Answer:
Approximately 4.5 square miles
Explanation:
(surrounding van Nuys and sepulveda stations) and one measuring approximately 2.85 square miles ( surrounding north Hollywood Station)
Final answer:
The 'orange line' referred to in the question likely represents data in a graph related to either the wavelength or frequency of orange light in the visible spectrum or a trend in statistical analysis. The exact answer depends on the specific context of the graph or data set from which the 'orange line' is derived.
Explanation:
The question 'The orange line is measuring the' seems to be related to analyzing data, most likely from a graph or chart that measures certain variables, potentially in the context of physics or mathematics dealing with light spectra or statistical analysis. Given the details provided, including mention of the visible light spectrum and specific wavelengths denoted by numbers, it appears that the 'orange line' may refer to a line on a graph that represents the wavelength or frequency of the color orange within the visible light spectrum.
However, the context provided is ambiguous. For instance, the chapter headings and the Conclusion provided suggests statistical analysis, perhaps in the field of queueing theory and how waiting times are affected by different systems, which implies a more mathematical or business-oriented question. In the field of mathematics, 'orange line' could also be a reference to a trend line or data line in a statistical plot.
[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the staring for t seconds. Determine t. Neglect friction and all masses except those of the four 3-kg spheres, which may be treated as particles.
Answer:
t = 5.89 s
Explanation:
To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.
Let us assume that the radius of the pulley ([tex]r_p[/tex]) = 0.4 m
Let the radius of the sphere (r) = 0.5 m
w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s
Tension (T) = 20 N
mass (m) = 3 kg each
[tex]\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw[/tex]
[tex]t = \frac{4r^2mw}{Tr_P}[/tex]
Substituting values:
[tex]t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s[/tex]
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value of t is [tex]t = 15.08 \ s[/tex]
Explanation:
From the question we are told that the angular speed is
The initial angular speed [tex]w_f = 150 rev/min = \frac{2 \pi }{60s} * 150 = 15.71rad /s[/tex]
The force is [tex]T = 20 N[/tex]
The radius is [tex]r = 400mm = \frac{400}{1000} = 0.4m[/tex]
The total mass of the four sphere is [tex]m_a = (4*3) = 12kg[/tex]
The initial velocity is [tex]v_i = 0m/s[/tex]
The radius of the pully is [tex]r_p = 100mm = \frac{100}{1000} = 0.10m[/tex]
The initial time is [tex]t_1 = 0s[/tex]
The final time is [tex]t = t[/tex]
Generally the final velocity of the sphere is mathematically represented as
[tex]v_f = r w_f[/tex]
[tex]v_f=15.7 r[/tex]
The angular impulse momentum principle can be represented methematically as
[tex](H_O)_i + \int\limits^{t_1}_{t_2}{(T \cdot r_{p})} \, dt = (H_O)_f[/tex]
[tex]r(m_a v_i ) + \int\limits^{t_2}_{t_1} {T \cdot r_p} \, dt = r(m_a v_f)[/tex]
[tex]r(m_a v_1 ) + T \cdot r_p (t_{2} -t_{1}) = r( m_a * 15.7 r )[/tex]
Substituting values
[tex]0.4 * (12* 0) + (20 *0.10 * (t-0)) = 0.40 * (12* 0.40 *15.7)[/tex]
=> [tex]0 + 2t = 30.16[/tex]
=> [tex]t = 15.08 \ s[/tex]
A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 77.9% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?
Answer:
(A) [tex]9.14\times 10^{-9}sec[/tex]
(B) [tex]6.20\times 10^{-3}A[/tex]
Explanation:
We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]
Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]
Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]
[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]
Battery voltage e = 16 volt
(a) It is given current becomes 79.9% of its final value
Current in RL circuit is given by
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
According to question
[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]e^{\frac{-t}{\tau }}=0.201[/tex]
[tex]{\frac{-t}{\tau }}=ln0.201[/tex]
[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]
[tex]t=9.14\times 10^{-9}sec[/tex]
(b) Current at [tex]t=\tau sec[/tex]
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]
[tex]i=6.20\times 10^{-3}A[/tex]
Consider four different oscillating systems, indexed using i = 1 , 2 , 3 , 4 . Each system consists of a block of mass mi moving at speed vi on a frictionless surface while attached to an ideal, horizontally fixed spring with a force constant of ki . Let x denote the displacement of the block from its equilibrium position. Order the systems from largest total mechanical energy to smallest.
a. m1= 0.5KG k2=500 N/m amplitude A = 0.02 m
b. m2= 0.6KG k2=300 N/m v2= 1 m/s . when passing through equilibrium
c. m3= 1.2KG k3=400 N/m v3= 0.5 m/s . when passing through x= -0.01 m
d. m4= 2 KG k4=200 N/m v4= 0.2 m/s . when passing through x=-0.05 m
Answer:
The order is 2>4>3>1 (TE)
Explanation:
Look up attached file
Write the ordered pair that represents vec yz .Then find the magnitude of vec yz Y(- 4, 12), (1, 19)
Answer with Explanation:
We are given that
Y=(-4,12)
Z=(1,19)
We have to find the ordered pair which represents the vector YZ and magnitude of vector YZ.
Vector YZ=Z-Y=<1,19>-<-4,12>
Vector YZ=<5,7>
Magnitude of vector r
[tex]r=xi+yj[/tex]
[tex]\mid r\mid=\sqrt{x^2+y^2}[/tex]
Using the formula
[tex]\vec{YZ}=\sqrt{5^2+7^2}[/tex]
[tex]\vec{YZ}=\sqrt{25+49}=\sqrt{74} units[/tex]
A car is moving down a flat, horizontal highway at a constant speed of 21 m/s when suddenly a rock dropped from rest straight down from a bridge crashes through the windshield. The rock strikes the windshield at a right angle, and the windshield makes a 60 degree angle with the horizontal.
a. Find the speed of the rock relative to the car when it hits the windshield.
b. Find the distance above the point of contact with the windshield from which the rock was dropped.
c. If the driver had seen what was about to happen just as the rock was being released, how much time would she have had to react before the rock strikes the windshield?
Answer:
a)9.8
Explanation:
because any object falling from any height is 9.8
(sorry that's the only one i know)
1. The speed of the rock relative to the car when it hits the windshield is [tex]\( {42.00 \, \text{m/s}} \).[/tex]
2. The distance above the point of contact with the windshield from which the rock was dropped is [tex]\( {67.5 \, \text{m}} \).[/tex]
3. The driver would have had [tex]\( {3.71 \, \text{s}} \)[/tex] to react before the rock strikes the windshield.
Let's solve each part of the problem step-by-step.
Part (a): Find the speed of the rock relative to the car when it hits the windshield.
Given:
- Car speed, [tex]\( v_{\text{car}} = 21 \, \text{m/s} \)[/tex]
- Windshield angle with the horizontal,[tex]\( \theta = 60^\circ \)[/tex]
Since the rock is dropped from rest, its initial vertical velocity is [tex]\( 0 \, \text{m/s} \).[/tex]
The rock falls under the influence of gravity, so its vertical velocity [tex]\( v_{\text{vertical}} \)[/tex] at any time t is given by:
[tex]\[ v_{\text{vertical}} = g t \][/tex]
where [tex]\( g = 9.8 \, \text{m/s}^2 \).[/tex]
The horizontal velocity of the rock is [tex]\( 0 \, \text{m/s} \)[/tex] with respect to the ground.
To find the speed of the rock relative to the car when it hits the windshield, we need to consider the components of velocity with respect to the car:
1. The car's horizontal velocity component.
2. The rock's vertical velocity component.
The velocity components with respect to the car:
- Horizontal component relative to the car, [tex]\( v_{\text{horizontal, relative}} = v_{\text{car}} \)[/tex]
- Vertical component, [tex]\( v_{\text{vertical}} = g t \)[/tex]
The speed of the rock relative to the car can be found using the Pythagorean theorem:
[tex]\[ v_{\text{relative}} = \sqrt{(v_{\text{horizontal, relative}})^2 + (v_{\text{vertical}})^2} \][/tex]
We need to find t when the rock strikes the windshield.
Since the rock strikes the windshield at a right angle, and the windshield makes a 60° angle with the horizontal, the relative speed of the rock [tex]\( v_{\text{relative}} \)[/tex] must have components that align with the 60° angle.
Thus:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{relative}} \sin(60^\circ) \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{relative}} \cos(60^\circ) \][/tex]
Given:
[tex]\[ v_{\text{relative, vertical}} = v_{\text{vertical}} = g t \][/tex]
[tex]\[ v_{\text{relative, horizontal}} = v_{\text{car}} = 21 \, \text{m/s} \][/tex]
From the component relationship:
[tex]\[ \frac{v_{\text{relative, vertical}}}{v_{\text{relative, horizontal}}} = \tan(60^\circ) = \sqrt{3} \][/tex]
[tex]\[ \frac{g t}{21} = \sqrt{3} \][/tex]
[tex]\[ g t = 21 \sqrt{3} \][/tex]
[tex]\[ t = \frac{21 \sqrt{3}}{9.8} \][/tex]
Calculate t:
[tex]\[ t \approx \frac{21 \cdot 1.732}{9.8} \approx \frac{36.372}{9.8} \approx 3.71 \, \text{s} \][/tex]
Now, we can find the vertical velocity [tex]\( v_{\text{vertical}} \)[/tex]:
[tex]\[ v_{\text{vertical}} = g t = 9.8 \times 3.71 \approx 36.36 \, \text{m/s} \][/tex]
Now, the speed of the rock relative to the car:
[tex]\[ v_{\text{relative}} = \sqrt{(21)^2 + (36.36)^2} \approx \sqrt{441 + 1322.49} \approx \sqrt{1763.49} \approx 42.00 \, \text{m/s} \][/tex]
Part (b): Find the distance above the point of contact with the windshield from which the rock was dropped.
The distance h from which the rock was dropped can be found using the kinematic equation:
[tex]\[ h = \frac{1}{2} g t^2 \][/tex]
Using [tex]\( t \approx 3.71 \, \text{s} \)[/tex]:
[tex]\[ h = \frac{1}{2} \times 9.8 \times (3.71)^2 \approx \frac{1}{2} \times 9.8 \times 13.76 \approx 67.5 \, \text{m} \][/tex]
Part (c): The time available for the driver to react is the time \( t \) it takes for the rock to fall, which we calculated in part (a):
[tex]\[ t \approx 3.71 \, \text{s} \][/tex]
(hrw8c13p51) A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as 0.11. What is the semimajor axis of the comet's orbit
Answer:
Semimajor axis of the comet's orbit is; α = 1.89 x 10^(13) m
Explanation:
From the question, we have a comet that was seen in April 574 by Chinese astronomers and was seen again in May 1994. That is a period of; T =(April)1994 - (May) 574 = 1420 years plus one month
Converting this to seconds, we have; T = [(1420 x 12) + 1] x 2628002.88 = 4.478 x 10^(10) seconds
Now, the square of the period of any planet is given by the formula;
T² = (4π²/GM)α³
Where;
G is gravitational constant and has a value of 6.67 x 10^(-11) m³/kg.s²
M is mass of sun which has a value of 1.99 x 10^(30) kg
α is the semi major axis of the comets orbit.
Thus, making α the subject, we have;
α = ∛(GMT²/4π²)
So, α = ∛{(6.67 x 10^(-11) x 1.99 x 10^(30) x (4.478 x 10^(10))²}/4π²)
α = 1.89 x 10^(13) m
The magnetic flux through a loop:
A. is maximized when the B field lines in the plane of the loop
B. is zero when the B field is perpendicular to the plane of the loop
C. doesn't depend on the direction of the B field
D. depends just on the B field going through the loop
E. is proportional to the number of magnetic field lines going through it
Answer:
The magnetic flux through a loop is zero when the B field is perpendicular to the plane of the loop.
Explanation:
Magnetic flux are also known as the magnetic line of force surrounding a bar magnetic in a magnetic field.
It is expressed mathematically as
Φ = B A cos(θ) where
Φ is the magnetic flux
B is the magnetic field strength
A is the area
θ is the angle that the magnetic field make with the plane of the loop
If B is acting perpendicular to the plane of the loop, this means that θ = 90°
Magnetic flux Φ = BA cos90°
Since cos90° = 0
Φ = BA ×0
Φ = 0
This shows that the magnetic flux is zero when the magnetic field strength B is perpendicular to the plane of the loop.
Answer:
D. depends just on the B field going through the loop
Explanation:
The magnetic flux trough a loop is given by the formula:
[tex]\Phi_B=\vec{B}\cdot \vec{A}[/tex]
[tex]\Phi_B=BAcos\theta[/tex]
Where B is the magnitude of the magnetic field and A is the area of the loop.
It is clear that the flux is maximum when the angle between the direction of B and the direction of the normal vector of A is zero (cos0 = 1).
Furthermore, we can notice that only the magnetic field that crosses the loop contributes to the flux.
Hence, the correct answer is:
D. depends just on the B field going through the loop
hope this helps!!
A 47 g particle undergoes SHM with an amplitude of 7.1 mm, a maximum acceleration of magnitude 5.4 x 103 m/s2, and an unknown phase constant φ. What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?
Answer:
Explanation:
check the picture attached for the explanation. I hope it helps . Thank you
Answer:
Explanation:
a) the period of the motion:
In order to calculate, we have the following formula.
f = 1/T whereas, T = 1/f. So, we need frequency first to calculate period of the motion.
In the question, we have been given amplitude and maximum acceleration. So, we can use the formula of maximum acceleration to calculate the frequency. Here's how:
Maximum acceleration = (2πf)² x A
where, A is amplitude.
a(max) = maximum acceleration
By rearranging the equation and making the frequency our subject and plugging in the values of given quantities then we have:
f = 138.80 hertz
Now, we can calculate, period of the motion by plugging in the value of frequency in the equation of period mentioned above.
Period of the motion = T = 1/f
T= 1/138.80
T = 7.204 x [tex]10^{-3}[/tex] seconds.
b) Maximum Speed of the particle
In order to calculate max. speed of the particle, we have to use following formula:
Max. Speed = 2πf x A
Max. Speed = ( 2 x 3.14 x 138.80 x 7.1x[tex]10^{-3}[/tex])
Max. Speed = 6.19 m/s
c) Total mechanical energy of the oscillator
Total mechanical energy of the oscillator is the sum of kinetic energy and potential energy and for which we have formula to calculate total mechanical energy of the oscillator:
Total Mechanical Energy = T.E = K.E + P.E
T.E = 2m[tex]\pi ^{2}[/tex][tex]f^{2}[/tex][tex]A^{2}[/tex]
we do have values of all the quantities and now by just plugging in the values we will get the total mechanical energy of the oscillator.
T.E = 2 x 0.047 x [tex]3.14^{2}[/tex] x [tex]138.80^{2}[/tex] x [tex]0.0071^{2}[/tex]
T.E = 0.90 J
d) Magnitude of the force on the particle when the particle is at its maximum displacement.
F = -KA here, amplitude is used because it is the maximum displacement from the mean position.
where K = -m ω² x A
ω = 2πf = 2 x 3.14 x 138.80
ω = 871.66
F = -(0.047 x 871.66² x 0.0071²)
F = - 1.8 N
e) when the displacement is at half
F = -Kx A/2
A/2 = 0.0071 / 2
A/2 = 0.0035
F = - m ω² x A/2
F = - (0.047 x 871.66² x 0.0035)
F = - 125 N
An automobile has a vertical antenna 1.45 m long. The automobile travels at 90 km/h on a horizontal road where the Earth's magnetic field is 59.0 µT directed to the north and downward at an angle of 65.5° below the horizontal. Calculate the maximum magnitude of the induced emf in the antenna.
Answer:
[tex]9.05\times 10^{-4} V[/tex]
Explanation:
We are given that
Length of antenna,l=1.45 m
Earth's magnetic field,[tex]B=59\mu T=59\times 10^{-6} T[/tex]
[tex]1\mu=10^{-6}[/tex]
[tex]\theta=65.5^{\circ}[/tex]
Speed,v=90 km/h=[tex]90\times \frac{5}{18}=25m/s[/tex]
[tex]1km/h=\frac{5}{18}m/s[/tex]
Magnitude of maximum induced emf in the antenna=[tex]E=Blvcos\theta[/tex]
Substitute the values
[tex]E=59\times 10^{-6}\times 1.45\times 25cos65[/tex]
[tex]E=9.05\times 10^{-4} V[/tex]