distributed load and adhering to the aspect ratio constraint in compliance with the allowable bending stress.
In a series RCL circuit the generator is set to a frequency that is not the resonant frequency. This nonresonant frequency is such that the ratio of the inductive reactance to the capacitive reactance of the circuit is observed to be 5.68. The resonant frequency is 240 Hz. What is the frequency of the generator
Answer:
Explanation:
Resonant frequency is 240
4π² x 240² = 1 / LC
230400π² = 1 / LC
Let the required frequency = n
inductive reactance = 2 πn L
capacitative reactance = 1 / 2 π n C
inductive reactance / capacitative reactance
= 4π² x n ² x LC = 5.68
4π² x n ² = 1 / LC x 5.68
= 230400π² x 5.68
4n ²= 230400 x 5.68
n ²= 57600 x 5.68
n ² = 327168
n = 572 approx
The generator's frequency in this RLC circuit, where the ratio of inductive to capacitive reactance is 5.68 and the resonant frequency is 240 Hz, is approximately 571 Hz.
To solve this problem, we need to use the relationship between inductive and capacitive reactance in an RLC circuit. Given that the ratio of the inductive reactance (XL) to the capacitive reactance (XC) is 5.68,
X_L / X_C = 5.68Reactance is frequency-dependent, with the following formulas for inductive and capacitive reactance:
X_L = 2πfLX_C = 1 / (2πfC)Given the resonant frequency as 240 Hz, we start by calculating the resonance reactances:
At resonance: X_L = X_C, so 2π(240)L = 1 / (2π(240)C)Using the non-resonant frequency, we use the reactance ratio:
2πfL / (1 / (2πfC)) = 5.68Rearranging gives us:
f² = 5.68 / (2π)²(LC)Since 240 Hz is the resonant frequency, substituting f:
f² = 5.68 * (240)²Simplifying yields:
f = 240 * √(5.68)f ≈ 571 HzTherefore, the generator frequency is approximately 571 Hz.
A string of mass m is under tension, and the speed of a wave in the string is v. What will be the speed of a wave in the string if the mass of the string is increased to 2m but with no change in the length or tension?
A) v/ sq. rt. of 2
B) v/2
C) 2v
D) v * sq. rt. of 2
E) 4v
Answer:
A) v/ sq. rt. of 2
Explanation:
The speed of the wave in the string is defined as:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
Where T is the tension on the string and [tex]\mu[/tex] is the linear density, that is, the mass per unit length:
[tex]\mu=\frac{m}{L}[/tex]
Where m is the mass of the string and L its length. We have [tex]m'=2m[/tex], [tex]T'=T[/tex] and [tex]L'=L[/tex]:
[tex]v'=\sqrt\frac{T'}{m'/L'}\\v'=\sqrt\frac{T}{2m/L}\\v'=\frac{1}{\sqrt2}\sqrt\frac{T}{m/L}\\v'=\frac{v}{\sqrt2}[/tex]
Explanation:
Below is an attachment containing the solution.
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A, how many turns must the solenoid have? Express your answer using two significant figures.
Answer:
33,458.71 turns
Explanation:
Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter
μ₀ = Permeability of free space = 4 π × 10 ⁻⁷
Solution:
We have B = μ₀ × n × I
⇒ n = B/ (μ₀ × I)
n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)
n = 90,428.94 turn/m
No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37
= 33,458.71 turns
Given Information:
Diameter of solenoid = d = 1.8 cm = 0.018 m
Length of solenoid = L = 37 cm = 0.37 m
Current = I = 4.4 A
Magnetic field = B = 0.50 T
Required Information:
Number of turns = N = ?
Answer:
Number of turns ≈ 33,498 or 33,458
Step-by-step explanation:
The magnetic field at the center of the solenoid is given by
B = μ₀NI/√ (L²+4r²)
N = B√ (L²+4r²)/μ₀I
Where L is the length and r is the radius of the solenoid, N is the number of turns and B is the magnetic field.
r = d/2 = 0.018/2 = 0.009 m
N = 0.50√ (0.37)²+(4*0.009²)/4πx10⁻⁷*4.4
N ≈ 33,498 Turns
Please note that we can also use a more simplified approximate model for this problem since the length of the solenoid is much greater than the radius of the solenoid
L = 0.37 >> r = 0.009
The approximate model is given by
B = μ₀NI/L
N = BL/μ₀I
N = 0.50*0.37/4πx10⁻⁷*4.4
N ≈ 33,458 Turns
As you can notice the results with the approximate model are very close to the exact model.
A knife thrower throws a knife toward a 300 g target that is sliding in her direction at a speed of 2.45 m/s on a horizontal frictionless surface. She throws a 22.5 g knife at the target with a speed of 36.0 m/s. The target is stopped by the impact and the knife passes through the target. Determine the speed of the knife (in m/s) after passing through the target.
Answer:
Explanation:
We shall apply conservation of momentum here because it is a case of inelastic collision
u₁ , u₂ initial velocity of knife and target . v₁ , v₂ be their final velocity.
m₁ u₁ + m₂u₂ = m₁v₁ + m₂v₂
u₂ will be negative as it is coming from opposite direction.
36 x 22.5 - 300 x 2.45 = 22.5 x v₁ + 0
810 - 735 = 22.5 x v₁
v₁ = 3.33 m /s
In each case the momentum before the collision is: (2.00 kg) (2.00 m/s) = 4.00 kg * m/s
1. In each of the three cases above show that momentum is conserved by finding the total momentum after the collision.
2. In each of the three cases, find the kinetic energy lost and characterize the collision as elastic, partially inelastic, or totally inelastic. The kinetic energy before the collision is (1/2)(2.00 kg)(2.00 m/s)^2 = 4.00 kg * m^2/s^2 = 4.00 J.
3. An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s. First show that this collision would satisfy conservation of kinetic energy and then explain briefly why it is an impossible result.
Answer:
Check Explanation.
Explanation:
Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s
a) Scenario A
After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s
Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s
Which is equal to the momentum before collision, hence, momentum is conserved.
Scenario B
They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s
Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s
This is equal to the momentum before collision too, hence, momentum is conserved.
Scenario C
Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s
Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s
This is equal to the momentum before collision, hence, momentum is conserved.
b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.
Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.
Scenario A
After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s
Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J
Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J
Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.
Scenario B
They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s
Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J
Kinetic energy lost = 4 - 4 = 0 J
Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.
Scenario C
Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s
Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J
Kinetic energy lost = 4 - 4 = 0 J
Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.
c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.
In this scenario,
Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J
This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.
But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.
Momentum after collision = (2+2)(1.414) = 5.656 kgm/s
which is not equal to the 4.0 kgm/s obtained before collision.
This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.
A mass is connected to a spring on a horizontal frictionless surface. The potential energy of the system is zero when the mass is centered on x = 0, its equilibrium position. If the potential energy is 3.0 J when x = 0.050 m, what is the potential energy when the mass is at x = 0.10 m?
Answer:
U = 12 J.
Explanation:
The potential energy in a spring is given by the following formula
[tex]U = \frac{1}{2}kx^2[/tex]
where k is the spring force constant and x is the displacement from the equilibrium.
If U = 3 J when x = 0.05 m, then k is
[tex]3 = \frac{1}{2}k(0.05)^2\\k = 2400~N/m[/tex]
Using this constant, we can calculate the potential energy at x = 0.10 m:
[tex]U = \frac{1}{2}(2400)(0.1)^2 = 12 ~J[/tex]
The potential energy when the mass is at [tex]\( x = 0.10 \)[/tex] m is 12 J.
The potential energy (U) of a mass-spring system is given by the equation:
[tex]\[ U = \frac{1}{2} k x^2 \][/tex]
Given that the potential energy is 3.0 J when \( x = 0.050 \) m, we can use this information to find the spring constant \( k \). Plugging the values into the equation, we get:
[tex]\[ 3.0 \, \text{J} = \frac{1}{2} k (0.050 \, \text{m})^2 \] \[ k = \frac{2 \times 3.0 \, \text{J}}{(0.050 \, \text{m})^2} \] \[ k = \frac{6.0 \, \text{J}}{0.0025 \, \text{m}^2} \] \[ k = 2400 \, \text{N/m} \][/tex]
Now that we have the spring constant, we can find the potential energy when [tex]\( x = 0.10 \)[/tex]m using the same formula:
[tex]\[ U = \frac{1}{2} k x^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.10 \, \text{m})^2 \] \[ U = \frac{1}{2} (2400 \, \text{N/m}) (0.010 \, \text{m}^2) \] \[ U = 1200 \, \text{N/m} \times 0.010 \, \text{m}^2 \] \[ U = 12 \, \text{J} \][/tex]
Therefore, the potential energy when the mass is at[tex]\( x = 0.10 \)[/tex]m is 12 J."
Star #1 is approaching the Earth with speed v. Star #2 is receding from the Earth with the same speed v. Measurements of the same spectral line from each of the stars show Doppler shifts in frequency. The light from which star will have the larger magnitude shift in frequency? a. star #1 b. Both stars will have the same shift. c. The value of the speed must be known before an answer can be found. d. star #2
Answer:
b. Both stars will have the same shift.
Explanation:
It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.97 10-6 W/m2 at a distance of 113 m from the explosion, at what distance from the explosion is the sound intensity half this value?
Answer:
160 m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, r, from the source.
[tex]I\propto \dfrac{1}{r^2}[/tex]
Hence,
[tex]I_1r_1^2 = I_2r_2^2[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{I_2}}[/tex]
From the question, [tex]I_2[/tex] is half of [tex]I_1[/tex]
[tex]r_2 = r_1\sqrt{\dfrac{I_1}{0.5I_1}}[/tex]
[tex]r_2 = r_1\sqrt{2}[/tex]
[tex]r_2 = 113\text{ m}\sqrt{2} = 160 \text{ m}[/tex]
Answer:
Distance ,d= 159.81m
Explanation:
The intensity, I, of the sound is inversely proportional to the square of the distance, d,from the source.
Using the equation d/do=sqrt2
d= dosqrt2
Where d=113m
d= 113sqrt2
d= 159.81m
Sound with frequency 1240 Hz leaves a room through a doorway with a width of 1.13 m . At what minimum angle relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraunhofer diffraction to apply. You can ignore effects of reflections. Express your answer in radians.
Given Information:
frequency = 1240 Hz
width = a = 1.13 m
speed of sound = c = 344 m/s
Required Information:
angle = θ = ?
Answer:
θ = 14.18 rad
Explanation:
We can find out the angle relative to the centerline perpendicular to the doorway by using the following relation
sin(θ) = λ/a
Where λ is the wavelength of the sound wave and a is width
λ = c/f
Where c is the speed of the sound and f is the frequency
λ = 344/1240
λ = 0.277
sin(θ) = λ/a
θ =sin⁻¹(λ/a)
θ =sin⁻¹(0.277/1.13)
θ =sin⁻¹(0.277/1.13)
θ = 14.18 rad
A projectile is fired with an initial speed of 230 m/s and an angle of elevation 60°. The projectile is fired from a position 100 m above the ground. (Recall g = 9.8 m/s². Round your answers to the nearest whole number.)
(a) Find the range of the projectile.
(b) Find the maximum height reached.
(c) Find the speed at impact.
Answer:
Explanation:....
The range of the projectile is 4675 m. The maximum height reached by the projectile is 2124 m. The speed of the projectile at impact is 234 m/s.
(a) Range is defined as the maximum horizontal distance covered by a projectile during its motion. It is given by the formula:
[tex]R = \frac{u^2 sin2\theta}{g}[/tex],
where u = initial speed of the projectile = 230 m/s
θ = angle of projection = 60°
g = acceleration due to gravity = 9.8 m/s²
Using these in formulas, we get:
[tex]R = \frac{(230 \hspace{0.8mm} m/s)^2 \hspace{0.8mm} sin2(60)}{9.8 \hspace{0.8mm} m/s^2}[/tex]
or, R = 4674.77 m ≈ 4675 m
(b) The maximum height reached by the projectile is given by the formula:
[tex]H = \frac{u^2 \hspace{0.8mm} sin^2\theta}{2g}[/tex]
using the numeric values, we get:
[tex]H = \frac{(230 \hspace{0.8mm}m/s)^2 sin^2(60)}{2 \times (9.8 \hspace{0.8mm} m/s^2)}[/tex]
or, H = 2024.23 m
However, since the projectile was fired 100 m above the ground, hence, the maximum height would be:
h = 100 m + H = 100 m + 2024.23 m = 2124.23 m ≈ 2124 m
(c) The speed of the projectile comprises of two parts, a horizontal velocity and a vertical velocity.
The horizontal velocity is given as [tex]v_x[/tex]. It is constant throught the motion and its magnitude can be determined by the formula:
[tex]v_x = v cos\theta = 230 \hspace{0.8 mm} m/s \times cos(60)[/tex]
or, [tex]v_x = 115 \hspace{0.8mm} m/s[/tex]
The vertical velocity is given as [tex]v_y[/tex]. It is not constant throught the motion. At the highest point, the magnitude of vertical velocity is zero. Using the kinetic equation,
[tex]v_y^2[/tex] = u² + 2gs, where s = h = 2124 m, we get:
[tex]v_y^2[/tex] = 0 + (2 × 9.8 m/s² × 2124 m)
or, [tex]v_y^2[/tex] = 41630.4 m²/s²
or, [tex]v_y[/tex] = 204.03 m/s ≈ 204 m/s
Now, the magnitude of speed will be equal to the resultant of both the horzontal and vertical velocities.
[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]
or, [tex]v = \sqrt{204^2 + 115^2} \hspace{0.8 mm} m/s[/tex]
or, v = 234.21 m/s ≈ 234 m/s
After the NEAR spacecraft passed Mathilde, on several occasions rocket propellant was expelled to adjust the spacecraft's momentum in order to follow a path that would approach the asteroid Eros, the final destination for the mission. After getting close to Eros, further small adjustments made the momentum just right to give a circular orbit of radius 45 km (45 × 103 m) around the asteroid. So much propellant had been used that the final mass of the spacecraft while in circular orbit around Eros was only 550 kg. The spacecraft took 1.04 days to make one complete circular orbit around Eros. Calculate what the mass of Eros must be.
Answer:
[tex]6.68\times 10^{15}\ kg[/tex]
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
R = Radius of orbit = 45 km
T = Time period = 1.04 days
Mass of Eros would be given by the following equation
[tex]M=\dfrac{4\pi^2R^3}{GT^2}\\\Rightarrow M=\dfrac{4\pi^2\times (45\times 10^3)^3}{6.67\times 10^{-11}\times (1.04\times 24\times 3600)^2}\\\Rightarrow M=6.68\times 10^{15}\ kg[/tex]
The mass of Eros is [tex]6.68\times 10^{15}\ kg[/tex]
For a two-level system, the weight of a given energy distribution can be expressed in terms of the number of systems, N, and the number of systems occupying the excited state, n1. What is the expression for weight in terms of these quantities
Answer:
W = N!/(n0! * n1!)
Explanation:
Let n0 = number of particles in the lowest energy state
n1 = number of particles in the excited energy state.
Using this, we can say that N = n0 + n1
From this we can then express the weight, W of the close system by finding the factorials of each particles
W = N!/(n0! * n1!)
Hence, the weight W is expressed as W = N!/(n0! * n1!)
Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and makes the equation true. (Simplify your answer completely. If no expression exists, enter DNE.)
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
Explanation:The given expression is 6m(2m + 18). First, we distribute 6m across the terms in the parentheses, resulting in 12m² + 108m. So, an expression that uses the variable 'm' and makes the equation true when the given expression is simplified would be 12m² + 108m. This answer is fully simplified, and the variable 'm' is assumed to be an integer.
The expression 6m(2m + 18) simplifies to 12m² + 108m. Therefore, the equivalent expression using the variable 'm' is 12m² + 108m.
Learn more about Simplifying expressions here:https://brainly.com/question/403991
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Europe and North America are moving apart by about 5 m per century. As the continents separate, new ocean floor is created along the mid-Atlantic Rift. If the rift is 5000 km long, what is the total area of new ocean floor created in the Atlantic each century?
Answer:
Area = 25km² in a century.
Explanation:
Given
Spread Rate = 5m/century
Length of rift = 5000km
Convert to metres
Length of rift = 5000 * 1000m
Length of rift = 5,000,000m
In one century, the additional area to Atlantis = (Rift Length) * (Spread rate) * 1 century
Atea = 5,000,000 m * 5m/century * 1 century
Area = 25,000,000m² in a century
----- Convert to km²
Area = 25,000,000km² * 1m²/1,000,000km²
Area = 25km² in a century.
Hence, the total area of new ocean floor created in the Atlantic each century is 25km²
Final answer:
The total area of new ocean floor created along the mid-Atlantic Rift each century is 5,000,000 m², or 25 km² of new oceanic crust formed every century.
Explanation:
The question asks how much new ocean floor is created in the Atlantic Ocean each century along the mid-Atlantic Rift, which is known to be 5000 km long, as Europe and North America drift apart by about 5 m per century.
To calculate the total area of the new ocean floor created, we consider the length of the rift (5000 km) and the rate of separation (5 m per century).
First, we need to convert kilometers to meters for uniform units:
5000 km * 1000 m/km = 5,000,000 m.
Now we can calculate the area:
Area = Length * Width
Area = 5,000,000 m * 5 m
Area = 25,000,000
So, every century, an area of 25,000,000 m2 of new ocean floor is created along the mid-Atlantic Rift. This is equivalent to 25 km2 of new oceanic crust, since 1 km2 = 1,000,000 m2.
5. From Gauss’s law, the electric field set up by a uniform line of charge is E S 5 a l 2pP0r b r^ where r^ is a unit vector pointing radially away from the line and l is the linear charge density along the line. Derive an expression for the potential difference between r 5 r1 and r 5 r2.
Answer:
ΔV = 2k λ ln (r₁ / r₂)
Explanation:
The electric potential is
ΔV = - ∫ E. ds
The expression for the electric field of a charge line is given
E = 2k λ / r
Let's replace and integrate
ΔV = - 2k λ ∫ dr / r
ΔV = -2 k λ ln r
Let's evaluate
ΔV = - 2k λ ln r₂- ln r₁
ΔV = -2k λ ln (r₂ / r₁)
ΔV = 2k λ ln (r₁ / r₂)
A rod of length L has a charge Q uniformly distributed along its length. The rod lies along the y axis with one end at the origin. (a) Find an expression for the electric potential as a function of position along the x axis.
Answer:
V = k Q/l ln [(l +√(l² + x²)) / x]
Explanation:
The electrical potential for a continuous distribution of charges is
V = k ∫ dq / r
Let's apply this expression to our case, define a linear charge density for the bar
λ = dq / dy
dq = λ dy
The distance from a point on the bar to the x-axis is
r = √ (x² + y²)
Let's replace
V = K ∫ λ dy /√ (x² + y²)
We integrate
V = k λ ln (y + √ (x² + y²))
Let's evaluate between y = 0 and y = l
V = k λ [ ln (l +√(x² + l²) - ln x]
We substitute the linear density
V = k Q/l ln [(l +√(l² + x²)) / x]
Final answer:
To calculate the electric potential at a point on the x-axis due to a uniformly charged rod, we use the concept of charge density and integrate the contributions from each infinitesimal charge element along the rod with respect to the position.
Explanation:
The student is asking about the electric potential due to a uniformly charged rod on the x-axis. To find an expression for the electric potential at a point on the x-axis, we can imagine dividing the rod into infinitesimally small segments of charge dq. Given that the charge Q is evenly distributed along the rod of length L, the linear charge density λ is Q/L. The potential dV due to a small element dq at a distance x from the rod is given by Coulomb's law as dV = k * dq / r, where k is Coulomb's constant and r is the distance from the charge element to the point on the x-axis.
To find the total potential V, we integrate this expression from one end of the rod to the other. The distance r varies along the rod, so we integrate with respect to y, the position along the rod. Setting up the integral, we have V = k * ∫_{-L/2}^{L/2} (dq / √((L/2 - y)^2 + x^2)) where the limits of integration account for the rod's position along the y-axis with one end at the origin and λ = Q/L. After substituting dq with λdy, performing the integration, and simplifying, we obtain the electric potential V(x) as a function of position along the x-axis.
In the picture below, a 7.00 kg piece of aluminum hanging from a spring scale is immersed in water. The total mass of the water is 3.00 kg, and it is contained in a beaker with a mass of 2.00 kg. The beaker sits on top of another scale. Find the readings on both scales. Take the density of aluminum to be 2700 kg/m3 and the density of water to be 1000 kg/m3.
Answer:
Upper scale reads 4.4kg while lower scale reads 7.6 kg
Explanation:
The buoyant force of the water on the aluminum piece would equal to the weight of the water displaced by the aluminum.
The mass of the water displaced by aluminum is
[tex]m_w = V_w \rho_w = \frac{m_a}{\rho_a}\rho_w = \frac{7}{2700}1000 = 2.6 kg[/tex]
As 7kg aluminum piece is supported by 2.6 kg buoyant force, the scale that the aluminum is hung on would read
Mass of aluminum - mass of water displaced
7 - 2.6 = 4.4 kg
This buoyant force would also created a reaction on the lower scale of 2.6 kg (according to Newton's 3rd law). So the lower scale would read:
Mass of water + mass of bleak + mass of buoyant reaction force
3 + 2 + 2.6 = 7.6 kg