Answer:
The wavelength is 503 nm
Explanation:
Considering constructive interference , this means that route(path) difference is equal to the product of order of fringe and wavelength of the light
i.e dsinθ = m[tex]\lambda[/tex]
Where [tex]\lambda[/tex] is the wavelength of light and m is the order of the fringe
Looking at θ to be very small , sin θ can be approximated to θ
and [tex]\theta \approx \frac{x}{l}[/tex]
Substituting this into the above equation
[tex]d[\frac{x}{l} ] =m\lambda[/tex]
making x the subject
[tex]x =\frac{m\lambda l}{d}[/tex]
This above equation will give the value of the distance of the [tex]m^{th}[/tex] order fringe of the wavelength [tex]\lambda[/tex] from the central fringe
Replacing with the value given in the question we have
[tex]\lambda[/tex] = 710 nm m = 2 d =0.66 mm , l = 2.0 m
[tex]x = \frac{(2)(710nm)(2.0m)[\frac{10^9}{1m} ]}{(0.66mm)(\frac{10^6}{1mm} )}[/tex]
[tex]=(4.303*10^6nm)[\frac{\frac{1}{10^6}mm }{1nm} ][/tex]
[tex]=4.303mm[/tex]
The separation of the second fringe from central maximum is 4,303 mm
To obtain the separation of the second order fringe of the unknown light from central maximum
[tex]x' = 4.303mm - 1.25 mm = 3.053mm[/tex]
Now to obtain the wavelength of this second source
from [tex]x = \frac{m\lambda l}{d}[/tex]
[tex]\lambda' = \frac{x'd}{ml}[/tex]
Now substituting 3,053 mm for [tex]x'[/tex] 2.0 mm for l , 0.66 mm for d and 2 for m in the above formula
[tex]\lambda' =\frac{(3.053mm)(0.66mm)}{(2)(2.0)(\frac{10^3mm}{1m} )}[/tex]
[tex]= (503.7*10^{-6}mm)(\frac{10^6nm}{1mm} )[/tex]
[tex]=503.7nm[/tex]
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h. When a second tugboat applies an additional constant force of magnitude F2 in the same direction, the speed increases by 11 km/h during a 10 s interval. How do the magnitudes of F1 and F2 compare? (Neglect the effects of water resistance and air resistance.)
Answer:
The magnitude of F₁ is 3.7 times of F₂
Explanation:
Given that,
Time = 10 sec
Speed = 3.0 km/h
Speed of second tugboat = 11 km/h
We need to calculate the speed
[tex]v_{1}=\dfrac{3.0\times10^{3}}{3600}[/tex]
[tex]v_{1}=0.833\ m/s[/tex]
The force F₁is constant acceleration is also a constant.
[tex]F_{1}=ma_{1}[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a_{1}=\dfrac{v}{t}[/tex]
[tex]a_{1}=\dfrac{0.833}{10}[/tex]
[tex]a_{1}=0.083\ m/s^2[/tex]
Similarly,
[tex]F_{2}=ma_{2}[/tex]
For total force,
[tex]F_{3}=F_{2}+F_{1}[/tex]
[tex]ma_{3}=ma_{2}+ma_{1}[/tex]
The speed of second tugboat is
[tex]v=\dfrac{11\times10^{3}}{3600}[/tex]
[tex]v=3.05\ m/s[/tex]
We need to calculate total acceleration
[tex]a_{3}=\dfrac{v}{t}[/tex]
[tex]a_{3}=\dfrac{3.05}{10}[/tex]
[tex]a_{3}=0.305\ m/s^2[/tex]
We need to calculate the acceleration a₂
[tex]0.305=a_{2}+0.083[/tex]
[tex]a_{2}=0.305-0.083[/tex]
[tex]a_{2}=0.222\ m/s^2[/tex]
We need to calculate the factor of F₁ and F₂
Dividing force F₁ by F₂
[tex]\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}[/tex]
[tex]\dfrac{F_{1}}{F_{2}}=3.7[/tex]
[tex]F_{1}=3.7F_{2}[/tex]
Hence, The magnitude of F₁ is 3.7 times of F₂
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]
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 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."
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
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.
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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.
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.
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 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.
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 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 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 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.
Suppose a person whose mass is m is being held up against the wall with a constant tangential velocity v greater than the minimum necessary. Find the magnitude of the frictional force between the person and the wall
The magnitude of the frictional force between the person and the wall is
[tex]\[ f = \frac{mv^2}{r} \][/tex]
To find the magnitude of the frictional force between the person and the wall, we can use the concept of centripetal force.
First, let's define the variables:
m = mass of the person
v = tangential velocity of the person
R = radius of the circular path (distance from the person to the axis of rotation, in this case, the distance from the person to the wall)
The centripetal force [tex](\( F_c \))[/tex] required to keep the person moving in a circular path is given by:
[tex]\[ F_c = \frac{{mv^2}}{R} \][/tex]
Since the person is held against the wall, the frictional force [tex](\( F_f \))[/tex] provides the centripetal force. Therefore,
[tex]\[ F_f = F_c \][/tex]
[tex]\[ F_f = \frac{{mv^2}}{R} \][/tex]
Therefore, the magnitude of the frictional force between the person and the wall is [tex]\( \frac{{mv^2}}{R} \)[/tex].
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.
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)
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
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.
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
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.
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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When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rate of 2.5 m>s 1150 m>min, or 500 ft>min2. What fraction of the engine power is being used to make the airplane climb
Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
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
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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What is the electric flux through one side of a cube that has a single point charge of -3.80 µC placed at its center? Hint: You do not need to integrate any equations to get the answer.
? N·m^2/C
Answer:
φ = -7.16 × 10⁴ Nm²/C
Explanation:
q = -3.80 × 10⁻⁶ C
ε₀ = 8.85 10⁻¹² C²/Nm²
By Gauss's law
φ = q/ε₀
as the cube has 6 faces so for one side the flux will
φ =1/6 ( q/ε₀)
φ = 1/6 ( -3.80 × 10⁻⁶ C / 8.85 10⁻¹² C²/Nm²)
φ = -71,563.088 Nm²/C
φ = -7.16 × 10⁴ Nm²/C
-7.00 x 10⁴Nm²/C
Explanation:Let the single point charge of -3.80µC be Q
i.e
Q = -3.80µC = -3.80 x 10⁻⁶C
From Gauss's law, the total electric flux,φ, through an enclosed surface is equal to the quotient of the enclosed charge Q, and the permittivity of free space, ε₀. i.e;
φ = Q / ε₀ ----------------------(i)
Where;
ε₀ = known constant = 8.85 x 10⁻¹²F/m
Now, If the total flux through the enclosed cube is given by equation (i), then the flux, φ₁, through one of the six sides of the cube is found by dividing the equation by 6 and is given as follows;
φ₁ = Q / 6ε₀ --------------------(ii)
Substitute the values of Q and ε₀ into equation (ii) as follows;
φ₁ = -3.80 x 10⁻⁶ / (6 x 8.85 x 10⁻¹²)
φ₁ = -0.07 x 10⁶
φ₁ = -7.00 x 10⁴
Therefore, the electric flux through one side of the cube is -7.00 x 10⁴ Nm²/C
Note:
The result (flux) above is negative to show that the electric field lines from the point charge points radially inwards in all directions.
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
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]
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.