So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, this picture is not quite complete! You are on the surface of the Earth, which is rotating. While answering the following questions you may ignore the Earth’s motion around the sun, galactic center, etc. (though that is another interesting question!) (a) What is the magnitude of the acceleration of a person sitting in a chair on the equator? (b) At the equator, is your mass times the gravitational acceleration of the Earth greater than, less than, or equal to the normal force exerted on you by the chair you are sitting on? Explain. (c) A classmate of yours asks you why we have ignored this acceleration for the whole first term of physics. "Is everything we’ve learned a lie?" they ask. Ease their fears by calculating the percentage difference between the normal force from the chair and your weight while sitting on equator. (d) The latitude of Corvallis is 44.4˚. What is your acceleration while sitting in your chair?
Answers
Answer:
a) [tex]a=33.73mm/s^{2}[/tex]
b) mg>N
c) [tex]\%_{change}=0.343\%[/tex]
d) [tex]a=24.07mm/s^{2}[/tex]
Explanation:
In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).
In this case, we know the centripetal acceleration is given by the following formula:
[tex]a_{c}=\omega ^{2}r[/tex]
where:
[tex]\omega=\frac{2\pi}{T}[/tex]
we know the period of rotation of the earth is about 24 hours, so:
[tex]T=24hr*\frac{3600s}{1hr}=86400s[/tex]
so we can now find the angular speed:
[tex]\omega=\frac{2\pi}{86400s}[/tex]
[tex]\omega=72.72x10^{-6} rad/s^{2}[/tex]
So the centripetal acceleration will be:
[tex]a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)[/tex]
which yields:
[tex]a_{c}=33.73mm/s^{2}[/tex]
b)
In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)
So we can do a sum of forces in equilibrium:
[tex]\sum F=0[/tex]
so we get that:
[tex]N-mg+ma_{c} = 0[/tex]
and solve for the normal force:
[tex]N=mg-ma_{c}[/tex]
In this case, we can clearly see that:
[tex]mg>mg-ma_{c}[/tex]
therefore mg>N
This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.
c)
So let's calculate our weight and normal force:
Let's say we weight a total of 60kg, so:
[tex]mg=(60kg)(9.81m/s^{2})=588.6N[/tex]
and let's calculate the normal force:
[tex]N=m(g-a_{c})[/tex]
[tex]N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})[/tex]
N=586.58N
so now we can calculate the percentage change:
[tex]\%_{change} = \frac{mg-N}{mg}x100\%[/tex]
so we get:
[tex]\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%[/tex]
[tex]\%_{change}=0.343\%[/tex]
which is a really small change.
d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).
There, we can see that the radius can be found by using the cos function:
[tex]cos \theta = \frac{AS}{h}[/tex]
In this case:
[tex]cos \theta = \frac{r}{R_{E}}[/tex]
so we can solve for r, so we get:
[tex]r= R_{E}cos \theta[/tex]
in this case we'll use the average radius of earch which is 6,371 km, so we get:
[tex]r = (6371x10^{3}m)cos (44.4^{o})[/tex]
which yields:
r=4,551.91 km
and now we can calculate the acceleration at that point:
[tex]a=\omega ^{2}r[/tex]
[tex]a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m[/tex]
[tex]a=24.07 mm/s^{2}[/tex]
In this exercise we have to use the knowledge of mechanics to solve the magnitude of the acceleration and the acceleration of gravity, in this way we find that:
a)[tex]a=33.73 mm/s^2[/tex]
b)[tex]mg>N[/tex]
c) [tex]Change=0.343%[/tex]
d) [tex]a=24.07 mm/s^2[/tex]
Then calculating from the information given in the text;
a)In this case, we know the centripetal acceleration is given by the following formula:
[tex]a_c=w^2r\\w=\frac{2\pi}{T} \\T=24*\frac{3600}{60} = 86400 s[/tex]
so we can now find the angular speed:
[tex]w=\frac{2\pi}{86400} \\w=72.72*10^{-6} rad/s^2[/tex]
So the centripetal acceleration will be:
[tex]a_c=(72.72*10^{-6}rad/s^2)^2(6478*10^3m)\\a_c=33.73mm/s^2[/tex]
b)So we can do a sum of forces in equilibrium:
[tex]\sum F=o\\N-mg+ma_c=0\\N-mg+ma_c=0\\N=mg-ma_c\\mg>mg-ma_c\\mg>N[/tex]
This exist cause the centripetal increasing speed happen pulling united states of america upwards, that will create the size of the normal force tinier than the result or goods created of the bulk times the increasing speed of importance.
c) So let's calculate our weight and normal force:
[tex]mg=(60)(9.81)=588.6N\\N=m(g-a_c)\\N=(60)(9.81-33.73*10^{-3})\\N=586.58 N[/tex]
So now we can calculate the percentage change:
[tex]\%change= \frac{mg-N}{mg}*100\% \\=\frac{588.6-586.58}{588.6}*100\% \\=0.343\%[/tex]
d) There, we can see that the radius can be found by using the cos function:
[tex]cos\theta=\Delta S/h\\cos\theta=r/R_E\\r=R_E cos\theta\\r=(6371*10^3)cos(44.4)\\r=4,551.91 km[/tex]
And now we can calculate the acceleration at that point:
[tex]a=w^2r\\a=(72.72*10^{-6})^2(4,551.91*10^3)\\a=24.07 mm/s^2[/tex]
See more about acceleration at brainly.com/question/2437624
Related Questions
Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she arrives at the mark in 20min?
Answers
I got this by multiplying 7.25(3) because I know 20 minutes is 1/3 of 1 he
At the surface of a certain planet, the gravitational acceleration g has a magnitude of 20.0 m/s^2. A 22.0-kg brass ball is transported to this planet.
What is (a) the mass of the brass ball on the Earth and on the planet, and (b) the weight of the brass ball on the Earth and on the planet?
Answers
Answer:
a, 22 kg and 22 kg
b, 215.8 N and 440 N
Explanation:
a
The mass of the ball remains constant and unchanged irrespective of where it has been to, need to go or is going. So, basically the mass of the ball on earth is as the same mass of the ball on the said planet, 22 kg
b
The weight of any object factors in the acceleration due to gravity of the said area(or planet).
W = mg, with m being the mass and g being the acceleration due to gravity.
On earth
W = 22 * 9.81 = 215.8 N
On the said planet,
W = 22 * 20 = 440 N
Do therefore, the weight is 215.8 N on earth and 440 N on the planet
Homeostasis refers to the ability of the body to maintain a stable internal environment despite changes in external conditions.
True
False
Answers
Answer:
True
Explanation:
The correct answer is 40 or - 40??
Answers
Answer:
40 cm
hope it's help you but you can choose if your ans. is -40 cm
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.
Answers
Answer:
The unknown force will be 18.116 lb.
Explanation:
Given that,
Three forces act on a flange as shown in figure.
The net force acting on the flange is a minimum.
[tex]\dfrac{dF_{net}}{df}=0[/tex]
We need to calculate the unknown force
Using formula of net force
[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]
Put the value into the formula
[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]
[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]
The magnitude of net force,
[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]
[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]
[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]
[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]
On differentiating w.r.to F
[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]
[tex]0=2F+36.232[/tex]
[tex]F=-\dfrac{36.232}{2}[/tex]
[tex]F=-18.116\ lb[/tex]
Negative sign shows the direction of force which is downward.
Hence, The unknown force will be 18.116 lb.
Following are the solution to the given question:
Calculating the Net force on flange:
[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]
Calculating the magnitude:
[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]
Differentiate the value
[tex]\to \frac{F_{net}}{df}=0 [/tex]
[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]
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is c20h12 organic or not organic
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When the torques on the wheel and axle equal zero, the machine has
_____
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Answer:
8
Explanation:
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rotational equilibrium
Explanation:
A north magnetic pole brought near a south magnetic pole will...
A)
cancel the south pole.
B)
attract the south pole.
C)
repel the south pole.
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The density of a solid or liquid material divided by the density of water is called
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Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
What does a light-year measure? A. volume B. weight C. circumference D. distance
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Answer:
Distance. A light-year is the distance a single photon of light can travel in empty space in a year.
Answer:
D-Distance
Explanation:
pls help, i have no idea where to start
Answers
Answer:
Use equation 1/2 * m * v²
Explanation:
where m = mass in kg
and v = velocity in m/s
plug accordingly.
What height would a 4 kg book need to be to have a potential energy of
235.2 J on earth?*
need help!!
Answers
Answer:
5.99 m = 6 m
Explanation:
PE = m*g*h
235.2 J = (4 kg)(9.81 m/s^2)(h)
h = (235.2 J)/(9.81*4)
h = 5.99 m
h = 6 m
1. When you talk into your paper cup telephone, the person on the other end can feel the bottom
of their telephone vibrate. Why do you think this happens?
Answers
Answer:
The correct answer is - the sound waves make vibration that travels through the string.
Explanation:
When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.
The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.
You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them
what is the force of gravitational attraction between an object with a mass of 0.5kg and another object that had a mass of 0.33kg and a distance between them of 0.002m
Answers
Answer:
1.1x10^-4 N
Explanation:
F = G[(m_1*m_2)/r^2]
G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)
m_1 = mass 1
m_2 = mass 2
r = radius between objects
F = G[(0.5kg*0.33kg)/(0.001m)^2]
F = 1.1x10^-4 N
Problem 1. A tugboat exerts a constant force of 4000N toward the right on a ship, moving it
a distance of 15m. What work is done?
Answers
Answer:
75,000 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 5000 × 15
We have the final answer as
75,000 JHope this helps you
1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answers
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
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A 5000kg truck and a 50 kg person have the same momentum. How is this possible?
Answers
Explanation:
The mass of a truck = 5000 kg
Mass of a person = 50 kg
The momentum of a body is given by :
p = mv
m is mass
If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.
[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]
So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.
Can yall help me please
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Answer:
physical
Explanation:
it is a real life model that was built
The x-axis of a position-time graph represents
Answers
Answer:
The y-axis represents position relative to the starting point, and the x-axis represents time.
Explanation:
Explain why crinoid fossils can be found on the shore of Lake Michigan, which is a fresh water lake.
Answers
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. Show all work including units.
A horizontal force of 15 N is applied to the object.
a. Draw a free body diagram with the forces to scale.
b. Determine the force of friction.
c. Determine the acceleration of the object.
Answers
When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,
∑ F = n - w = 0 → n = w = mg = 112.5 N ≈ 113 N
where m = 12.5 kg and g = 9.80 m/s².
The minimum force F needed to overcome maximum static friction f and get the object moving is
F > f = 0.50 n = 61.25 N ≈ 61.3 N
which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.
Answers
Answer:
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Explanation:
Given
[tex]Height (h) = 1.70m[/tex]
Required
Determine the difference in the blood pressure from feet to top
This is calculated using Pascal's second law.
The second law is represented as:
[tex]P_2 = P_1 + pgd[/tex]
Subtract P1 from both sides
[tex]P_2 - P_1 = pgd[/tex]
Where
[tex]p = blood\ density = 1.06 * 10^3kg/m^3[/tex]
[tex]g = acceleration\ of\ gravity = 9.8N/kg[/tex]
[tex]d =height = 1.70m[/tex]
P2 - P1 = Blood Pressure Difference
So, the expression becomes:
[tex]P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70[/tex]
[tex]P_2 - P_1 = 17659.6Pa[/tex]
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Hence, the difference in blood pressure is approximately [tex]1.8 * 10^4\ Pa[/tex]
Which of the following best describes the charge of the nucleus of an atom?
A. The nucleus can have a positive, neutral, or negative charge.
B. The nucleus always has a positive charge.
c. Jahe nucleus always has a neutral charge.
D. The nucleus always has a negative charge.
SUBMIT
Answers
Example
A 1.8 m tallman stand in an elevator accelerating upward at 12 m/s?,
what is the blood pressure in the brain and foot.
Take the height difference between the heart and the brain to be
0.35 m?
13.3 x 103 Pa] & [Pblood = 1060 kg.m-3]
Note: [: P Heart
Answers
Answer:
Explanation:
From the given information; the diagram below shows a clearer understanding.
The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]
= 13300 - 1060 (12-9.81) 0.35
= 13300 - 1060 (2.19) 0.35
= 13300 - 812.49
= 12487.51 Pa
The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]
= 13300 + 1060 (12 + 9.81) 1.45
= 13300 + 1060( 21.81 ) 1.45
= 13300 + 33521.97
= 46821.97 Pa
Answer:
The blood pressure in the brain = [tex]12487.51 pa[/tex]The blood pressure in the feet = [tex]46821.97pa[/tex]Explanation:
[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]
[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]
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1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Answers
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.
What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth
Answers
Answer:
c
Explanation:
I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST
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Answer:
u will need 1/2 cells
Explanation:
What is the net force?
60N left
4ON, right
60N, right
0N
Answers
Answer:
60n left
Explanation:
Do
Which waves are electromagnetic and can travel through a vacuum?
Light and heat waves
Longitudinal and transverse waves
Sound waves
Surface waves
Answers
Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.
Answers
Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]