Describe a vibration that is not periodic. NO LINKS PLEASE

Answers

Answer 1

Answer:

1)The position change of almost any manually operated room light switch.

2) Sunlight striking a point on the ground on a partly cloudy and windy day

Explanation:


Related Questions

True or False: When looking at kinetic vs. thermodynamic products the kinetic product predominates at low temperature.

Answers

Answer:

true answer this question

Which statement describes the particles in a liquid?
A. They do not move.
B. They are far apart.
C. They are close together.
D. They are locked in position.

Answers

I think it’s C- hope that helps

A car slams on its brakes creating an acceleration of -4.7 m/s^2. It comes to rest after traveling a distance of 235 m. What was its velocity before it began to accelerate?

Answers

Answer:

Explanation:

v² = u² + 2as

0² = u² + 2(-4.7)(235)

u² = 2209

u = 47 m/s

A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m

Answers

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.


1. Wha' is the relationship between potential and kinetic energy?
As potential energy increases, kinetic energy increases.
b. As potential energy increases, kinetic energy decreases.
C. As potential energy decreases, kinetic energy decreases.
d. Potential and kinetic energy are two separate things and have no
relationship.

Answers

Answer:

B

Explanation:

The kinetic energy in an object is converted into potential energy. This makes the kinetic decrease, while the potential increases.

22 Newton force is working on a 1,901 gram object. What is the acceleration in
meter/s^2 unit

Answers

Answer:

11.573

Explanation:

f = m*a

where f is the force in Newtons, m is the mass of the object (in kg) and a is the acceleration

so, we solve for a

a = f/m

a = 22/1.901

a = 11.573

For northern hemisphere observers, which celestial object would be above the horizon for the greatest
amount of time: one that is on the celestial equator, one that is 30° above the celestial equator, one that is
70° above the celestial equator, or one that is 40" below the celestial equator? Which one would be above
the horizon the greatest amount of time for southern hemisphere observers? Explain your answer.

Answers

Answer:

Explanation:

For a person at about 20° North latitude, an object 70° above the celestial equator would never set. It's arc path would touch the horizon be never sink below it. Observers north of 20° see it all night. Observers south of 20° an object 70° above the celestial equator would spend the greatest amount of time above the horizon.

For southern hemisphere observers, the object 40" below the celestial equator will spend the most time above the horizon. Nearly 12 hours per day. Did you mean 40°? 40 seconds is very close to the equator itself. However, the result is the same.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

What is the equator?

The Equator is an imaginary line passing through the middle of a globe. It is equidistant from the North Pole and the South Pole, Its is a horizontal line residing at 0 degrees latitude.

For northern hemisphere observers, the celestial object that is 70° above the celestial equator would be above the horizon for the greatest amount of time.

One that is 40" below the celestial equator would be above the horizon for the greatest amount of time for southern hemisphere observers.

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I can learn new things but I cannot change how good I am at math. A strongly agree B. agree C disagree D. strongly disagree​

Answers

Mathematics is a broad subject that anyone who constantly practices by learning from first principle and from example will grow in perfection of the subject as time goes by. Hence, I strongly disagree

During my schools days, I struggled learning mathematics, at the time it was difficult, I began practising and learning from text examples, with time I got a hang of it and my grade and confidence level in the subject increased.

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What activity in orienteering can improve a player's math skills?

Answers

homework

or just play some 2012 math game

Answer:

I think calculating distance (if thats an option)

Explanation:

Really hope this helped!

What is the approximate value of k when 30 = e^5k?

Answers

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, how much time does it take the puck to come to rest?

Answers

Here’s my work to your question. I used Newton’s Second Law and a kinematics equation to arrive at the answer.

At the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11, time it take the puck to come to rest is  4.51 sec.

What is speed?

The speed of an item, which is a scalar quantity in everyday usage and kinematics, is the size of the change in that object's position over time or the size of the change in that object's position per unit of time.

Given in the question at the local hockey rink, a puck with a mass of 0.12kg is given an initial speed of 5.3m/s. If the coefficient of friction between the puck and ice is 0.11,

force = ma = μmg, putting the value,

a = - 1.175 m/sec²

now using equation of motion

v = u + at

t = 4.51 sec

So, time taken by the puck to come to rest is 4.51 sec.

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please help 9.2.1 project in science just ned an example​

Answers

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

K
Mission CG9: Weightlessness
Consider the several locations along a roller coaster
track. In which location(s) would the riders feel less
than their normal weight? Select all that apply.
Location A
Location B
Location C
a
=-10 m/s/s, dn
--2 m/s/s, up
a--6 m/s/s, dn
Location D
Location E
x=-12 m/s/s, dn
---6 m/s/s, up

Answers

The locations where the riders feel less than their normal weight are Location A, Location C and Location D.

The given parameters;

Location A, a = 10 m/s² downLocation B, a = 2 m/s² upLocation C, a = 6 m/s² downLocation D, a = 12 m/s² downLocation E, a = 6 m/s² up

The normal weight of the riders is calculated by applying Newton's second law of motion as follows;

W = mg

W = 9.8m

The apparent weight of the riders for the upward acceleration is calculated  as follows;

[tex]R = m(g + a)[/tex]

The apparent weight of the riders for the downward acceleration is calculated  as follows;

[tex]R = m(g - a)[/tex]

The apparent weight of the riders at location A is calculated as follows;

[tex]R_ A = m(9.8 - 10)\\\\R_ A = -0.2 m[/tex]

The apparent weight of the riders at location B is calculated as follows;

[tex]R_B = m(9.8 + 2)\\\\R_B = 11.8 m[/tex]

The apparent weight of the riders at location C is calculated as follows;

[tex]R_C = m(9.8 - 6)\\\\R_C = 3.8 m[/tex]

The apparent weight of the riders at location D is calculated as follows;

[tex]R_D = m(9.8 - 12)\\\\R_D = -2.2 m[/tex]

The apparent weight of the riders at location E is calculated as follows;

[tex]R_E = m(9.8 + 6)\\\\R_E = 15.8 m[/tex]

Thus, the locations where the riders feel less than their normal weight are;

Location ALocation CLocation D.

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This is an image of a satellite traveling around Earth. Explain what are the two forces that are keeping the satelite around Earth without flying off or hitting the ground.​

Answers

Answer:

One force will be gravity & inertia.

Explanation:

Bioth are combine to keep Earth in orbit around the sun, and the moon in orbit around Earth

The more matter in an object, the _____.(1 point)


lower its weight and mass

lower its weight and mass


more likely it is to have gravity

more likely it is to have gravity


greater its gravitational attraction to Earth

greater its gravitational attraction to Earth


less likely it is to have weight on the moon

Answers

Answer:

more likely it is to have gravity

how far from the lens is the image of the house if the house is 16 ft from the thin convex lens with a focal length of 8 ft

Answers

This question involves the concepts of the thin lens formula, focal length, and image distance.

The image of the house is "16 ft" away from the lens.

According to the thin lens formula:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}[/tex]

where,

f = focal length = 8 ft

p = object distance = 16 ft

q = image distance = ?

Therefore,

[tex]\frac{1}{8\ ft}=\frac{1}{16\ ft}+\frac{1}{q}\\\\\frac{1}{q}=\frac{1}{8\ ft}-\frac{1}{16\ ft}\\\\\frac{1}{q}=0.125\ ft^{-1}-0.0625\ ft^{-1}\\\\q=\frac{1}{0.0625\ ft^{-1}}\\\\[/tex]

q = 16 ft

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Why does the earth stay in orbit around the sun instead of drifting away from it into space?
A Electric force between the Sun and Earth
B Magnetic force between the Sun and Earth
Gravitational force between the Sun and Earth

Answers

Answer:

b it is b part answer i think so

What is work? A. What happens when energy is destroyed B. The number of molecules in a substance C. The total size of an object divided by its mass D. What happens when a force causes an object to move​

Answers

Answer:

Which extended definition would be most helpful to add to this body paragraph?

Laughter Yoga, a new form of yoga, is becoming increasingly popular.

Laughter Yoga, which is becoming more and more popular, was created in the 1990s.

Laughter Yoga is modeled after traditional yoga but includes laughter exercises.

Laughter Yoga is a form of yoga that was created by Dr. Madan Kataria in the 1990s.

Explanation:

c

Question No. 1 Marks = = 5 +5 +2 = 12

The driver of a 2.0 × 103 kg red car traveling on the highway at 45m/s slams on his brakes to avoid striking a second yellow car in front of him, which had come to rest because of blocking ahead as shown in above Fig. After the brakes are applied, a constant friction force of 7.5 × 103 N acts on the car. Ignore air resistance.
(a) Determine the least distance should the brakes be applied to avoid a collision with the other vehicle?
(b) If the distance between the vehicles is initially only 40.0 m, at what speed would the collision occur?
(c) Write your conclusive observations on the result obtained from this numerical. i.e. the importance of Physics in daily life.



Question No. 2 Marks = 8
Consider an automobile moving at v mph that skids d feet after its brakes lock. Calculate how far it would skid if it was moving at 2v and the brakes were locked.

Answers

The kinematics and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1.

     a) The stopping distance is: x = 270 m

     b) The initial velocity is: v₀ = 17.3 m / s

     c) Concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

The stopping distance is: x = 4d

Given parameters

Mass of the red carriage m1 = 2,0 10³ kg Red car speed vo = 45 m / s Friction force fr = 7.5 10³ N.

To find

Question 1.

    a) Minimum braking distance.

    b) If the distance is x = 40.0 m, what speed should the vehicles have?

    c)  Conclusive importance of physics in daily life.

Question 2.

The distace to stop.

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

         v² = v₀² - a2 x

Where v and v₀ are the current and initial velocity, respectively, at acceleration and x the distance traveled.

Newton's second law states that the net force is proportional to the mass and the acceleration of the body.

          F = ma

Where F is force, m is mass and acceleration.

In the attachment we see a diagram of the forces in the system. Let's look for the acceleration of the body

        fr = m a

        a =[tex]\frac{fr}{m}[/tex]  

        a = [tex]\frac{7.5 \ 10^3}{2.0 \ 10^3 }[/tex]  

        a = 3.75 m / s²

This acceleration is in the opposite direction to the speed.

Let's find the distance needed to stop, the final speed is zero.

          0 = v₀² - 2 ax

           x = [tex]\frac{v_o^2 }{ 2a}[/tex]  

Let's calculate.

          x = [tex]\frac{45^2 }{2 3.75}[/tex]  

          x = 270 m

This is the minimum distance that the two vehicles must separate to avoid a collision.

b) We look for speed.

        v₀ = [tex]\sqrt{2ax}[/tex]  

        v₀ = [tex]\sqrt{2 \ 3.75 \ 40.0}[/tex]  

        v₀ = 17.3 m / s

c) The concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

2) They indicate that the initial velocity is v and the distance traveled to stop is d, let's find the acceleration.

           0 = v₀² - 2ax

Let's substitute.

             a = [tex]\frac{v^2}{2d}[/tex]  

They ask the distance traveled if this car traveled from an initial speed 2v.

             0 = v² - 2 a x

             x = [tex]\frac{v^2}{2a}[/tex]  

We substitute

            x = [tex]\frac{(2v)^2 }{2} \ (\frac{2d}{v^2})[/tex]  

            x = 4 d

In conclusion, using the kinematic relations and Newton's second law we can find the results for the questions about the braking movement of the car are;

Question 1

       a) The stopping distance is: x = 270 m

       b) The initial velocity is: v₀ = 17.3 m / s

        c) concepts of kinematics and Newton's second law show us the expressions to make safe trips and avoid accidents on the roads.

Question 2.

 The stopping distance is x = 4d

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A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

Answers

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

Length of the string, L = 100 cm

The wavelengths of the constituent travelling waves is calculated as follows;

[tex]L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}[/tex]

for first mode: n = 1

[tex]\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm[/tex]

for second mode: n = 2

[tex]\lambda = \frac{2L}{2} = L = 100 \ cm[/tex]

For the third mode: n = 3

[tex]\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm[/tex]

For fourth mode: n = 4

[tex]\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm[/tex]

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

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Shorter the vibrating part more will be the pitch. How?​

Answers

Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.

How much power does it take to lift 30.0 N 10.0 m high in 10.00 s?

Answers

Answer:

60w

Explanation:

The power required is 30 Watt.

Let us recall that power is defined as the rate of doing work. Hence, we can write as follows;

Power = Work done/ time taken

Now;

work done =  Force × distance

Force = 30.0 N

Distance = 10.0 m

work done = 30.0 N × 10.0 m = 300 J

The power expended = 300 J/10.00 s = 30 Watt

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If the velocity and frequency of a wave are both doubled, how does the wavelength change?

Answers

The wavelength will remain unchanged.

Explanation:

The velocity [tex]v[/tex] of a wave in terms of its wavelength [tex]\lambda[/tex] and frequency [tex]\nu[/tex] is

[tex]v = \lambda\nu[/tex] (1)

so if we double both the velocity and the frequency, the equation above becomes

[tex]2v = \lambda(2\nu)[/tex] (2)

Solving for the wavelength from Eqn(2), we get

[tex]\lambda = \dfrac{2v}{2\nu} = \dfrac{v}{\nu}[/tex]

We would have gotten the same result had we used Eqn(1) instead.

Answer:

the wavelength increases

Explanation:

The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive toward this parked car, pass it, and drive away, you observe the frequency to change by 69.5 Hz. At what speed are you driving

Answers

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

The speed of car driving is 15.5 m/s as the car is parked and drive away.

What is speed?

Speed is defined as a measurement of the length of time it takes for an object to travel a certain distance. You can determine an object's speed if you know how far it moves in a given amount of time. Time does not move, hence there is no concept of a speed of time. Time refers to how we move through the temporal realm. Speed is a unit of measurement for how quickly something is moving. A change in velocity results in a change in speed.

To calculate the speed we use the formula

f₀ = f (v + vo) / (v + vs)

Δf = f approach - f depart

69.5 = (769(343 + vo) / (343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

vo = 15.5 m/s

Thus, the speed of car driving is 15.5 m/s as the car is parked and drive away.

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The qualitative equivalent of external validity is:
A- Credibility

B- Dependability

C- Transformability

D- Confirmability

Answers

c transformability i think

A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.​

Answers

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

A cyclist rides in a circle with speed 8.1 m/s. What is his centripetal
acceleration if the circle has a radius of 27 m?

Answers

Explanation:

We know that the tangent velocity is 8.1 m/s. We also know that the tangent velocity can be written in the following way:

Vt = ωr with ω being the angular velocity.

We now calculate ω:

ω = Vt/r = 8.1 m/s / 27m = 0.3 rad/s

Now that we have ω we can calculate the centripetal aceleration:

a = ω^2 * r = ( 0.3 )^2 * 27 = 2.43 m/s^2

the turns ratio for a transformer with 225 turns of wire in its primary winding and 675 turns in the secondary is: n

Answers

The ratio of the primary turns to the secondary turns is 1/3

The correct answer to the question is Option A. 1/3

From the question given above, the following data were obtained:

Primary turn (Nₚ) = 225 turnsSecondary turn (Nᵣ) = 675 turns Ratio of primary to secondary =?

Ratio = Nₚ/Nᵣ

Nₚ/Nᵣ = 225 / 675

Nₚ/Nᵣ = 1/3

Therefore, the ratio of the primary turns to the secondary turns is 1/3

Complete question:

See attached photo

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Scenario 3: You are driving through a rain storm talking to your family but you can only hear every other word. What 2 medias are the waves passing in this scenario?

Answers

The sound waves are passing through the sound of the rain and the air around you

Help me outtttt jejjejejeje

Answers

Answer:

do it got a picture

Explanation:

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