The temperature of water affects the solubility of material the ability of a material, the solute, to combine with another substance, the solvent, is known as solubility in chemistry.
The contrary is appropriate for insoluble except in extremely rare circumstances, a solid or liquid cannot be "dissolved" in gas without first becoming gaseous composition of the solute and solvent, including their pH and the presence of other dissolved substances, as well as temperature and pressure, are the key determinants of solubility. The dependence is frequently explicable in terms of interactions between the atoms, molecules, or ions of the two substances as well as in terms of thermodynamic notions like enthalpy and entropy.
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Question 2 of 16
A Elag for Review
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Scientists today are studying tidal power as an alternative energy source for generating electricity.
Some scientists conclude that tidal power is a good alternative source of energy because it uses a natural process, does not rely on fossil fuels, and does not
release greenhouse gases. These scientists recommend building tidal power plants around the United States in locations with high tidal power potential.
Other scientists conclude that tidal power is not a good alternative energy source because tidal power plants can negatively impact the surrounding ecosystem by
killing marine animals, restricting fish migration, reducing the natural flow of water, and causing silt buildup in waterways. These scientists recommend not building
tidal power plants in U.S. waterways.
Both conclusions are based on valid data and scientific reasoning. How can both conclusions be valid?
OA. The two groups of scientists believe different theories about the process of generating electricity.
OB. The two groups of scientists are looking at different specific aspects of the same scientific problem.
OC.
The two groups of scientists collect and analyze data about all types of possible alternative energy sources.
OD. The two groups of scientists are on opposite sides of the debate on whether the United States needs to find clean alternative energy sources.
Answer:
Explanation:
The short answer is B.
Both answers are correct if you omit the concerns of the other group. The second group is primarily concerned about the effects of tidal power on the environment. They are worried about marine animals, fish, silt buildups.
Contrary to this, the first group see the need for power that is relatively clean.
That's what B points out.
A bullet is moving at a speed of 350 m/s when it embeds into a lump of moist clay. the bullet penetrates for a distance of 0.070 m. determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration)
Acceleration of the bullet is -87500m/s2
The rate at which the velocity of an object varies in relation to time is known as acceleration.
Acceleration is a vector quantity and the direction of the net force applied on an item determines the direction of its acceleration.
Acceleration is the rate at which the speed and direction of a moving object vary over time. When anything moves faster or slower moving straight, it is considered to be accelerating. Even though the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects influence to the acceleration for all other motions.
Since acceleration has both a magnitude and a direction, it is considered as a vector quantity. The velocity vector change during a time interval divided by the time interval is the definition of acceleration.
The limit of the ratio of the variation in velocity during a specified time period to the time period as the time period approaches zero determines the instantaneous acceleration (at a specific time and location).
Formula:
[tex]v^2 = u^2 + 2aS[/tex]
[tex]a = (v^2 - u^2) / 2S[/tex]
[tex]a = -350^2 / 2 * 0.07[/tex]
[tex]a = -122500 / 0.14[/tex]
[tex]a = -87500 m/s^2[/tex]
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What has been the affect of agricultural price supports
Overproduction is caused by price supports. Governments assist farmers to increase production by subsidising prices above the point at which the market clears. Farmers use more inputs per acre in order to produce more. Additionally, they compete with one another for the limited supply of farmland, driving increasing the cost of it.
What is agricultural price supports ?Agricultural assistance is defined as the annual dollar amount of gross consumer and taxation transfers to agriculture resulting from governmental policies that support agriculture, independent of the goals or effects on the economy.
The link between price and supply is generally positive since farmers will typically desire to increase the quantities they bring to market. It demonstrates that goods manufacturers are willing to create more as the price rises.Learn more about Agricultural price supports here:
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Three bitmap charges are placed on the y-axis as in
the figure if the Outcome of the electric forces is
(q1) equal to 4.2 N in the negative y-axis direction
selects the charge type q3 and calculate its amount.
The magnitude of the third charge is determined as -1.37 μc.
Magnitude of charge q3The magnitude of the charge on q3 is calculated as follows;
F = q12 + q23 + q13
F = kq₁q₂/r₂ + kq₂q₃/r² + kq₁q₃/r²
4.2 = (9 x 10⁹ x 2 x 10⁻⁶ x 3 x 10⁻⁶)/(0.1²) + (9 x 10⁹ x 3 x 10⁻⁶q₃)/(0.2²) + (9 x 10⁹ x 2 x 10⁻⁶ x q₃)/0.3²)
4.2 = 5.4 + 675,000q₃ + 200,000q₃
-1.2 = 875,000q₃
q₃ = -1.2/875000
q₃ = -1.37 x 10⁻⁶ C
q₃ = -1.37 μC
Thus, the magnitude of the third charge is determined as -1.37 μc.
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A car drives horizontally off a 63-m-high cliff at a speed of 29 m/s . Ignore air resistance.
1. How long will it take the car to hit the ground? (in seconds)
a) 2.9
b) 4.0
c) 3.6
d) 1.9
e) 5.2
2. How far from the base of the cliff will the car hit? (in meters)
a) 54
b) 190
c) 100
d) 210
e) 250
I already know the answers to these questions, but if I could have an explanation as to how to solve these problems I would greatly appreciate it!!
The car's vertical position [tex]y[/tex] at time [tex]t[/tex] is
[tex]y = 63\,\mathrm m - \dfrac12 gt^2[/tex]
since it starts 63 m above the ground, and after leaving the cliff it accelerates downward due to gravity.
Its horizontal position [tex]x[/tex] is
[tex]x = \left(29\dfrac{\rm m}{\rm s}\right) t[/tex]
since the car leaves the cliff horizontally at 29 m/s, and is not influenced by any other acceleration in this plane.
1. Solve for [tex]t[/tex] such that [tex]y=0[/tex].
[tex]63\,\mathrm m - \dfrac12 gt^2 = 0 \implies t = \sqrt{\dfrac{126\,\rm m}g} \approx \boxed{3.6}\,\rm s[/tex]
2. Solve for [tex]x[/tex] at this value of [tex]t[/tex].
[tex]x = \left(29\dfrac{\rm m}{\rm s}\right) \sqrt{\dfrac{126\,\rm m}g} \approx 104\,\rm m \approx \boxed{100}\,\rm m[/tex]
A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.38 m/s, and 13.3 s later the particle is moving in the positive x direction at a speed of 7.95 m/s.
Part A: What is the particle’s acceleration in m/s2?
Part B: What is the particle’s velocity, in m/s, 13.3 s before it was moving in the negative x direction at a speed of 4.38 m/s?
Part C: When is the particle at rest? Express this answer as a time in seconds elapsed since it was moving in the negative x direction at a speed of 4.38 m/s?
solve all parts please
The particle’s acceleration is 0.268 m/s².
The initial velocity of the particle before 13.3 seconds is 3.56 m/s.
The time when the particle is at rest is 32.68 seconds.
Acceleration of the particleThe acceleration of the particle is calculated as follows;
a = Δv/Δt
a = (7.95 m/s - 4.38 m/s) / 13.3 s
a = 0.268 m/s²
Initial velocity of the particle before 13.3 secondsu = at
u = 0.268 m/s² x 13.3 s
u = 3.56 m/s
Time when the particle is at rests = ut + ¹/₂at²
0 = -4.38t + ¹/₂(0.268)t²
0.134t² = 4.38t
0.134t = 4.38
t = 32.68 seconds
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'A drag racer starts from rest and accelerates uniformly to a speed of 76.5 m/s in 8.5 seconds. what is the distance traveled by the car in the 8.5 s?
Answer:
325.125m
Explanation:
We can use the kinematics equations to solve this problem.
Given:
u = Initial Velocity = 0 m/s
v = Final Velocity = 76.5 m/s
t = 8.5 s
We can use the first kinematics equation to find the acceleration:
v = u + at
76.5 = 0 + 8.5a
a = 76.5 ÷ 8.5
[tex] = 9\frac{m}{ {s}^{2} } [/tex]
Since we now have the acceleration, we can use the third kinematics equation to find the distance travelled:
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = (0)(8.5) + \frac{1}{2} (9) {(8.5)}^{2} \\ = 325.125m[/tex]
A fireworks shell is accelerated at constant acceleration from rest to a velocity of 61 m/s over a distance of 2 m.
Part A. How long, in seconds, does the acceleration last?
Part B. Calculate the acceleration, in meters per second squared.
The acceleration is 930 m/s^2 and the time taken is 0.07 s.
What is the acceleration?We know that the acceleration is the rate of change of velocity with time. We can find the acceleration in this case by the use of kinematic equations of motion as we shall demonstrate below.
Let use the formula;
v^2 = u^2 + 2as
Where
v = final velocity
u = initial velocity
a = acceleration
s = distance covered
u = 0 m/s because the motion started from rest
v^2 = 2as
a = v^2/2s
a = (61)^2/2 * 2
a = 930 m/s^2
The time taken is;
v = u + at
u = 0 m/s
v = at
t = v/a
t = 61 m/s/930 m/s^2
t = 0.07 s
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did you know that before pencil erasers were added
Answer:
Yes, The original erasers were bread. Until the 1770's humanity's preferred way of erasing graphite marks relied on bread that had been de-crusted, moistened and balled up. The first pencil only came out in 1795.
Explanation:
Answer:
Until the 1770s, humanity's preferred way of erasing errant graphite marks relied on bread that had been de-crusted, moistened and balled up. While these erasers were cheap and plentiful, they had a distinct disadvantage: They were, you know, made of bread.
Riley was standing on the balcony and dropped a paper rocket. How far below did the rocket land
when it hit the ground assuming the total air time was 4.3 seconds?
The rocket landed 92.45m below when it hit the ground.
When a body is dropped from a height, its initial velocity is 0. A body executing free fall moves with constant acceleration i.e., acceleration due to gravity
Given data -
time = t = 4.3 secpaper rocket is dropped so u ( initial velocity ) = 0 and a = g ( acceleration due to gravity )Distance = s = ?By using 2nd equation of motion:
[tex]s = ut + \frac{1}{2} at^2[/tex]
Equation of motion under gravity:
[tex]s = ut + \frac{1}{2} gt^2[/tex]
[tex]= 0 \times 4.3 \sec + \frac{1}{2} \times 10 \times (4.3)^{2}[/tex] ( assuming g = 10 m/s²)
[tex]= \frac{1}{2} \times 10 \times 18.49[/tex]
= 5 × 18.49
= 92.45 m
Thus we can conclude that the rocket landed 92.45m below when it hit the ground.
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what is projectiles
A CD has diameter of 110 millimeters. When playing audio, the angular speed varies to keep the linear speed constant where the disc is being read. When reading along the outer edge of the disc, the angular speed is about 180 RPM (revolutions per minute). Find the linear speed.
The linear velocity of the CD is 1.03 m/s.
What is the angular velocity?We know that the angular velocity and the linear velocity are related by the use of the formula;
v = rω
v = velocity of the body (linear)
r = radius of the body
ω = angular velocity
Given that we have to convert from rpm to rad/s
1 RPM = 0.10472 rad/s
180 RPM = 180 RPM * 0.10472 rad/s/1 RPM
= 18.7 rad/s
We know that radius = diameter/2 = 110 millimeters/2 = 55 millimeters or 55 * 10^-3 m
Substituting values;
v = 55 * 10^-3 m * 18.7 rad/s
v = 1.03 m/s
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a fluorescent lamp rated at 100 kw has an efficiency of 20% how much energy is dissipated as heat if the lamp is used for 5 hours?
Energy is dissipated as heat from the lamp is
400 kw
It is given that
power of lamp = 100 kw
efficiency = 20%
time = 5 hour
We have to find the energy of lamp
Efficiency is the measure of how completely work input is connected to work output.
Efficiency = work(out)/work(in) * 100%
Efficiency of visible light = 20%
Energy converted to heat = 80%
Energy = Power * time
Energy = 100 kw x 80% x 5 = 400 kw
Hence, Energy dissipated as heat is 400 kw
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The speed of light in a vacuum is 300,000.0 km/s. How long does it take light to travel 150,000,000 km from the Sun to Earth?
It takes light to travel 150,000,000 km from the Sun to Earth in 500 seconds.
Light travels quicker in a vacuum than in another medium. this is because there's no obstruction in the vacuum for the propagation of mild and accordingly, the refractive index of the vacuum is the lowest. the rate of mild in a vacuum is 3×108 ms. daylight takes approximately 8 minutes to reach the Earth. thru the vacuum of space, irrespective of what their strength is, they always travel at the rate of light.
It would not remember how fast you chase after or run closer to light, both that velocity you view it journeying at will continually be equal. The thing that shifts, in place of its velocity, can be the mild's electricity. Albert Einstein's special principle of relativity famously dictates that no regarded object can journey faster than the speed of mild in a vacuum, which is 299,792 km. This speed restriction makes it not going that human beings will ever be able to ship spacecraft to discover beyond our nearby vicinity of the Milky Way.
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A space shuttle orbits earth at a speed of 21,000 km/he. how far will it travel in 1.4hours? Assume the shuttle is traveling on a straight path (it is actually orbiting the Earth but for this time period the path is relatively straight).
The space shuttle will travel as far as 29400 km in 1.4 hours
What is Speed ?Speed is the distance travelled per time taken. It is a scalar quantity.
Given that a space shuttle orbits earth at a speed of 21,000 km/hr. Since it is assumed that the shuttle is traveling on a straight path, to know how far it will travel in 1.4 hours, we will apply the speed formula.
Speed = distance / time
21000 = distance / 1.4
distance = 21000 x 1.4
distance = 29400 Km
Therefore, the space shuttle will travel as far as 29400 km in 1.4 hours
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motorcycle accelerate
Which of these reactions is a combustion reaction?
O A. burning wood
O B. decomposition of water
O c.
production of food by plants with the help of the Sun's energy
O D.
rotting of fruits and vegetables
Answer:
Option B. Burning of wood.
Help with #2. Check my answer
The total distance traveled for the entire trip is 30 meters and the total time consumed in the trip would be 4 seconds so the correct answer for the second problem is 4 seconds.
What is displacement?Displacement describes this shift in location.
Displacement is a vector quantity.
As given in the problem we have to find out from the graph the total time consumed in the trip,
As we can observe from the graph the position of the object is changing up to 4 seconds so we can conclude that the entire trip would have taken a total of 4 seconds.
Thus, the total time of the trip would be 4 seconds.
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A dog weighing 18 kg runs into a park at a velocity of .63 m/s. What is the momentum of the dog? (1 point)
O.035 kg*m/s
O 17.37 kg*m/s
O 28.5 kg*m/s
O 11.34 kg* m/s
The dog has a momentum of 11.34 kg*m/s.
The correct answer is D.
What exactly is momentum?Product of a particle's mass and velocity. A vector quantity, momentum has both a magnitude and a direction. The force exerted on a particle is equated to the time rate of change of momentum by Isaac Newton's second equation of motion.
How does momentum work?The sum of the units of mass and velocity yields the unit of momentum. The momentum is expressed in kilogram meters per second (kgm/s) in SI units if the mass and velocity are both expressed in kilograms and meters per second, respectively.
Briefing:P = momentum
m = mass
v = velocity
m = 18 kg
v = .63 m/s
P = m × v
P= 18 × .63
p = 11.34 kg*m/s.
The dog has a momentum of 11.34 kg*m/s.
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The complete question is -
A dog weighing 18 kg runs into a park at a velocity of .63 m/s. What is the momentum of the dog? (1 point)
A-035 kg*m/s
B-17.37 kg*m/s
C- 28.5 kg*m/s
D- 11.34 kg* m/s
Answer:
The dog has a momentum of 11.34 kg*m/s.
You lift a 50.0 kg object distance of 0.500m how much work did you do
The amount of work done in lifting the 50.0 Kg to a distance of 0.5 m is 245
What is work done?This simply defined as the product of force and distance moved in the direction of the force. mathematically it is expressed as:
Work done (Wd) = force (F) × distance (d)
Wd = fd
Where height is involved, the workdone is given as as:
Work done (Wd) = mass (m) × acceleration due to graity (g) × height (h)
Wd = mgh
With the above formula, we can obtain the work done. details below.
From the question given above, the following data were obtained:
Mass (m) = 50 KgHeight (h) = 0.5 mAcceleration due to gravity (g) = 9.8 m/s²Work done (Wd) =?Wd = mgh
Wd = 50 × 9.8 × 0.5
Work done = 245 J
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According to the Doppler effect, what will happen to the pitch when a car with a blowing horn is approaching?
O It will increase.
O It will decrease.
-O It will stay the same.
Answer:
Compared to the emitted frequency, the received frequency is higher during the approach, identical at the instant of passing by, and lower during the recession.
Explanation:
Princess Leia once fell out of the spaceship. She came back with the Force. From the farthest spot ( zero speed) back to the spaceship was 10 meters, and it took 7.1 seconds. She weighed 45kg. Assume constant force during this trip. The force needed was __ N.
The force needed by Princess Leia is 17.87 N.
Acceleration of Princess Leia
The acceleration of Princess Leia is calculated as follows;
s = u + ¹/₂at²
where;
u is the initial velocity = 0s = ¹/₂at²
2s = at²
a = 2s/t²
a = (2 x 10) / (7.1²)
a = 0.397 m/s²
Force needed by Princess LeiaThe force needed by Princess Leia is calculated as follows;
F = ma
F = 45 x 0.397 m/s²
F = 17.87 N
Thus, the force needed by Princess Leia is 17.87 N.
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Characters used: 88/ 15000
Part E
From your observations about the horizontal and vertical component of position, velocity and acceleration, do you think taking
"Legs" of the snowboarder as reference point for plotting the motion is a good option for describing the snowboarder's
projectile motion?
BIUX² X₂ 15px
Characters used: 0/ 15000
A
!!!
≡ ≡ ≡ ≡
A projectile's horizontal velocity is constant (a never changing in value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion.
For relatively small displacements, the force of gravity has little effect on an object's horizontal motion. This indicates that while the vertical velocity is accelerating, the horizontal velocity is remaining constant. The cause is that perpendicular vectors behave differently from one another. A projectile's vertical velocity is zero at its maximum height. There is now only the horizontal velocity.
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A runner is jogging in a straight line at a steady v, 3.5 km/hr. When the runner is L= 3.1 km from the finish line, a bird begins flying straight from the runner to the finish line at v= 17.5 km/hr (5 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner. finish line What cumulative distance does the bird travel? Even though the bird is a dodo, assume that it occupies only one point in space (a "zero" length bird), travels in a straight line, and that it can turn without loss of speed.Answer in units of km.
The cumulative distance traveled by the bird is determined as 2.63 km.
Time of motion of the birdThe time of motion of the bird can be determined from the principle of relative velocity as shown below.
(Vb + Vr)t = D
where;
Vb is the velocity of the birdVr is the velocity of the runnert is time of motion of the birdD is the distance between the bird and the runner(3.5 + 17.5)t = 3.1
21t = 3.1
t = 3.1/21
t = 0.15 hr
Cumulative distance traveled by the birdD = vt
D = 17.5 km/hr x 0.15 hr
D = 2.63 km
Thus, the cumulative distance traveled by the bird is determined as 2.63 km.
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Can you help with this ?
Answer
25 miles/hour
Explanation
Given that,
A pilot can travel with wind = 416miles
A pilot can travel against wind = 336miles
A pilot can speed without wind = 235miles/hour
let wind speed = v miles/hr
pilot's speed with the wind = 235+ v
pilot's speed against the wind = 235 - v
Time = distance / speed
416 / (235 + v) = 336 / (235 - v)
416(235 - v) = 336(235+v)
97760 - 416v = 78960 + 336v
97760 - 78960 = 416v + 336v
18800 = 752v
v = 18800/752
v = 25 miles/hour
wind speed = 25 mph
How fast should you move away from a 6.0 × 1014 Hz light source to
observe
waves with a frequency of 4.0 × 1014 Hz?
We can move away from a 6.0 × 1014 Hz light source to observe waves with a frequency of 4.0 × 1014 Hz with a speed of 0.38c.
What is Frequency?Physics refers to the frequency as the quantity of waves passing a fixed point in a unit of time. Additionally, it is the quantity of vibrations or cycles that a body in periodic motion experiences in a certain period of time. One cycle or vibration is defined as the movement of a body in periodic motion across a sequence of occasions or locations before returning to its initial state.
According to the question,
6.0× 1014 Hz can be written as 6 A° similarly,
4.0×1014 Hz can be written as 4 A°
By using Doppler's effect, the wavelength of light will be :
[tex]\lambda\\[/tex]=[tex]\lambda[/tex]₀[tex]\sqrt{1+(v/c)/1-(v/c)}[/tex]
6.0 A°= (4.0 A°) [tex]\sqrt{1+(v/c)/1-(v/c)}[/tex]
v=0.38c
Hence, it will move with 0.38c.
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Can someone give an outline for what to write?
write a lab report for this lesson’s lab. Be sure that your report: includes all major elements of a lab report. meets your teacher’s content and format expectations. is clearly organized and formatted. demonstrates strong scientific reasoning and writing. While writing, you can revisit previous parts of the lesson by returning to the course map. Be sure to refer to the lab’s student guide, which you can find on the first page of the lab experiment activity. You may also find it helpful to refer to the remaining pages of this guide, which provide general guidelines for writing lab reports. You can upload your completed report with the upload tool in formats such as OpenOffice.org, Microsoft Word, or PDF. Alternatively, your teacher may ask you to turn in a paper copy of your report or to use a web-based writing tool.
An outline for how to approach writing your lab report is as follows
The parts
"TITLE," "INTRODUCTION" "PROCEDURE" "RESULTS" "DISCUSSION/CONCLUSION"This is further explained below.
How to write a lab report?Generally, A lab report is composed of eight different components, which are as follows: the title, the abstract, the introduction, the techniques and materials, the findings, the discussion, and the conclusion.
In conclusion, It is important that the experiment is adequately described in the title of the lab report, and that the title also accurately represents the data obtained from the experiment.
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In what time would a Current
Current of
IA transfer
Change of 30c
a Change
A current of 1 Ampere would transfer a charge to 30 coulombs in 30 seconds.
Current is one of the fundamental quantities of physics that are used to derive the other quantities. It is the number of charges in coulombs flown through an area of cross-section per unit time. Its S.I. unit is Ampere. It is denoted by the capital letter 'I'.
Given in the question
Current = I = 1 A
Charge Transferred = 30 Coulombs
Using the Formula: Current = Charge/ Time
Time = Charge/Current
Put in the value
Time = 30/ 1
Time = 30 seconds
Therefore. a current of 1 ampere would take a time of 30 seconds to transfer a charge equal to 30 coulombs.
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2. How would the horizontal range change if the height from the ground was
doubled? Explain.
As the height from the ground was doubled the horizontal range increases.
As we increases the height from the ground, then the time ball will remain in air, means time of flight will increase which will ultimately increase the horizontal range.
The maximum horizontal range of an object may go when shot as a projectile from any point is referred to as the range of a projectile motion.Numerous pieces of military equipment heavily rely on the projectile's range of motion. Calculating the precise distance a firearm may travel in a combat zone can greatly aid in gaining tactical advantages.In the field of astronomy as well, a projectile's range is crucial. The projectile motion equations can be used to calculate the precise distance that an astronomical object can travel. Furthermore, by making exact estimates of the trajectory and angle of inclination, the idea is utilized to launch a spacecraft into orbit.To know more about horizontal range visit : https://brainly.com/question/23827445
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Particle 1 is moving on the x-axis with
an acceleration of 3.37 m/s² in the positive
x-direction. Particle 2 is moving on the y-
axis with an acceleration of 5.18 m/s² in the
negative y-direction. Both particles were at
rest at the origin at t = 0 s.
Find the speed of particle 1 with respect to
particle 2 at 2.83 s.
Answer in units of m/s.
The speed of particle 1 with respect to particle 2 at 2.83 s is determined as 17.5 m/s.
What is relative velocity?
The relative velocity of an object 1 relative to another object 2 is the velocity that object 1 would appear to have to an observer moving with 2.
Velocity of particle 1 after 2 .83 seconds,V(1) = 3.37 m/s² x 2.83 s
V(1) = 9.54 m/s
Velocity of particle 2 after 2 .83 seconds,V(2) = 5.18 m/s² x 2.83 s
V(2) = 14.66 m/s
Relative velocity of particle 1 with respect to particle 2V(12) = √(9.54² + 14.66²)
V(12) = 17.5 m/s
Thus, the speed of particle 1 with respect to particle 2 at 2.83 s is determined as 17.5 m/s.
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