Two objects charged with an equal amount of charge interact with each other with a force
of 3.62 Newton when separated by a distance of 16.9 cm. Determine the magnitude of
the charge on each object.

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

Answer:

≈933.3kg/m^3

Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

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 .

Answer:

0.03 Joules have been converted into other forms of energy as the direct result of the collision.

Explanation:

Let's start studying the conservation of momentum for the system:

[tex]P_i=P_f\\(0.25\,kg)\,{0.6\,m/s)+(0.5\,kg)\,(0\,m/s)=(0.25\,kg+0.5\,kg)\, v_f \\\\\\ 0.15\,kg\,m/s=0.75\,kg\,\,v_f\\v_f=0.15/0.75\,\,m/s\\v_f=0.2\,\,m/s[/tex]

Now that we know the speed of the newly created object, we can calculate how the final kinetic energy differs from the initial one:

[tex]K_i=\frac{1}{2} (0.25)\,(0.6)^2+\frac{1}{2} (0.5)\,(0)^2=0.045\,\,J\\ \\K_f=\frac{1}{2} (0.75)\,(0.2)^2=0.015\,\,J\\[/tex]

Then, when we subtract one from the other, we can estimate how much kinetic energy has been converted into other forms of energy in the collision:

0.045 J - 0.015 J = 0.03 J

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:

These 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.

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.

Answer:

the water is going to boil and the mercury ill melt and shoot the cork out the bottom of the tube

Explanation:

Answer:

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Explanation:

Puesto que la bola gira en un círculo vertical, existe claramente una diferencia entre las tensiones debido a la influencia de la gravedad y la tensión que resulta de la aceleración centrípeta experimentada por la masa. La máxima tensión ocurre cuando la bola se encuentra en el nadir (o la sima) del trayecto circular, la cual se describe por la Segunda Ley de Newton:

[tex]T_{max} - m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

En cambio, la mínima tensión aparece cuando la bola se encuentra en el cénit (o la cima) del trayecto circular, descrita por la misma ley de Newton:

[tex]T_{min} + m\cdot g = m\cdot \frac{v^{2}}{L}[/tex]

Donde:

[tex]T_{min}[/tex], [tex]T_{max}[/tex] - Tensiones mínima y máxima, medidas en newtons.

[tex]m[/tex] - Masa de la bola, medida en kilogramos.

[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.

[tex]L[/tex] - Distancia con respecto al eje de rotación, medida en metros.

[tex]v[/tex] - Rapidez tangencial, medido en metros por segundo.

Se elimina la aceleración centrípeta de ambas expresiones por igualación:

[tex]T_{min} + m\cdot g = T_{max} - m\cdot g[/tex]

Ahora, la diferencia entre las tensiones máxima y mínima es:

[tex]T_{max} - T_{min} = 2\cdot m \cdot g[/tex]

Si [tex]m = 1\,kg[/tex] y [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], entonces:

[tex]T_{max} - T_{min} = 2\cdot (1\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]T_{max}-T_{min} = 19.614\,N[/tex]

La diferencia entre las tensiones máxima y mínima es de 19.614 newtons.

Answer is given below

Explanation:

Answer:

Yes it is normal

Explanation:

When you exercise, your heart beat goes up, resulting in people saying that their heart feels like it is "beating out of their chests".

Answer:

Explanation:

We shall apply the concept of impulse which is given as follows .

Impulse = force x time

Impulse = change in momentum

If u be the initial velocity of golf ball and v be the final velocity , m be the mass

change in momentum

= mu - ( - mv )

= mu+ mv

If F be the force applied and t be the duration of touch with the ball

Impulse = F x t

F x t = mu + mv

mv = Ft - mu

For given mu , greater the value of t , greater will be the value of v

so v is increased when t is increased .

Increased value of v will help in achieving greater distance attained by

golf ball

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

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..

Answer:

a. Convert  120 × 10⁸ g to i mg = 1.2 × 10¹³ mg ii. to g = 1.2 × 10⁷ kg

b. Convert 200000 × 10³ m to i. micrometers = 0.2 × 10³ μm  ii. to cm = 2 × 10⁶ cm iii. to km = 2 × 10⁵ km

Explanation:

a. i. To convert the mass = 120 × 10⁸ g to mg, We know that 1000 mg = 10³ mg = 1 g, Since we are converting to mg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10³ mg = 120 × 10¹¹ mg = 1.2 × 10² × 10¹¹ mg = 1.2 × 10¹³ mg

ii. To convert the mass = 120 × 10⁸ g to kg, We know that 1000 g = 10³ g = 1 kg, 1 g = 10⁻³ kg. Since we are converting to kg, 120 × 10⁸ g = 120 × 10⁸ × 1g = 120 × 10⁸ × 10⁻³ kg = 120 × 10⁵ kg = 1.2 × 10² × 10⁵ kg = 1.2 × 10⁷ kg

b. i.To convert the length = 200000 × 10³ m to micrometers, We know that 1/1000000 μm = 10⁻⁶ mg = 1 m, Since we are converting to micrometers, μm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000000 μm = 200000/1000000 × 10³ μm = 0.2 × 10³ μm

ii. To convert the length = 200000 × 10³ m to cm, We know that 100 cm = 10² cm = 1 m, 1 m = 10⁻² cm = 1/100 cm. Since we are converting to cm, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ × 1/100 cm = 200000/100 × 10³ cm = 2000 × 10³ cm  = 2 × 10³ × 10³ cm = 2 × 10⁶ cm

iii. To convert the length = 200000 × 10³ m to km, We know that 1000 m = 10³ m = 1 km, 1 m = 10⁻³ km = 1/1000 km Since we are converting to km, 200000 × 10³ m = 200000 × 10³ × 1 m = 200000 × 10³ ×  1/1000 km = 200000/1000 × 10³ km = 200 × 10³ km = 2 × 10² × 10³ km = 2 × 10⁵ km

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))

Answer:

See the attachment below.

Explanation:

Best Regards!

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.

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.

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.

Answer:

Liquids

Explanation:

Liquids take up the shape of the container it is poured into but will never change its volume.

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

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..

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).

Answer:

option d is answer because pd is measured in volt.

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

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].

Answer:

Explanation:

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

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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