Answer:
It can easily be transformed from high voltages to low voltages
How does the periodic table organize atoms of elements with the same number of valence electrons?
1.in cells
2.in columns
3.in diagonals
4.in rows
Answer:
2.in columns
Explanation:
The elements of the periodic table is distribute as in 7 periods and 18 groups. The element is distribute on base on increasing atomic number. The period is called horizontal row and the group is called vertical column. in periodic number of sphere around the nucleus remain the same, but as we move from left to right, the valence electrons continue to increase.so correct option is 2.in columns
Sonia was experimenting with electric charges. She tied two inflated balloons together, held them next to each other, and rubbed both with a piece of wool.
What did Sonia observe, and why?
Answer: They will repel each other.
Explanation:
Two inflated balloons when rubbed with woolen cloth will lead to repeal each other because of the similar charges on both the balloons.
Rubbing both the balloons together by the woolen cloth will introduce negative charge in the balloons.
As, we know that the same charges repeal each other both of the balloons with be apart from each other.
This is due to the static electricity, the negatively charged particles jump to the positive one. When balloons are rubbed they become negatively charged.
Answer:
Sample Response: Sonia observed that the two balloons repelled each other. This is because both balloons acquired the same charge when she rubbed them with the piece of wool, and like charges repel each other.
Explanation:
Did it on Egde 2020
Explain why you cannot charge one end of a steel rod and leave the other end uncharged
Answer:
Electrons can be made to move from one object to another. However, protons do not move because they are tightly bound in the nuclei of atoms.Static charge occurs when electrons build up on an object. Static charge:
can only build up on objects which are insulators, eg plastic or wood
cannot build up on objects that act as conductors, eg metals
Conductors allow the electrons to flow away, forming an electric current.
When a static charge on an object is discharged, an electric current flows through the air. This can cause sparks. Lightning is an example of a large amount of static charge being discharged.
Explanation:
I Science Sem 1
1 7.1.2 Exam: Semester 1 Exam
Question 18 of 30
3 Points
How much force is required in Newtons) to accelerate a 12-kg bicycle, along
with its 50-kg rider, at 2 m/s2?
Give your answer as a number.
Answer here
Answer:
Force, F = 124 N
Explanation:
We have,
Mass of bicycle is 12 kg and mass of rider is 50 kg
Total mass of the system is 12 kg + 50 kg = 62 kg
Acceleration of the system is 2 m/s²
It is required to find the force required to accelerate the system. The force acting on an object is given by :
[tex]F=ma\\\\F=62\ kg\times 2\ m/s^2\\\\F=124\ N[/tex]
So, the force of 124 N is acting on the system.
4. What is anti node?
Answer:
The position of maximum displacement in a standing wave system.
Why will a change in velocity have a greater affect on KE than a change in mass?
Answer: the answer is corect of what everone else said
Explanation:
Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equationF=Gm1m2r2where F is the magnitude of the gravitational attraction on either body, m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units of force are kgâ‹…m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G must be which of the following?Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes the moon to orbit the earth. The force of gravitational attraction is represented by the equation,where is the magnitude of the gravitational attraction on either body, and are the masses of the bodies, is the distance between them, and is the gravitational constant. In SI units, the units of force are , the units of mass are , and the units of distance are . For this equation to have consistent units, the units of must be which of the following?A. kg3mâ‹…s2B. kgâ‹…s2m3C. m3kgâ‹…s2D. mkgâ‹…s2
Answer:
The S.I unit of G must be m³/kg.s² to keep the equation consistent.
Explanation:
We have the equation:
F = Gm₁m₂/r²
where,
F = Gravitational Force between Two Bodies
G = Gravitational Constant
m₁ = Mass of 1st Body
m₂ = Mass of 2nd body
r = Distance between the Bodies
The S.I Units of the quantities are:
F = Newton = kg.m/s²
m₁ = kg
m₂ = kg
r = meter = m
G = ?
Therefore, to find the units of Gravitational Constant (G), we substitute the known units in the formula:
kg.m/s² = G(kg)(kg)/m²
G = m³/kg.s²
In this example we will see how a transformer can be used to change the voltage output of a wall receptacle. A friend returns from Europe with a device that she claims to be the world's greatest coffee maker. Unfortunately, it was designed to operate from a 240 V line, standard in Europe. At this rms voltage, the coffee draws 1500 W of power. (a) If your friend wants to operate the coffee maker in the United States, where the rms line voltage is 120 V, what turns ratio does her transformer need to have for the coffee maker to work
Answer:
Explanation:
In order to use a device rated for use in 240 volt for use in 120 V , we have to use a step down transformer . In step down transformer no of turn in secondary coil is less than that in primary coil .
For transformer the relation is as follows
V₂ / V₁ = N₂ / N₁ = turn ratio
Where V₂ and V₁ are volts in secondary and primary coil and N₂ and N₁ is no of turns in secondary and primary coil.
Here V₂ = 120V , V₁ = 240 V
turn ratio = N₂ / N₁ = V₂ / V₁ = 120 / 240 = .5
turn ratio = .5 .
ASAP pls answer right and I will mark brainiest
Answer: question one: i cannot tell Question 2: waxing crescent Question 3: waxing gibbous Question 4: B
Explanation
What is the linear diameter (in meters) of an object that has an angular diameter of 110 arcseconds and a distance of 25,000 m?
Answer:
Explanation:
angle θ = 110 arcsec
= 110 /60 arcmin
= 1.8333 arcmin
= 1.8333 / 60 degree
= .030555 degree
= .030555 x π / 180 radian
= 5.33 x 10⁻⁴ radian .
linear diameter = angular diameter in radian x distance
= 5.33 x 10⁻⁴ x 25000
= 13.325 m
linear diameter of object = 13.325 m .
Fill in the word that completes each statement.
Current cannot flow through a circuit when the switch is
Current flows through a circuit and all of the bulbs are lit only when the switch is
Answer:
Current cannot flow through a circuit when the switch is OPENCurrent flows through a circuit and all of the bulbs are lit only when the Switch is CLOSEExplanation:
In a series connected electrical circuit as also seen in all electrical Circuit the switch allows the flow of Electricity through out the circuit therefore causing the lighting of bulbs and also other electrical appliances to receive electricity.
When the switch is Open the flow of electricity in the circuit is disrupted while when the Switch is closed it allows the flow of electricity seamlessly through out the circuit. there are different types of switches but the most important is the MAIN SWITCH because it is used to shutoff electricity completely from the entire circuit during emergencies or repairs.
Answer:
Current cannot flow through a circuit when the switch is OPEN
Current flows through a circuit and all of the bulbs are lit only when the switch is CLOSED
Explanation:
What do you think are the key consideration for machining parts and joints when designing self-assembly furniture like what you see in Ikea?
Answer:
then, that IKEA would be the authority on flat-pack ... Hardware for the Hopen Bed, via Swedish Furniture Parts.
Explanation:
A string, fixed at both ends and 2.14 m long, is vibrating in its second harmonic. It excites a 0.5-m long pipe closed on one end into its third harmonic (first overtone). Use a speed of sound in air of 345 m/s if needed. (a) Sketch the normal modes given above for both the string and the closed-open pipe. [8] (b) What is the wave speed on the string? [7] (c) Write down the wave function y(x,t) for the 2nd harmonic wave on the string. Assume an arbitrary amplitude. [10] (d) If the mass density of the string is 0.55 g/m, what is the tension in the string? [5]
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Throughout the reflection, make sure you have a copy of the Student Guide, PDF or Word, and your data tables. Use the drop-down menus to complete the statements.
This activity was a multistep laboratory. Think about each step and identify the variables.
In all three parts, the dependent variable was .
The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was .
In Part II involving the inflated and deflated footballs, the independent variable was .
In Part III involving the baseball, the independent variable was .
Answer:
-In all three parts, the dependent variable was density.
-The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was mass.
-In Part II involving the inflated and deflated footballs, the independent variable was volume.
-In Part III involving the baseball, the independent variable was material.
Explanation:
correct on edge! hope it helps :)
Answers:
in all three parts, the dependent variable was. density
The independent variable is the one that the scientist intentionally manipulates. In Part I involving the golf ball and table tennis ball, the independent variable was. mass
In Part II involving the inflated and deflated footballs, the independent variable was. volume
In Part III involving the baseball, the independent variable was .material
Explanation:
Some students conduct an experiment to find out how long different rubber balls would keep bouncing before they stop, when they dropped from a fixed height. which unit would be the most suitable for measuring the results of their experiment?
A. hours
B. kilograms
C. metres
D. seconds
TIME REMAINING
58:50
Shelly tells a friend that her computer needs to be fixed because it has been producing a lot of heat and is smoking. Where
is the error in her statement?
• Heat cannot be described as having been produced.
Thermal energy cannot be produced by electronics.
Her computer should be producing a lot of heat and does not need to be fixed.
Her computer is producing thermal energy, not heat.
Mark this and return
Save and Exit
Next
Submit
Answer:
Her computer is producing thermal energy, not heat.
Explanation:
Answer:
Her computer is producing thermal energy,not heat
Explanation:
A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can reach.
Answer:
h = 16.9 m
Explanation:
When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:
Kinetic Energy Lost by Ball = Potential Energy Gained by Ball
(0.5)m(Vf² - Vi²) = mgh
h = (0.5)(Vf² - Vi²)/g
where,
Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)
Vi = Initial Speed of Ball = 18.2 m/s
g = acceleration due to gravity = - 9.8 m/s² ( negative for upward motion)
h = maximum height the ball can reach = ?
Therefore, using values in the equation, we get:
h = (0.5)[(0 m/s)² - (18.2 m/s)²]/(-9.8 m/s²)
h = 16.9 m
Humans living in highly populated areas are more inclined to
a. violence
b. socially isolate
C.appetite loss
d. all of the above
Answer:
they're more inclined to be violent so A
It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is 0.0154 (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastic having a density such that each square meter has a mass of 4.60 . We inflate them with a very light gas whose mass we can neglect. So far I found the following: What should be the radius of these balloons so they just hover above the surface of Mars? Radius of the balloon = /896 m What should be the mass of these balloons so they just hover above the surface of Mars? Mass of balloon = 4.64*10^-2 kg If we released one of the balloons from part A on earth, where the atmospheric density is 1.20 , what would be its initial acceleration assuming it was the same size as on Mars? If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
Answer:
the radius of the balloon r = 0.896 m and mass m = 4.64 × 10⁻² kg
the initial acceleration is a = 753.47 m/s²
the an instrument package could they carry a required mass of 4.64 kg
Explanation:
a) What should be the radius of these balloons so they just hover above the surface of Mars?
Given that :
The density of the Martian atmosphere, ρ = 0.0154 kg/m³
The volume of the sphere, V = (4/3)πr³
The area of the sphere, A = 4πr²
The mass of the balloon is m = (4.60 g/m²)A
m = (4.60×10⁻³ kg/m²)(4πr²)
The formula for the buoyant force is expressed as :
F = ρVg
m×g = ρ×V×g
m = ρ×V
Now;
(4.60×10⁻³ kg/m²)(4πr²)= ρ(4/3)πr³
r = 3(4.60×10⁻³ kg/m²)/ ρ
r = 3(4.60×10⁻³ kg/m²)/ 0.0154 kg/m³
r = 0.896 m
Thus; the radius of the balloon r = 0.896 m
The mass of the balloon is (4.60×10⁻³ kg/m²)(4πr²)
m = (4.60×10⁻³ kg/m²)(4π×0.896²)
m = 4.64 × 10⁻² kg
b) what would be its initial acceleration assuming it was the same size as on Mars?
The density of the air on earth, ρ = 1.20 kg/m³
The volume of the balloon is V = (4/3)(π)(0.896 m)³
V = 3.01156 m³
Considering the net force acting on the balloon ; we have
ΣF = ρVg - mg = ma
However; making the initial acceleration a of the balloon the subject ; we have:
a = (ρVg - mg)/m
a = (1.20 kg/m³)(3.01156 m³)(9.8 m/s²) - (4.64×10⁻² kg)(9.8 m/s²)]/(4.64×10⁻² kg)
a = 753.47 m/s²
c) If on Mars these balloons have five times the radius found in part A, how heavy an instrument package could they carry?
The volume of the total system is V' = (4/3)π(5r)³
V' = (4/3)π(5)³(0.896 m)³
V' =376.446 m³
The mass of the total system is m = (4.60×10⁻³ kg/m²) (4π(5r)²)
m = [4.60×10⁻³ kg/m²][4π][25](0.896 m)²
m = 1.159587 kg
We can then say that the buoyant force is equals to the weight of the total mass (balloon+load) and is expressed as:
F = (m + m')g
ρV'g = (m + m')g
ρV' = (m + m')
Thus; the required mass m' is = ρV' - m
m' = ρV' - m
m' = (0.0154 kg/m³)(376.446 m³) - (1.159587 kg)
m' = 4.64 kg
Thus; the an instrument package could they carry a required mass of 4.64 kg
Which multiplier does the word kilo represent? A. 100 B. 1,000 C. `(1)/(100)` D. `(1)/(10)`
Answer:
A. 1,000
Explanation:
Kilo is the prefix for thousands, thus, the multiplier that kilo represents is 1,000
A. 1,000
Answer:
B. 1,000
Explanation:
plato
What is the kinetic energy of a 633.0-kg car moving at a speed of 11.2 m/s?
Answer:
39.7 kJ
Explanation:
The kinetic energy formula applies:
KE = (1/2)mv^2 = (1/2)(633.0 kg)(11.2 m/s)^2 = 39,701.76 J
The kinetic energy is about 39.7 kJ.
What an athlete do to increase his stability when he lands on the balance beam from a flip?
Answer:
A series of body movements in which a person leaps into the air and then rotates one or more times while airborne is referred to as a flip.
When an athlete flips and lands on the beam, to increase his or her balance, he or she must spread out their hands-on either sides.
In doing this, they redistribute their forces (centre of gravity) to gain equilibrium create inertia to dampen the kinetic energy during the flip.
Cheers!
In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their components. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors
A =(1,0,−3), B =(−2,5,1), and C =(3,1,1). Calculate the following, and express your answers as ordered triplets of values separated by commas.
a. A- B=______
b. B-C=_______
c. -A +B- C=______
d. 3A- 2C=______
e. -2A+3B- C=_____
f. 2A- 3(B-C)
Answer:
a)A-B =(1,0,-3) +(2,-5,-1) =(3, -5, -4);
b)B-C =(-2-3, 5-1, 1-1) = (-5,4,0);
c)-A +(B –C) = (-1,0,3) +(-5,4,0) =(-6,4,3);
d) =(3,0,-9) –(6,2,2) = (-3, -2, -11);
e) = (-2,0,6) +(-6,15,3) +(-3,-1,-1) =
d) = (2,0,-6) –(-5*3, 4*3, 0) =
Explanation:
1)C=k*A = (k*a1, k*a2, k*a3);
2)C=-A = -1*A;
3)C= A+B = (a1+b1, a2+b2, c2+c2);
4)C=A-B =A +(-B);
5)A+B=B+A;
6)A+B+C =(A+B)+C =A+(B+C
A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.1 kg · m2 . A student tosses a 1.5-kg mass with a speed of 2.7 m/s to the professor, who catches it at a distance of 0.40 m from the axis of rotation. What is the resulting angular speed of the professor and the stool? (Assume that when the professor catches the mass, their arm is extended along a line radially outward from the axis of rotation, and the velocity of the mass is perpendicular to that line.)
Answer:
[tex]\omega=0.37 [rad/s][/tex]
Explanation:
We can use the conservation of the angular momentum.
[tex]L=mvR[/tex]
[tex]I\omega=mvR[/tex]
Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.
So we will have:
[tex](I_{proffesor - stool}+mR^{2})\omega=mvR[/tex]
Now, we just need to solve it for ω.
[tex]\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}[/tex]
[tex]\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}[/tex]
[tex]\omega=0.37 [rad/s][/tex]
I hope it helps you!
An object sits at rest at some position to the left of the origin. Draw what you expect the position vs time, velocity vs time, and acceleration vs time graphs to look like.
Answer:
see below
Explanation:
The position graph will be a horizontal line (of constant position) located at the negative value representing the position to the left of the origin.
The velocity graph will be a horizontal line at 0, since the object is at rest.
The acceleration graph will be a horizontal line at 0, since the object's velocity is not changing.
Batman and Robin are attempting to escape that dastardly villain, the Joker, by hiding in a large pool of water (refractive index nwater = 1.333). The Joker stands gloating at the edge of the pool. (His makeup is watersoluble.) He holds a powerful laser weapon y1 = 1.49 m above the surface of the water and fires at an angle of θ1 = 27◦ to the horizontal. He hits the Boy Wonder squarely on the letter "R", which is located y2 = 3.77 m below the surface of the water. θ x y y 1 1 2 R J Batplastic surface Mirrored Surface water B How far (horizontal distance) is Robin from the edge of the pool? (Fear not, Batfans. The "R" is made of laser-reflective material.) Answer in units of m.
Answer:
x_total = 4.29m
Explanation:
To solve this exercise we must work in parts. Let's use the law of refraction to find the angle of the refracted ray and trigonometry to find the distances.
Let's start by looking for the angles that the laser refracts
n₁ sin θ₁ = n₂ sin θ₂
where n₁ is the air refraction compensation n₁ = 1, n₂ the water refractive index n₂ = 1,333
θ₂ = sin⁻¹ (n₁ sin θ₁/n₂)
θ₂ = sin⁻¹ (1 sin 27 / 1,333)
θ₂ = sin⁻¹ 0.34057
θ₂ = 19.9º
now let's find the distance from the edge of the pool to the point where the ₂lightning strikes the water
tan θ₁ = y₁ / x₁
x₁ = y₁ / tan θ₁
x₁ = 1.49 / tan 27
x₁ = 2,924 m
Now let's look for the waterfall in the water as far as Robin
tan θ₂₂ = y₂ / x₂
x₂ = y₂ / tan θ₂
x₂ = 3.77 / tan 19.9
x₂ = 1,364
the distance from the edge of the pool to Robin is
x_total = x₁ + x₂
x_total = 2,924 + 1,364
x_total = 4.29m
Which best explains why making a pancake from batter is an example of a chemical change?
-The pancake that forms is a different state of matter.
-The change from batter to pancake can be reversed.
-A new substance forms when the batter is cooked.
-The batter changes shape when it is cooked.
The correct answer is C. A new substance forms when the batter is cooked.
Explanation:
When a chemical change occurs the properties, and composition of substances change. This means atoms in the substance re-arrange to form a new substance. This only occurs when there is a chemical change, but not when physical changes occur, indeed a physical change only affects the state of the matter, shape, size, etc.
In the case of the pancake, this is an example of a chemical change because though the process of cooking the pancake changes its composition. Due to this, the properties of the cooked pancake, and the butter are not the same as a new substance forms. Also, in this and most chemical changes, reversibility is not possible, that is why you cannot reverse the process and make the cooked pancake batter once again.
Answer:
C. on edge
Explanation:
A lightbulb manufacturer makes bulbs with different "color temperatures," meaning that the spectrum of light they emit is similar to a blackbody with that temperature. Assuming the emitting areas of the filaments in two bulbs with color temperatures of 2,000 K and 4,000 K are the same, which of the two is the brighter?
Answer:
The bulb with higher temperature(4000 K) will be brighter
Explanation:
From the question we are told that
The color temperature for first bulb is [tex]T_1 = 2000K[/tex]
The color temperature for second bulb is [tex]T_2 = 4000K[/tex]
Generally the emission power of black body radiation is mathematically represented as
[tex]E = \sigma T^4[/tex]
Where [tex]\sigma[/tex] is the Stefan-Boltzmann constant with a value [tex]\sigma = 5.67 * 10^{-8} W m^{-2} K^{-4.}[/tex]
Now for [tex]T_1 = 2000K[/tex]
[tex]E_1 = 5.67*10^{-8} * (2000)^4[/tex]
[tex]E_1 = 907.2 \ W/m^2[/tex]
At [tex]T_2 = 4000K[/tex]
[tex]E_2 = 5.67*10^{-8} * 4000[/tex]
[tex]E_2 = 14515.2 \ KW/m^2[/tex]
Looking at the result we got we see that the emission power for the higher temperature bulb is higher, this means that its power to emit in the visible spectrum range would be higher
So the bulb with higher temperature will be brighter
what bo you nmean by ABC rule ?
A spring with a 6-kg mass and a damping constant 7 can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t with the general form c1eαtcos(βt)+c2eγtsin(δt)
Answer:
The value for [tex]c^2 - 4mk[/tex] is : [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]
Explanation:
The spring constant is :
[tex]F = kx \\ \\ k = \frac{F}{x} \\ \\ k = \frac{1.5}{0.5} \\ \\ k = 3 \ N/m[/tex]
The value for [tex]c^2 - 4mk[/tex] is :
= [tex]7^2 - 4(6)(3)[/tex]
= [tex]49 - 27[/tex]
= [tex]\mathbf{-23 \ m^2kg^2/sec^2}[/tex]
The differential equation for this system is :
[tex]m \frac{d^2x}{dt^2}+c \frac{dx}{dt}+kx = 0[/tex]
[tex]6 \frac{d^2x}{dt^2}+7\frac{dx}{dt}+3x = 0[/tex]
and the auxiliary equation for this differential equation is :
[tex]6m ^2+ 6m + 3 = 0[/tex]
using the quadratic formula :
[tex]\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}[/tex]
[tex]\frac{-6 \pm \sqrt{6^2 - 4(6)(3)} }{2(6)}[/tex]
= [tex]\frac{-6 \pm \sqrt{-36} }{12}[/tex]
= [tex]-\frac{1}{2} \pm \frac{\sqrt{36} }{12}i[/tex]
The general solution is :
[tex]x(t) = e^{-\frac{1}{2}t}}(c_1 cos {\frac{\sqrt{36} }{12}} t + c_2 sin \frac{\sqrt{36} }{12} t})[/tex]
Initial conditions : [tex]x(0) = 1 \ m , x' (0) = 0\\[/tex]
[tex]x(0) = ( c_1 cos 0 + c_2 sin 0)[/tex]
[tex]1 = c_1[/tex]
[tex]x't = e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin \frac{\sqrt{36} }{12}t + c_2 \frac{\sqrt{36} }{12} cos \frac{\sqrt{36} }{12}t) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_1cos \frac{\sqrt{36} }{12} t +c_2 sin \frac{\sqrt{36} }{12} t)[/tex]
[tex]x'(0)= e^{- \frac{1}{2}t}}}(- c_1 \frac{\sqrt{36} }{12} sin0 + c_2 \frac{\sqrt{36} }{12} cos 0) - e^{- \frac{1}{2}t}}} (\frac{1}{2}) (c_10+c_2 sin0)[/tex]
[tex]x'(0) = c_2 \frac{\sqrt{36} }{12}-c_1 \frac{1}{2}[/tex]
replacing 1 for [tex]c_1[/tex]
[tex]0 = c_2 \frac{\sqrt{36} }{12} -\frac{1}{2}[/tex]
[tex]c_2 \frac{\sqrt{36} }{12} = \frac{1}{2}[/tex]
[tex]c_2 = \frac{6}{\sqrt{36} }[/tex]
The position of the mass (m) after t seconds is:
[tex]\mathbf{x(t) = e^{\frac{1}{2}t }( cos \frac{\sqrt{36}}{12}t + \frac{6}{\sqrt{36}} sin \frac{\sqrt{36}}{12}t)}[/tex]