PLEASE ANSWER THIS I AM SO STUCK The increasing trend of energy requirements in the U.S.: a) is directly proportional to population growth b) is inversely proportional to population growth c) exceeds the rate of population growth
Answer:
It should be A
Explanation:
the amount of energy a population has should be based on how many people there are. if more people come into the population, the more energy it should have. if people leave the population, the less energy it should have.
A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0.15 s
Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
Magnetic field lines exit out of the . Magnetic field lines enter into the . Magnetic field lines travel around a bar magnet in
Answer:
Magnetic field lines exit out of the North pole . Magnetic field lines enter into the South pole. Magnetic field lines travel around a bar magnet in closed loops.
Explanation:
Magnetic field lines shows the direction of a magnetic force and how it acts, it gives the direction of the magnetic field at that point in time.
For a bar magnetic, the magnetic field lines runs from the north pole to the south pole, i.e. it exits the north pole and enters into the south pole. This magnetic field lines also go through the magnet forming closed loops without ends.
Answer:
Magnetic field lines exit out of the
✔ north pole
.
Magnetic field lines enter into the
✔ south pole
.
Magnetic field lines travel around a bar magnet in
✔ a closed loop
.
Explanation:
a child kicks a soccer ball across the grounds the ball travels 10.0 meters in 0.56 seconds. what is the speed of the ball
What kind of substance can you pour from one container into another without a change in volume
Answer:
Liquids
Explanation:
Liquids take up the shape of the container it is poured into but will never change its volume.
Will give brainliest for the person who answers first
A 75W television is on for 400s, how much energy is transferred, leave your answer in Joules
Answer:
30,000 J
Explanation:
Energy (Joules) = Power (Watt) × Time (seconds)
Energy (J) = 75 × 400
Energy (J) = 30,000
30,000 Joules of energy will be transferred.
A pendulum at position A is released and swings through position B to position Con the other side.
B
1. Describe the total mechanical energy at each of the following positions. (3)
A.
B.
C
Explanation:
Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),
the conditions can be summarized bellow
point A
initial velocity= 0
final velocity=0
P.E= Max
K.E= 0
point B
initial velocity= maximum
final velocity=maximum
P.E=K.E
point C
initial velocity= min
final velocity=min
P.E= 0
K.E= max
An object moves along a straight line path from P to Q under the action of a force (4 3 3 ) N. I j k − + If the coordinates of P and Q, in metres, are (3, 3, -1) and (2, -1, 4) respectively, then the work done by the force is
Answer:
1 Nm
Explanation:
Given;
Force = F = (4, 3, 3)N
Position 1 = P = (3, 3, -1)m
Position 2 = Q = (2, -1, 4)m
The object moves along a straight line path from P to Q, therefore, the distance vector (d) is given by;
d = Q - P
d = (3, 3, -1) - (2, -1, 4)
d = (1, 4, -5)m
Now the work done (W) by the force (F) to move through the distance (d) is the dot product of the two vectors: F and d. i.e
W = F . d
For clarity, let's write vectors F and d in vector unit notation as follows;
F = 4 i + 3 j + 3 k
d = 1 i + 4 j - 5k
Therefore,
W = (4 i + 3 j + 3 k ) . (1 i + 4 j - 5k)
W = (4 + 12 - 15)
W = 1
Therefore, the workdone by the force is 1 Nm
How many atoms of oxygen in the chemical formula 2 Ca(ClO2)2?
Answer:
8
Explanation:
Ca(ClO2)2 - 2*2 = 4 Oxygen atoms
2 Ca(ClO2)2 - 2*4 = 8 Oxygen atoms
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
[tex]C=2\pi r[/tex] (1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
[tex]r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m[/tex]
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
[tex]L=Lo[1+\alpha \Delta T][/tex] (2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
[tex]L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m[/tex]
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
[tex]r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m[/tex]
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m
Explanation:
Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las personas junto a él respiran para calentarlo con su aliento y aumentar su temperatura 1 grado Celsius. El tubo se hace más largo. También ya no queda ajustado. ¿A qué distancia sube sobre sobre el nivel del suelo? (solo tomar en cuenta la expansión radial al centro de la tierra, y aplicar la fórmula geométrica que relaciona la circunferencia C con el radio r: C= 2πr).
Un contenedor de 1800 N está en reposo sobre un plano inclinado a 28°, el coeficiente de fricción entre el contenedor y el plano es de 0.4. Calcule la fuerza P paralela al plano y dirigida hacia arriba de éste que hará que el contenedor se mueva hacia arriba con una velocidad constante.
Answer:
F = 1480.77N
Explanation:
In order to calculate the required force to push the container with a constant velocity, you take into account the the sum of force on the container is equal to zero. Furthermore, you have for an incline the following sum of forces:
[tex]F-Wsin\alpha-F_r=0\\\\F-Wsin\alpha-N\mu cos\alpha=0\\\\F-Wsin\alpha-W\mu cos\alpha=0[/tex] (1)
F: required force = ?
W: weight of the container = 1800N
N: normal force = weigth
α: angle of the incline = 28°
g: gravitational acceleration = 9.8m/s^2
μ: coefficient of friction = 0.4
You solve the equation (1) for F and replace the values of the other parameters:
[tex]F=W(sin\alpha+\mu cos\alpha)\\\\F=(1800N)(sin28\°+(0.4)cos28\°)=1480.77N[/tex]
The required force to push the container for the incline with a constant velocity is 1480.77N
state the observations made in this experiment
Answer:
the water is going to boil and the mercury ill melt and shoot the cork out the bottom of the tube
Explanation:
Two identical resistors are connected in parallel across a 26-V battery, which supplies them with a total power of 7.1 W. While the battery is still connected, one of the resistors is heated so that its resistance doubles. The resistance of the other resistor remains unchanged. Find (a) the initial resistance of each resistor, and (b) the total power delivered to the resistors after one resistor has been heated.
Answer:
A) R = 190.42 Ω
B) P = 5.325 W
Explanation:
We are given;
Total power;P_tot = 7.1 W
Voltage;V = 26 V
A)We are told that while the battery is still connected, one of the resistors is heated, so that its resistance doubles.
Thus, the power is doubled.
Now, formula for power is;
P = IV
Thus, since power is doubled, we have;
P = 2(IV)
Now, formula for current is; I = V/R
So, P = 2V²/R
Making R the subject, we have;
R = 2V²/P
In this question, P is p_total = 7.1 W
Thus;
R = (2 × 26²)/7.1
R = 190.42 Ω
B) Now, the resistance of the resistors are R and 2R.
Formula for power in this context is;
P = V²/R
Thus,
Total power delivered to the resistors is;
P = V²/R + V²/2R
P = 3V²/2R
P = (3 × 26²)/(2 × 190.42)
P = 5.325 W
What did classical physics predict would happen to the light given off by an
object as its temperature increased?
A. The wavelength of the light would decrease from visible light into
the ultraviolet range.
B. The color of the light would fade to black as the brightness
decreased.
C. The color of the light would change to white, and the brightness
would decrease.
D. The wavelength of the light would decrease from visible light into
the infrared range.
Answer:
The Wavelength of the light would decrease from visible light into ultraviolet range A P E X
Explanation:
Classical physics predict that as the temperature is increased, the wavelength of the light would decrease from visible light into more energetic ultraviolet range.
What is classical physics?When an electric charge vibrates, an electromagnetic wave is produced. Remember that heat is simply the kinetic energy of random motion. In a hot object, electrons vibrate in random directions and produce light as a result.
A hotter object produces more energetic vibrations, and thus emits more light —- it glows brighter. So far, so good. However, classical physics was unable to explain the shape of the blackbody spectrum.
As per the classical physics concepts, as the temperature increases, the wavelength of light becomes shorter and shifts to higher energy regions. Therefore, option A is correct.
To find more on blackbody radiation, refer here:
https://brainly.com/question/16749080
#SPJ5
What can happen if a body moves through speed of light
As waves travel into the denser medium, they slow down and wavelength decreases.
Explanation:
The denser the medium the slower the waves (speed of light) travels.
◦•●◉✿When the body approaches the speed of light, the body's length appears to contract in the direction of travel, and its mass appears to increase from the point of view of a stationary observer. Only photons move to light velocity. They don´t have length.✿◉●•◦
A wire carries a steady current of 2.80 A. A straight section of the wire is 0.750 m long and lies along the x axis within a uniform magnetic field, = 1.50 T. If the current is in the positive x direction, what is the magnetic force on the section of wire?
Answer:
The magnetic force in the wire is 3.15NExplanation:
Given
current I= 2.80 A.
length of conductor L= 0.75 m
Magnetic field, B = 1.50 T
∅=90
according to Fleming's left hand rule the conductor will observe a force perpendicular to it
Applying the formula
[tex]F= BIL* sin(90)[/tex]
[tex]F=1.50* 2.80* 0.750* sin(90)\\\F= 3.15N[/tex]
Note: sine(90)= 1
Un depósito de gran superficie se llena de agua hasta una altura de 0,3 m. En el fondo del depósito hay un orificio de 5 cm2 de sección por el que sale el agua con un chorro continuo. A) ¿Qué cantidad de líquido saldrá del depósito expresada en m3/s?
Answer:
a) El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
Explanation:
a) Asúmase que el tanque se encuentra a presión atmósferica y que la sima del tanque tiene una altura de 0 metros. La rapidez de salida del chorro del depósito se determined a partir del Principio de Bernoulli, cuya línea de corriente entre la cima y la sima del tanque queda descrita por la siguiente ecuación:
[tex]\Delta z = \frac{v_{out}^{2}}{2\cdot g}[/tex]
Donde:
[tex]\Delta z[/tex] - Diferencia de altura, medida en metros.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
Se despeja la rapidez de salida del chorro:
[tex]v_{out} = \sqrt{2\cdot g \cdot \Delta z}[/tex]
Si [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] y [tex]\Delta z = 0.3\,m[/tex], entonces la rapidez de salida del chorro es:
[tex]v_{out} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.3\,m)}[/tex]
[tex]v_{out} \approx 2.426\,\frac{m}{s}[/tex]
Ahora, la cantidad de líquido que sale del depósito por unidad de tiempo se obtiene al multiplicar la rapidez de salida del chorro por el área transversal del orificio. Esto es:
[tex]\dot V_{out} = v_{out}\cdot A_{t}[/tex]
Donde:
[tex]v_{out}[/tex] - Rapidez de salida del chorro, medida en metros por segundo.
[tex]A_{t}[/tex] - Área transversal del orificio, medido en metros cuadrados.
[tex]\dot V_{out}[/tex] - Caudal de salida del chorro, medido en metros cúbicos por segundo.
Dado que [tex]v_{out} = 2.426\,\frac{m}{s}[/tex] y [tex]A_{t} = 5\,cm^{2}[/tex], el caudal de salida del chorro es:
[tex]\dot V_{out} = \left(2.426\,\frac{m}{s} \right)\cdot (5\,cm^{2})\cdot \left(\frac{1}{10000}\,\frac{m^{2}}{cm^{2}} \right)[/tex]
[tex]\dot V_{out} = 1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex]
El caudal de salida del chorro es [tex]1.213\times 10^{-3}\,\frac{m^{3}}{s}[/tex].
Which types of materials have a molecular structure that makes them good conductors? Explain how understanding this relationship can help people design more efficient technologies
Answer:
Most conductors are made from elements metallic elements
Metallic elements are those elements grouped as alkali metals, alkaline earth metals, basic metals, rare earth elements, actinides, and basic metals
Conductors of electricity carry electricity by means of the movable charged particles present in the conducting material
The movable charged particle are the electrons which are most mobile in metals because of their crystalline structure and available valence electron, which are the electrons in the outermost orbit of an atom and hence freest to move about within the material mass
As such the availability of free electrons determine conductivity of materials
Explanation:
Understanding the relationship between electrical conductivity and the structure of atoms and molecules will contribute to the design of efficiency and reliability of technologies deployed to remote locations such as the Moon, by ensuring the best possible output from the electric control inputs and the redesigning of existing electrical installations using efficient amount of manufacturing materials, thereby saving the environment and costs.
Free electo
Electric conductors possess movable electrically charged particles, referred to as "electrons" in metals. When an electric charge is applied to a metal at certain points, the electrons will move and allow electricity to pass through. Materials with high electron mobility are good conductors and materials with low electron mobility are not good conductors, instead referred to as "insulators."
Silver is the best conductor of electricity because it contains a higher number of movable atoms (free electrons). For a material to be a good conductor, the electricity passed through it must be able to move the electrons; the more free electrons in a metal, the greater its conductivity. However, silver is more expensive than other materials and is not normally used unless it is required for specialized equipment like satellites or circuit boards. Copper is less conductive than silver but is cheaper and commonly used as an effective conductor in household appliances. Most wires are copper-plated and electromagnet cores are normally wrapped with copper wire. Copper is also easy to solder and wrap into wires, so it is often used when a large amount of conductive material is required.
Una carga de 4 uC penetra perpendicularmente en un campo magnetico de 0.4 T con una velocidad de 7.5x10 4 m/s. Calcular la fuerza que recibe la carga
Answer:
F_B = 0.12N
Explanation:
In order to calculate the magnetic force on the charge, you use the following formula:
[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex] (1)
q: charge of the particle = 4μC = 4*10^-6 C
v: speed of the charge = 7.5*10^4 m/s
B: magnitude of the magnetic field = 0.4T
The direction of the motion of the charge is perpendicular to the direction of the magnetic field. Then, the magnitude of the magnetic force is:
[tex]F_B=qvBsin90\°\\\\F_B=(4*10^{-6}C)(7.5*10^4 m/s)(0.4T)=0.12N[/tex]
The magnetic force on the charge is 0.12N
What type of image does a concave lens form? A. real image B. magnified image C. virtual image D. reverse image
The skier starts from rest. The total distance travelled by the skier during the descent is 2800 m. The average resistive force on the skier is 220 N. Calculate the work done against the resistive force
Answer:
Explanation:
Force equal to resistive force will be applied for movement . So force applied
F = 220 N .
displacement = 2800 m
work done against resistive force
= force x displacement
= 220 x 2800 J
= 6.16 x 10⁵ J .
What is the best explanation of work?
Answer:
Work done by the force acting on a body is defined as the product of force and displacement of the body in the direction of the force. It is a scalar quantity. Force acting on the body must produce a displacement for the work is to be done by the force. Thus, for the work to be done, the following conditions must be fulfilled:
A force must be applied and the applied force must produce a displacement in any direction except perpendicular to the direction of the force.
Mathematically,
[tex]work = force \times \: displacement[/tex]
( in the direction of force)
i.e W= F•D
The SI unit of force is Newton (N) and that of displacement of a metre (M). So the unit of work is Newton metre(Nm) which is joule (J).
Thus, one joule work is said to be done when one Newton force displaces a body through one metre in its own direction.
Hope this helps...
Good luck on your assignment..
On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.
Answer:
ω = √(2T / (mL))
Explanation:
(a) Draw a free body diagram of the mass. There are two tension forces, one pulling down and left, the other pulling down and right.
The x-components of the tension forces cancel each other out, so the net force is in the y direction:
∑F = -2T sin θ, where θ is the angle from the horizontal.
For small angles, sin θ ≈ tan θ.
∑F = -2T tan θ
∑F = -2T (Δy / L)
(b) For a spring, the restoring force is F = -kx, and the frequency is ω = √(k/m). (This is derived by solving a second order differential equation.)
In this case, k = 2T/L, so the frequency is:
ω = √((2T/L) / m)
ω = √(2T / (mL))
10 advantages of friction
Answer:
1. it helps to change the direction.
2. it helps us to walk on ground.
3. it helps the vechils to break while moving.
4. helps in changing one form of enegry to another form. eg when we rub our hands we feel heat energy.
5. it opposites the force.
6. it helps us to change shape of objects.eg we roll the dough to make it roti.
7. it changes the state of body from rest motion.eg when we push any obj from inclined plane it moves.
i all know is just 7..
Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry, yet he is able to move the bricks. How is this possible?
Answer is given below
Explanation:
This is happen because here when Rick walks with full loaded wheelbarriow of bricks, he able to move it because Rick lifts the wheelbarrow handle So, most of the weight of full loaded wheelbarrow's load goes on that's wheel and due to friction force between wheel and surface it can easy to move He uses force to rotate the wheel, much more than the force applied to the rim of the wheel on the axis of rotation or torqueAnswer part (d) please
Answer:
MARK me brainliest please and follow my page
Explanation:
All you have to do to get the average speed is to calculate the total distance covered and divide it by the total time taken
= 16/18 = 0.88m/s
Average speed = (distance covered) / (time to cover the distance)
For the full 18 seconds described by the graph . . .
Average speed = (16 meters) / (18 seconds)
Average speed = (16 / 18) m/s
Average speed = 0.89 m/s
Guys I'm in kind of a PICKLE!!!!!! I know people say it a lot but I will give Brainiest to the best explained answer. Determine the net force charge acting at q1 (+ 2.0 × 10^-5C), caused by q2 (-4.0 × 10-5 C) and q3 (-4.0 × 10^-5C). They create a right angles triangle, where q1 is at the 90° corner
Determine the net electric field acting at q1
Answer:
E≅1.2×10^7 N/C
Explanation:
First off I'd like to say that I'm taking "net electric field" to mean that they don't want this answer to be put into vector component form and instead want magnitudes. Sometimes the wording of these questions throws me off, so sorry ahead of time if that's what they want from you!
Edit: I ended up adding it anyways ;P
Since we are observing the net electric field acting at q1, we need to use the formula: [tex]E=k\frac{q}{r^{2} }[/tex]
And since we are observing the effects of multiple charges at once...
E=ΣE, which just means wee need to add all the observed electric fields together:
ΣE= [tex]k\frac{q2}{r^{2} } +k\frac{q3}{r^{2} }[/tex]
Since we are observing [static] electric fields here, we don't actually need q1's charge. (Though if you wanted to find the net force you would.) Now, before we start plugging values in, let's acknowledge what we know. We know that:
q2=q3they are the same distance from q1These are actually really nice to have, because now we can simplify our expression to:
[tex]E=k\frac{2q}{r^{2} }[/tex]
Now let's plug in our values and get an answer out.
E= 2(8.99×10^9)(4×10^-5)/(0.24)
Plugging all that in, I get:
E≅1.2×10^7 N/C
If you end up needing the net force, F=(q1)(E). That is, you just multiply the electric field by the value of q1. And again, if your teacher wants the answer in vector component form, then the answer will look different.
Let me know what doesn't make sense, or if I got something wrong. Good luck with AP Phy.!
Edit: I put the component form for my answer in the attachment. I also noticed a small calculator related error in my original answer. I updated that to match the new one.
Which of these processes is NOT reduced by insulation?
conduction
distillation
convection
radiation
Answer:
I'm thinking Distillation
Explanation:
I'm not sure, but convection and conduction are insulators. It's between radiation and distillation if im not mistaking.
A ball has a mass of 0.25 kg and is moving to the right at 1.0 m/s. It hits a ball of mass 0.15 kg that is initially at rest. After the collision, the 0.15 kg ball moves off to the right with a velocity of 0.75 m/s. What is the final velocity of the 0.25 kg ball? 0.42 m/s to the right 0.42 m/s to the left 0.55 m/s to the right 0.55 m/s to the left
Answer:
C-0.55 m/s to the right
Explanation:
edg
Answer:
0.55 to the right
Explanation:
A mass of 1 slug is suspended from a spring whose spring constant is 9 lb/ft. The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of Ï3 ft/s. Find the times at which the mass is heading downward at a velocity of 3 ft/s.
Answer:
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Explanation:
To solve the problem/ we first write the differential equation governing the motion. So,
[tex]m\frac{d^{2} x}{dt^{2} } = -kx \\ m\frac{d^{2} x}{dt^{2} } + kx = 0\\\frac{d^{2} x}{dt^{2} } + \frac{k}{m} x = 0[/tex]
with m = 1 slug and k = 9 lb/ft, the equation becomes
[tex]\frac{d^{2} x}{dt^{2} } + \frac{9}{1} x = 0\\\frac{d^{2} x}{dt^{2} } + 9 x = 0[/tex]
The characteristic equation is
D² + 9 = 0
D = ±√-9 = ±3i
The general solution of the above equation is thus
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6. So,
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
The time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.
Given data:
The mass suspended from spring is, m = 1 slug.
The spring constant is, k = 9 lb/ft.
The magnitude of upward velocity is, v = 3 ft/s.
The magnitude of downward velocity is, v' = 3 ft/s.
The given problem can be resolved by framing a differential equation that governs the motion of spring. The differential equation governing the motion of spring is,
[tex]m \dfrac{d^{2}x}{dt^{2}}=-kx\\\\\\\dfrac{d^{2}x}{dt^{2}}+\dfrac{k}{m}x=0[/tex]
With m = 1 slug and k = 9 lb/ft, the equation becomes
[tex]\dfrac{d^{2}x}{dt^{2}}+\dfrac{9}{1}x=0\\\\\\\dfrac{d^{2}x}{dt^{2}}+9x=0[/tex]
Now, the characteristic equation is,
D² + 9 = 0
D = ±√-9 = ±3i
And the general solution of the above equation is,
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂²
= √[(-1)² + (-√3/3)²]
= √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3
and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ
where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6.
t = 5π/18 + 2nπ/3
here, n is a natural number.
Thus, we can conclude that the time interval at which the mass will head downward at the velocity of 3 ft/s is t = 5π/18 + 2nπ/3.
Learn more about the differential equation here:
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