Answer:
ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]
Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.
As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]
Contents
1 Definitions in American context
2 History
2.1 Faster inter-city trains: 1920–1941
2.2 Post-war period: 1945–1960
2.3 First attempts: 1960–1992
2.4 Renewed interest: 1993–2008
2.5 Plans for 2008–2013
3 Current state and regional efforts
3.1 The Northeast
3.1.1 Northeast Corridor: Next Generation High-Speed Rail
3.1.1.1 Proposed routes
3.1.2 Northeast Maglev proposal
3.1.3 New Jersey–New York City upgrades
3.1.4 New York
3.1.5 Pennsylvania
3.2 Western States
3.2.1 California
3.2.2 Pacific Northwest
3.2.3 Arizona
3.3 Mid-Atlantic and the South
3.3.1 Florida
3.3.2 Southeast
3.3.3 Texas
3.4 Midwest
3.4.1 Illinois and the Midwest
3.5 The Southwest
4 Federal high-speed rail initiatives
4.1 American Recovery and Reinvestment Act of 2009
4.1.1 Strategic plan
4.2 2009 federal grant funding
4.3 2010 allocation
4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida
4.4 2011 and 2012 proposals and rejections of funding
5 See also
6 Notes
7 Further reading
8 External links
Explanation:
5. In Investigation 2, if everything stays the same, except the diameter of the outer ring is doubled, how does the electric field change?
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
There is a change in the electric field by this factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]E(r)_1 = [\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}[/tex]
Now when the outer diameter is doubled, the radius(b) is also doubled
So
[tex]E(r)_2 = [\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}[/tex]
Now
[tex]\frac{E(r)_2}{E(r)_1} = \frac{\frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r}}{\frac{V_o}{ln(2b) -ln(a)} ] * \frac{1}{r}}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} = \frac{V_o}{ln(b) -ln(a)} ] * \frac{1}{r} * \frac{ ln(2b) -ln(a)}{V_o} ] * \frac{r}{1}[/tex]
=> [tex]\frac{E(r)_2}{E(r)_1} =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
[tex]=> E(r)_2 =\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]} }{E(r)_1}[/tex]
Here we see that the electric field changes by the factor [tex]\frac{ln[\frac{b}{a} ]}{ln[\frac{2b}{a} ]}[/tex]
a motor boat is traveling at 25 knots towards 340 degree on a river flowing at 20 knots towards 175 degrees. What is the actual speed of the boat as seen by a helicopter piolet hovering above?
Answer:
Vbx = 25 * cos 340 = 23.5 knot x-component of boats speed
Vrx = 20 cos 175 = -19.9 knot x-component of rivers speed
Vx = 3.58 knot x-component of boat and river speed
Vby = 25 sin 340 = -8.55 knot y-component of boat speed
Vry = 20 sin 175 = 1.74 knot y-component of river speed
Vy = -6.81 knot y-component of boat and river speed
V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots
How far will a 600 kg boat travel in 12 s if there is a constant 900 N force on it and it starts from rest?
Answer:
108 metres
Explanation:
Given
[tex]Force = 900N[/tex]
[tex]Mass = 600kg[/tex]
[tex]Time = 12s[/tex]
Required
Determine the distance moved
First, we need to determine the acceleration.
[tex]Force = Mass * Acceleration[/tex]
[tex]900N = 600kg * a[/tex]
Solve for a
[tex]a = 900/600[/tex]
[tex]a = 1.5m/s^2[/tex]
Next, we determine the distance using:
[tex]S = ut + \frac{1}{2}at^2[/tex]
Since it starts from rest,
[tex]u = 0[/tex]
[tex]t = 12[/tex]
[tex]a = 1.5[/tex]
So:
[tex]S = 0 * 12 + \frac{1}{2} * 1.5 * 12^2[/tex]
[tex]S = \frac{1}{2} * 1.5 * 144[/tex]
[tex]S = \frac{1}{2} * 216[/tex]
[tex]S = 108m[/tex]
The volume of water in a water bottle, is about 398
g
cm
km/hr
Kg
g/mL
ml
km
m/s
Answer:
milliliters (ml)
Explanation:
millileters is the correct measurement for liquids
A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m
Answer:
35.35 m
Explanation:
The following data were obtained from the question:
Initial velocity (u) = 20 m/s
Angle of projection (θ) = 30°
Acceleration due to gravity (g) = 9.8 m/s²
Range (R) =.?
The range (i.e how far away) of the ball can be obtained as follow:
R = u² Sine 2θ /g
R = 20² Sine (2×30) / 9.8
R = 400 Sine 60 / 9.8
R = (400 × 0866) / 9.8
R = 346.4 / 9.8
R = 35.35 m
Therefore, the range (i.e how far away) of the ball is 35.35 m
Matching type. Send help please. ASAP!
Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
A hiker walks 11 km due north from camp and then turns and walks 11 km due east.
What is the total distance walked by the hiker?
What is the displacement (on a straight line) of the hiker from the camp?
please answer all questions
What is the force of a 12 kg object that is accelerating 6 m/s
We are given:
Mass of object (m) = 12 kg
acceleration (a) = 6 m/s²
Solving for the Force:
From newton's second law of motion:
F = ma
replacing the variables
F = 12*6
F = 72N
A cyclist and his bicycle have a combined mass of 88 kg and a combined
weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he
reaches a speed of 2.4 m/s. What is his momentum at this point?
A. 36.7 kg m/s
B. 359.3 kg:m/s
C. 105.6 kg-m/s
D. 211.2 kg:m/s
The cyclist accelerates at 1.2 m / s² after 2 seconds he reaches a speed of 2.4 m / s, then the momentum at this point would be 211.2 kg-m/s, therefore the correct answer is option D.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle, and the momentum of any particle is expressed in Kg m/s unit.
As given in the problem a cyclist and his bicycle has a combined mass of 88 kg and a combined weight of 862.4 N. The cyclist accelerates at 1.2 m/s2. After 2 seconds he reaches a speed of 2.4 m/s.
The momentum of the cyclist = 88 × 2.4
= 211.0 kgm/s
Thus, the momentum of the cyclist would be 211.0 kgm/s.
To learn more about momentum from here, refer to the link given below;
brainly.com/question/17662202
#SPJ2
correct me if im wrong
1. According to its computer, a rocket launched
traveled 1200 m, had an average speed of 100.0
m/s. How-long did the trip take?
Answer:
I think it's = 12 seconds
Explanation:
the formula for speed is:
speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:
time=[tex]\frac{distance}{speed}[/tex]
(sub the numbers in the formula)
distance=1200m, speed=100m/s
time=[tex]\frac{1200}{100}[/tex]
=12 seconds
A pair of glasses are dropped from the top of a 32.0 m high stadium. A pen is dropped 2.00 s later. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = -g = -9.81 m/s2.)
Answer:
Explanation:
the distance have the following relation:
d = (1/2)gt2
D=32.0 m
t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s
it take 2.56s from the glasses to hit the ground
when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s
x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m
the pen only travel 1.54m
so the pen is above the ground 32.0m - 1.54m = 30.46m
The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.
What is the height?Height is a numerical representation of the distance between two objects or locations on the vertical axis.
The height can refer to a physical length or an estimate based on other factors in physics or common use. |
The given data in the problem is;
h is the height from the top of a stadium = 32.0 m
t is the time period when the pen is dropped later = 2.00 s
x is the height above the ground
a is the air resistance. a = -g = -9.81 m/s²
From the second equation of motion;
[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]
When the glasses fall to the ground, the pen only travels a short distance;
[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]
So the pen travel the distance;
[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]
The pen above the ground is found as;
[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]
Hence the pen is 30.46m above the ground. when the glasses hit the ground.
To learn more about the height refer to the link;
https://brainly.com/question/10726356
A car traveling at 32.4 m/s skids to a stop in 4.55 s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
x = 73.71 [m]
Explanation:
In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.
[tex]v_{f }= v_{i}-(a*t)[/tex]
where:
Vf = fnal velocity = 0
Vi = initial velocity = 32.4 [m/s]
t = time = 4,55 [s]
a = acceleration or desacceleration [m/s^2]
0 = 32.4 - (a*4.55)
a = 7.12 [m/s^2]
Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.
Now using the following equation:
[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]
where:
xo = initial distance = 0
x = final distance [m]
Therefore we have:
x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)
x = 73.71 [m]
A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?
Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
A car with a mass of 2,000 kg travels a distance of 400m as it moves from one stoplight to the next. At its fastest , the car travels 18m/s. What is the kinetic energy at this point ?
Answer: KE= 324,000
Explanation: I hope that this helps! -_-
Answer:
324,000
Explanation:
the radius of the earth social
calculating light in physics
if a ramp is 12 meters long has a mechanical advantage of 6 whats its height brainly HELPP!
Answer:
h = 2 m
Explanation:
Mechanical advantage of a ramp is given by :
MA = length of incline/height of incline
Length of the ramp, l = 12 m
MA = 6
We need to find the height of the ramp.
So,
height of ramp = length of incline/MA
h = 12/6
h = 2 m
So, the height of the ramp is 2 m.
In a spy movie, the hero, James, stands on a scale that is positioned horizontally on the floor. It registers his weight as 810 N . Unknown to our hero, the floor is actually a trap door, and when the door suddenly disappears, James and the scale fall at the acceleration of gravity, down towards an unknown fate. As James falls, he looks at the scale to see his weight. What does he see
Answer:
His weight would be zero on the scale i.e he is weightless at that instance.
Explanation:
weight = mg
where m is the mass of the object, and g is the acceleration of gravity.
⇒ 810 = mg
During free fall, the weight of an object can be determined by:
W = mg - ma (provided that acceleration of gravity is greater than acceleration of the object)
where a is the acceleration of the object.
But since James fall at the acceleration of gravity, then:
g = a
mg = ma = 810 N
So that;
W = 810 - 810
= 0 N
Therefore though the weight of James is 810 N, but the scale reads 0 N. this condition is referred to as weightlessness.
why do some athletes get injuries before and after the game?
Answer:
they don't strech so they tear a muscle when they perform
Explanation:
how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?
Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
Victor has 100 trading cards, each with a distinct power level between 101 and 200 inclusive. Whenever he heads out, he always randomly selects 21 cards to bring with him just in case he meets a fellow collector. Prove that no matter which 21 cards he brings, Victor will always be able to select 4 of those cards that exhibit the following property:
Let the average power level of all 4 cards bep. The cards can be split into two pairs, each of which also has an average power level of p.
Answer:
Victor will always be able to select 4 of those cards with the following property
Explanation:
Number of trading cards = 100
victor selects 21 cards
let the 4 cards be labelled : A,B,C and D
The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P
let the two pairs be : ( A + B ) and ( C + D )
note: average power of each pair = P and this shows that
( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property
we have to check out the possible choices of two cards out of 21 cards yield distinct sums.
= C(21,2)=(21x20)/2 = 210.
from the question the number of distinct sums that can be created using 101 through 200 is < 210 .
hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards
What is the approximate weight of a 400 kg object?
Answer:
881.84905 LBS
Explanation:
ThErE :p
3922.66 newtons.
This is an exact amount, to get newtons form kg, multiply by 9.8, or in this case, 10.
This gives you 4000 newtons
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
The car alarm on a stationary car emits sound waves with a frequency of 450 Hz. If you are moving away from the stationary car at 20 m/s, what alarm frequency do you perceive
Answer:
Explanation:
The frequency of wave is directly proportional to velocity
f = kV
k = f/V
f1/V1 = f2/V2
Given
f1 = 450Hz
V1 = 343m/s
f2 = ?
V2 = 20m/s
Substitute into the formula
450/343 = f2/20
Cross multiply
343f2 = 450×20
343f2 = 9000
f2 = 9000/343
f2 = 26.24Hz
A fountain shoots a jetof water straight up. The nozzle is 1 cm in diameter and the speed of the water exiting the nozzle is 30 m/s. What is the force exerted by the water jet
Answer:
Explanation:
mass of water coming out per second = A x v where A is area of cross section of the nozzle and v is velocity of water
A = 3.14 x .005²
= 785 x 10⁻⁷ m²
mass of water coming out per second = 785 x 10⁻⁷ x 30 = 23.55 x 10⁻⁴ kg
momentum of this mass = 23.55 x 10⁻⁴ x 30 = 706.5 x 10⁻⁴ kg m /s .
Rate of change of momentum = 706.5 x 10⁻⁴
Let force be F
F - mg = 706.5 x 10⁻⁴
F = mg + 706.5 x 10⁻⁴
F = 23.55 x 10⁻⁴ x 9.8 + 706.5 x 10⁻⁴
= 937.3 x 10⁻⁴ N .
An object, initially at rest, is subject to an acceleration of 45 m/s^2. How long will it take that object to travel 1000m? Round to one decimal place.
Answer:
6.7 seconds
Explanation:
d=(1/2)at^2
equation
1000=(1/2)45t^2.
substitute
2000=45t^2.
multiply by 2 for both sides
44.44=t^2.
divide both sides by 45
6.7=t
take the square root of both sides
A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.
Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.
Answer:
The value is [tex]H = 18*10^{2} \ Atom / sec [/tex]
Explanation:
From the question we are told that
The atom fraction of metal A at point G is [tex] A = 0.30 \ m[/tex]
The atom fraction of metal A at a distance 5000nm from G is [tex]A_2 = 0.35[/tex]
The number of atoms per [tex]m^3[/tex] is [tex]N_h = 9 * 10^{28}[/tex]
The diffusion coefficient is [tex]D = 2* 10^{-14 } m^2/s[/tex]
Generally of the concentration of atoms of metal A at G is
[tex] N_A = A * N_h [/tex]
=> [tex] N_A = 0.3 * 9 * 10^{28}[/tex]
=> [tex] N_A = 2.7 * 10^{28} 2.7 atoms/m^3[/tex]
Generally of the concentration of atoms of metal A at a distance 5000nm from G is
[tex]D = 0.35 *9 * 10^{28}[/tex]
=> [tex]D = 3.15 * 10^{28} \ atoms / m^3[/tex]
The concentration gradient is mathematically represented as
[tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = 9 * 10^{20} / m^4[/tex]
Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as
[tex]J = -D* \frac{d N_A}{dx}[/tex]
=> [tex]J = -2* 10^{-14 * 9 * 10^{20} [/tex]
=> [tex] J = 18*10^{6}\ atoms\ crossing\ /m^2 s [/tex]
Generally if the cross-section area is [tex] a = 1 cm^2 = 10^{-4} \ m^2[/tex]
Generally the number of atom crossing the above area per second is mathematically is
[tex]H = 18*10^{6} * 10^{-4} [/tex]
=> [tex]H = 18*10^{2} \ Atom / sec [/tex]
Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster
Answer: The options are not given.
Here are the options.
a) There is an additional force lifting up on you.
(b) At the top you continue going straight and the seat moves out from under you.
(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (
d) Both b and c are correct.
(e) a, b, and c are correct.
The correct option Is D.
B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.
Explanation:
At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.
That is the scenarion that occur...