Answer:
The speed will be decrease
True.or false A railroad track runs southwest to northeast.
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:
Please provide an explanation.
Thank you!!
Answer:
(a) 22 kN
(b) 36 kN, 29 kN
(c) left will decrease, right will increase
(d) 43 kN
Explanation:
(a) When the truck is off the bridge, there are 3 forces on the bridge.
Reaction force F₁ pushing up at the first support,
reaction force F₂ pushing up at the second support,
and weight force Mg pulling down at the middle of the bridge.
Sum the torques about the second support. (Remember that the magnitude of torque is force times the perpendicular distance. Take counterclockwise to be positive.)
∑τ = Iα
(Mg) (0.3 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L)
F₁ = ½ Mg
F₁ = ½ (44.0 kN)
F₁ = 22.0 kN
(b) This time, we have the added force of the truck's weight.
Using the same logic as part (a), we sum the torques about the second support:
∑τ = Iα
(Mg) (0.3 L) + (mg) (0.4 L) − F₁ (0.6 L) = 0
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.4 L)
F₁ = ½ Mg + ⅔ mg
F₁ = ½ (44.0 kN) + ⅔ (21.0 kN)
F₁ = 36.0 kN
Now sum the torques about the first support:
∑τ = Iα
-(Mg) (0.3 L) − (mg) (0.2 L) + F₂ (0.6 L) = 0
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.2 L)
F₂ = ½ Mg + ⅓ mg
F₂ = ½ (44.0 kN) + ⅓ (21.0 kN)
F₂ = 29.0 kN
Alternatively, sum the forces in the y direction.
∑F = ma
F₁ + F₂ − Mg − mg = 0
F₂ = Mg + mg − F₁
F₂ = 44.0 kN + 21.0 kN − 36.0 kN
F₂ = 29.0 kN
(c) If we say x is the distance between the truck and the first support, then using our equations from part (b):
F₁ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L − x)
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (x)
As x increases, F₁ decreases and F₂ increases.
(d) Using our equation from part (c), when x = 0.6 L, F₂ is:
F₂ (0.6 L) = (Mg) (0.3 L) + (mg) (0.6 L)
F₂ = ½ Mg + mg
F₂ = ½ (44.0 kN) + 21.0 kN
F₂ = 43.0 kN
Answer:
a. Left support = Right support = 22 kNb. Left support = 36 kN Right support = 29 kNc. Left support force will decrease Right support force will increase.d. Right support = 43 kNExplanation:
given:
weight of bridge = 44 kN
weight of truck = 21 kN
a) truck is off the bridge
since the bridge is symmetrical, left support is equal to right support.
Left support = Right support = 44/2
Left support = Right support = 22 kN
b) truck is positioned as shown.
to get the reaction at left support, take moment from right support = 0
∑M at Right support = 0
Left support (0.6) - weight of bridge (0.3) - weight of truck (0.4) = 0
Left support = 44 (0.3) + 21 (0.4)
0.6
Left support = 36 kN
Right support = weight of bridge + weight of truck - Left support
Right support = 44 + 21 - 36
Right support = 29 kN
c)
as the truck continues to drive to the right, Left support will decrease
as the truck get closer to the right support, Right support will increase.
d) truck is directly under the right support, find reaction at Right support?
∑M at Left support = 0
Right support (0.6) - weight of bridge (0.3) - weight of truck (0.6) = 0
Right support = 44 (0.3) + 21 (0.6)
0.6
Right support = 43 kN
i need help, for physics
How long does it take a P-wave to travel 7,000 km? ______ minutes b) How long does it take an S-wave to travel 7,000 km? ______ minutes c) How long does it take a Love wave to travel 7,000 km? ______ minutes d) How long does it take a Rayleigh wave to travel 7,000 km? ______minutes
Answer:
A. 8.64 secs.
B. 14.58 secs.
C. 26.002 secs.
D. 33.46secs.
Explanation:
A. P wave would travel 7000km
p-wave travels on a speed of 13.5km/s
= 7000km/13.5km/s
= 8.64 secs.
B. S-wave time to travel 7000km
s-wave travels on a speed of 8km/s
= 7000km/8km/s
= 14.58 secs.
C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).
= 7000km to miles
= 4349.598m/2.788m/s
= 26.002 secs.
D. Rayleigh wave to travel 7,000 km
10,000m/s ( 2.1667 m/s ).
= 7000km to miles
= 4349.598m/2.1667m/s
= 33.46secs.
How much work is done lifting a 5 kg ball from a height of 2 m to a height of 6 m? (Use 10 m/s2 for the acceleration of gravity.)
A) 100 J B) 200 J C) 300 J D) 400 J
Answer:
B) 200 [J]
Explanation:
In order to solve this problem we must remember the definition of work which tells us that it is equal to the product of force by a distance, in this case, the force is the weight of the ball. The distance traveled is 4 [m] since 6-2 = 4[m]
F = m*g
where:
m = mass = 5 [kg]
g = gravity acceleration = 10 [m/s^2]
F = 5*10 = 50 [N]
w = F*d
where:
F = force = 50 [N]
d = 4 [m]
w = 50*4 = 200 [J]
Which of the organisms in the food web above is the top level carnivore
Answer:
apex consumers
Explanation:
they are top
A tower crane has a hoist motor rated at 159 hp. If the crane is limited to using 72.0 % of its maximum hoisting power for safety reasons, what is the shortest time in which the crane can lift a 5550 kg load over a distance of 89.0 m
Answer:
The value is [tex]t = 56.68 \ s [/tex]
Explanation:
From the question we are told that
The rating of the hoist motor is [tex]k = 159hp = 159 *746 =118614 \ W[/tex]
The percentage of it power used is [tex]z = 0.72 * 118614=85402.08 \ W[/tex]
The mass of the load is m = 5550 kg
The distance is h = 89.0 m
The potential energy required to lift the load through that distance is
[tex]E = m * g * h[/tex]
=> [tex]E = 5550 * 9.8 * 89.0[/tex]
=> [tex]E = 4840710 \ J[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{E}{ z}[/tex]
=> [tex]t = \frac{4840710}{ 85402.08}[/tex]
=> [tex]t = 56.68 \ s [/tex]
An airplane flies with a constant speed of 600 km/h. to the west. How far can it travel in 1/4 hour?
Answer:
d = 150 km
Explanation:
Speed of an airplane is 600 km/h
We need to find how far it travel in 1/4 hour.
We know that the speed of an object is given by distance travelled divided by the time taken. Let d is the distance covered in 1/4 hour. So,
[tex]v=\dfrac{d}{t}\\\\d=v\times t\\\\d=600\ km/h \times \dfrac{1}{4}\ h\\\\d=150\ km[/tex]
So, it will cover 150 km.
what is the meaning of the word physics
Answer:
the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.
Explanation:
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Two charged objects are separated by distance, d. The first charge has a larger magnitude (size) than the second charge. Which one exerts the most force?
What is the volume of an object if it has a mass of 10 grams and a density of 87 g/ml
Answer:
The answer is 0.115 mLExplanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\[/tex]
From the question
mass = 10 g
density = 87 g/ml
We have
[tex]volume = \frac{10}{87} \\ = 0.114942528...[/tex]
We have the final answer as
0.115 mLHope this helps you
What is the maximum torque on a 150-turn square loop of wire 18.0 cm on a side that carries a 50.9 A current in a 1.60 T field
Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
how far will a brick starting from rest fall freely in 3.0 seconds?
Answer: It will be about 44.1m
Explanation:
please answer this question with an explanation
will mark brainiest
Answer:
jenny
Explanation:
A car travels at a velocity of 80 m/s, accelerates than stops in 250 second. What is the acceleration of the truck?
(I’ll give brainliest please someone help me on this and explain it with work I don’t get it)
A marathon is 26 mi and 385 yd long. Estimate how many strides would be required to run a marathon. Assume a reasonable value for the average number of feet/stride.
Answer:
According to the University of Iowa, the average length of a stride is 5ft.
Now, the total distance of the marathon is:
26mi and 385yd.
Let's transform that distance into ft.
1mi = 5280ft
Then:
26mi = 26*5280ft = 137,280ft
1yd = 3ft
then:
385yd = 385*3ft = 1,155ft.
Then the total distance of the marathon, in ft, is:
D = 137,280ft + 1,155ft = 138,435 ft.
Now the average number of strides needed will be equal to the quotient between the total distance of the marathon and the distance traveled in each stride.
N = 138,435ft/5ft = 27,687.
Find the linear velocity of a point moving with uniform circular motion, if the point covers a distance s in the given amount of time t. s
Answer:
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
Explanation:
From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:
[tex]v = r\cdot \omega[/tex] (Eq. 1)
Where:
[tex]r[/tex] - Radius of rotation of the particle, measured in meters.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]v[/tex] - Linear velocity of the point, measured in meters per second.
But we know that angular velocity is also equal to:
[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)
Where:
[tex]\theta[/tex] - Angular displacement, measured in radians.
[tex]t[/tex] - Time, measured in seconds.
By applying (Eq. 2) in (Eq. 1) we get that:
[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)
From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:
[tex]s = r\cdot \theta[/tex]
[tex]v = \frac{s}{t}[/tex]
The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]
am I right? be honest
Answer:
I chose c because it is the greater slope at point c
When electrical energy is used what type of energy is also produced and considered to be waste energy?
Radiant energy
Thermal energy
Mechanical energy
Nuclear energy
Answer:
Thermal Energy
Explanation:
The diagram shows two forces acting on the dog. What are these two forces
Answer:
kenietic and potential i guess
Explanation:
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:
21. A toy car starts from rest and begins to accelerate at 11.0 m/s2. What is the toy
car's final velocity after 6.0 seconds?
Answer:
Explanation:
Given parameters:
Initial velocity = 0
Acceleration = 11m/s²
Time = 6s
Unknown:
Final velocity = ?
Solution:
From the given parameters, we use one of the appropriate equations of motion to solve this problem.
V = U + at
V is the final velocity
U is the initial velocity
a is the acceleration due to gravity
t is the time taken
Input the parameters and solve;
V = 0 + 11 x6
V = 66m/s
The final velocity is 66m/s
HELP PLS7. A steel ball is dropped from a height of 100 meters. Which velocity-time graph best describes the
motion of the ball?
Answer:
Option C.
Explanation:
To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:
1. Determination of the velocity.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Final velocity (v) =.?
v² = u² + 2gh
v² = 0² + (2 × 9.8 × 100)
v² = 0 + 1960
v² = 1960
Take the square root of both side.
v = √(1960)
v = 44.27 m/s
2. Determination of the time taken.
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) = 100 m
Time (t) =.?
h = ½gt²
100 = ½ × 9.8 × t²
100 = 4.9 × t²
Divide both side by 4.9
t² = 100 / 4.9
Take the square root of both side
t = √(100 / 4.9)
t = 4.52 s
From the above illustration,
Initial time (t1) = 0 s
Final time (t2) = 4.52 s
Initial velocity (u) = 0 m/s
Final velocity (v) = 44.27 m/s
Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.
what is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
Density ρ is mass m per unit volume v, or
ρ = m / v
Solving for v gives
v = m / ρ
So the given object has a volume of
v = (130 g) / (65 g/cm³) = 2 cm³
correct me if im wrong
One airplane is approaching an airport from the north at 181 kn/hr. A second airplane approaches from the east at 278 km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 30 km away from the airport and the westbound plane is 15 km from airport.
Answer:
The value is [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
Explanation:
From the question we are told that
The speed of the airplane from the north is [tex]\frac{dN}{dt} = -181 \ km /hr[/tex]
The negative sign is because the direction is towards the south
The speed of the airplane from the east is [tex]\frac{dE}{dt} = -278 \ km/hr[/tex]
The negative sign is because the direction is towards the west
The distance of the southbound plane from the airport is [tex]N = 30 \ km[/tex]
The distance of the westbound plane is [tex]E = 15 \ km[/tex]
Generally the distance between the plane is mathematically represented using Pythagoras theorem as
[tex]R^2 = N^2 + E^2[/tex]
Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes
So
[tex]2R\frac{dR}{dt} = 2N \frac{dN}{dt} + 2E\frac{dE}{dt}[/tex]
Here
[tex]R = \sqrt{N^2 + E^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R = \sqrt{30^2 + 15^2}[/tex]
=> [tex]R =33.54 \ m [/tex]
[tex]2(33.54) * \frac{dR}{dt} = 2( 30)*(-181) + 2*15*(-278)[/tex]
=> [tex] 67.08 * \frac{dR}{dt} = -19200[/tex]
=> [tex] \frac{dR}{dt} = -286.2 \ km/hr [/tex]
The rate of change of the distance between the planes is 286.23 km/hr.
The given parameters;
speed of the airplane from North, dn/dt = 181 Km/hspeed of the airplane from the East, de/dt = 278 km/hnorth distance, n = 30 kmeast distance, e= 15 kmThe distance between the two planes is calculated by applying Pythagoras theorem as shown below;
[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]
The rate of change of the distance between the planes is calculated as follows;
[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]
Thus, the rate of change of the distance between the planes is 286.23 km/hr.
Learn more here:https://brainly.com/question/11488002
Newton's first law states that objects do not change their motion unless acted upon by a net force. What does the word 'net' mean in this context?
A woven net, such as a fishing net or basketball net
B To catch or ensnare
C Remaining or left over after everything has been accounted for
D To cover, such as with mosquito netting
when hydrogen shares electrons with oxygen the outermost shell of the hydrogen atoms are full with how many electrons? and oxygens valence shell is full with how many electrons? because the valence shells of these atoms are full,the atoms are stable.
Answer:
2 and 8
Explanation:
please mark me brainiest I would really appreciate it.
You have made a simple circuit with one bulb. If you wanted to add an
extra bulb without the first bulb dimming. What would you need to
design?
A. A series circuit
B. A complex circuit
C. A parallel circuit
D. An incomplete circuit
Answer:
[tex]A. \: A \: series \: circuit[/tex]
Explanation:
♨Rage♨
Answer:
C. A parallel circuit
Explanation:
Adding a bulb in parallel with the existing bulb will apply the same voltage to both bulbs. They will light equally bright.
You would design a parallel circuit.
_____
In a series circuit the same current would flow in both bulbs, but that current would be at half the original current level. Both bulbs would be dimmer than the first bulb was.
It is difficult to create a "complex" circuit with only two components. An "incomplete" circuit would result in no light at all.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
This question is incomplete, the complete question is;
A gong makes a loud noise when struck. The noise gradually gets less and less loud until it fades below the sensitivity of the human ear. The simplest model of how the gong produces the sound we hear treats the gong as a damped harmonic oscillator. The tone we hear is related to the frequency f of the oscillation, and its loudness is proportional to the energy of the oscillation.
If the loudness drops to 90 % of its original value in 5.0 s , what is the time constant of the damped oscillation
Answer: the time constant of the damped oscillation is 47.44s
Explanation:
Given that;
t = 5.0s
Lets say Ao is the amplitude of initial loudness and later A(t) = 0.9 Ao
EXPRESSION for amplitude is A(t) = Ao e^-t / T
t is time while T is time constant
so
0.9Ao = Ao e^-t / T
0.9 = e^ -t/T
So we take the natural log of both the sides
ln (0.9) = -t/T
-0.1054 = -t/T
0.1054 = t/T
WE now substitute our value of t
0.1054 = t/T
0.1054T = 5.0
T = 5 / 0.1054
T = 47.44s
therefore the time constant of the damped oscillation is 47.44s