A uniform string of length 0.50 m is fixed at both ends. Find the
wavelength of the fundamental mode of vibration. If the wave
speed is 300 mis, find the frequency of the fundamental and next
possible modes.

Answers

Answer 1

Answer:

configuration of string:

Node - Antinode - Node    or N-A-N

This is 1/2 wavelength since a full wavelength is N-A-N-A-N

f (fundamental) = V / wavelength

F0 = 300 m/s / 1 m = 100 / sec

F1 = 300 m/s / .5 m = 600 / sec

Each increase is a multiple of the fundamental since the wavelength

increases by 1/2 wavelength to keep nodes at both ends of the string


Related Questions

Evaluate tan(249).
O A. 1.36
B. 0.45
C. 0.41
D. 0.91

Answers

Answer:

tan 249 = 2.61

tan 249 = tan (249 - 180) = tan 69 = 2.61

Which theory of emotion explains the startle response

Answers

Answer:

In the present study, the startle blink reflex is used as a measure of emotion regulation to effective picture stimuli. Based on the aphasic theory of emotion, it is hypothesized that the startle response will be largest in magnitude in the presence of negative emotional stimuli (Varanasi, Spence, & Lang, 1988).

A rock falls to the ground with a force of 300N and air resistance pushes back up on the rock with a
force of 45N. Which is true of the rock?
A There are balanced forces acting on the rock and the net force is 255N down
B. There are unbalanced forces acting on the rock and the net force is 255N down
C. There are balanced forces acting on the rock and the net force is 345N down
. There are unbalanced forces acting on the rock and the net force is 345N down

Answers

Answer:

c

Explanation:

When the pushing force is increased to 88.2 N, the box just begins to move. What is the Friction Force if it is moving at a constant velocity? What is the coefficient of friction between the ground and box?​

Answers

In order to calculate frictional force look below..

The formula given by

[tex]\\ \sf\longmapsto F_f=\mu N[/tex]

Or

[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]

u is coefficient of friction

N is normal reaction.

25. Which of the following cannot be broken down into smaller parts through ordinary chemical means?
a. Nitrogen
b. Protein
C. Salt
d. Sugar

Answers

Answer:

the correct answer is sugar

I think it’s Sugar!!!!!

A stimulus that increses the occurance of a response is called

Answers

Answer:

reinforcing stimulus

Explanation:

Think its correct

Answer:

reinforcing stimulus

Explanation:

because it increases the occurence of a response

If Brad drives 100 meters forwards and then 100 meters backwards, what is her
distance? Her displacement?

Answers

Answer:

0 because she went to exact same place as before

Jonathan wants to separate stones, insects and other unwanted materials in his mixture of grains and corn. What technique of separating mixture is appropriate

A. Winnowing
B. physical manipulation
C. Filtering
D. Magnetism​

Answers

Answer:

What is B physical manipulation

Explanation:

Physical manipulation means fertilizers that are manufactured, blended, or mixed, or animal manures or compost that have been changed from their initial physical state by manipulations such as drying, cooking, chopping, grinding, shredding, ashing, or pelleting.

Please allow me to know if my answer helped you with a thank you!

Miss Hawaii

chứng minh mặt trời là nguồn gốc của tất cả nguồn năng lượng

Answers

An Excerpt from “Optimism”

by Helen Keller

1 Could we choose our environment, and were desire in human undertakings synonymous with

endowment, all men would, I suppose, be optimists. Certainly most of us regard happiness as

the proper end of all earthly enterprise. The will to be happy animates alike the philosopher, the

prince and the chimney-sweep. No matter how dull, or how mean, or how wise a man is, he feels

that happiness is his indisputable right.

2 It is curious to observe what different ideals of happiness people cherish, and in what singular

places they look for this well-spring of their life. Many look for it in the hoarding of riches, some

in the pride of power, and others in the achievements of art and literature; a few seek it in the

exploration of their own minds, or in the search for knowledge.

3 Most people measure their happiness in terms of physical pleasure and material possession.

Could they win some visible goal which they have set on the horizon, how happy they would be!

Lacking this gift or that circumstance, they would be miserable. If happiness is to be so

measured, I who cannot hear or see have every reason to sit in a corner with folded hands and

weep. If I am happy in spite of my deprivations, if my happiness is so deep that it is a faith, so

thoughtful that it becomes a philosophy of life,—if, in short, I am an optimist, my testimony to

the creed of optimism is worth hearing....

4 Once I knew the depth where no hope was, and darkness lay on the face of all things. Then

love came and set my soul free. Once I knew only darkness and stillness. Now I know hope and

joy. Once I fretted and beat myself against the wall that shut me in. Now I rejoice in the

consciousness that I can think, act and attain heaven. My life was without past or future; death,

the pessimist would say, “a consummation devoutly to be wished.” But a little word from the

fingers of another fell into my hand that clutched at emptiness, and my heart leaped to the

rapture of living. Night fled before the day of thought, and love and joy and hope came up in a

passion of obedience to knowledge. Can anyone who has escaped such captivity, who has felt

the thrill and glory of freedom, be a pessimist?

5 My early experience was thus a leap from bad to good. If I tried, I could not check the

momentum of my first leap out of the dark; to move breast forward is a habit learned suddenly

at that first moment of release and rush into the light. With the first word I used intelligently, I

learned to live, to think, to hope. Darkness cannot shut me in again. I have had a glimpse of the

shore, and can now live by the hope of reaching it.

6 So my optimism is no mild and unreasoning satisfaction. A poet once said I must be happy

because I did not see the bare, cold present, but lived in a beautiful dream. I do live in a

beautiful dream; but that dream is the actual, the present,—not cold, but warm; not bare, but

furnished with a thousand blessings. The very evil which the poet supposed would be a cruel

6) Read the last sentence from the text.

Only by contact with evil could I have learned to feel by contrast the beauty of truth and love and goodness.

Explain how Helen Keller develops this idea in the text. Use specific details to

support your answer.

HELPPPP

The maximum force of sliding friction between a 10 kg rubber box and the concrete
floor is 64 N. How much force should a worker push on the box with if he wants it to
move at a constant velocity?
1) A little less than 64 N
2)A little more than 64 N
3)Exactly 64 N.
4)Exactly 640 N

Answers

The force that will move the box at constant velocity must be a little more than 64 N.

The coefficient of sliding friction is obtained from the formula;

μ= F/R

Where;

F = frictional force

μ = coefficient of sliding friction

R = Normal reaction

It is necessary to note that the force that will move the body must be greater than the frictional force acting between the body and the surface in order to move the body. Hence, the force that will move the box at constant velocity must be a little more than 64 N.

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A baseball is traveling with a velocity of 12 m/s at an angle of 45 above
horizontal. What is the velocity of the ball after two seconds?

Answers

Focus on the ball. Do you see you? Keep following it. Put all your energy into watching it and then unlock your third eye. With that you'll be able to determine your question.

please help me
7. If microsecound = 0.5, how much force must be applied to a spring (spring constant of 0.8 N/m) which is attached to a block of wood (mass = 4.0 kg) in order to just begin to move the block?​

Answers

Answer:

Explanation:

Your question is quite confusing, particularly the information about microsecond = 0.5.

I'm going to ASSUME that you mean coefficient of static friction μs = 0.5

unfortunately typing a subscript "s" is very difficult and probably leads to such confusion.

I will also ASSUME that the block, and spring, and force vector are all horizontal.

If the force is slowly increased until the block slips, the spring will compress until the force on each end equals the maximum static friction force. As we are not concerned with the compression distance, only the force, we can ignore the spring constant information and simply find the maximum available static friction force.

F = μN

F = μmg

F = 0.5(4.0)(9.8)

F = 19.6 N

Not that it matters, but the spring will have extended or compressed 19.6/0.8 = 24.5 m, which is a very long and very light spring

In hockey activities, a warm hockey puck and a frozen hockey puck has a different coefficient of restitution: 0.5 for a warm hockey puck, and 0.35 for a frozen one. NHL requires the frozen pucks to be used in games. To make sure the puck can be used in the game, the referee drops the puck on its side from a height of 2.5 m. How high should the puck bounce if it is a frozen puck

Answers

If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

Given the data in the question;

Since the hockey puck was initially in the referee's hands

Initial velocity; [tex]u = 0m/s[/tex]Distance or height from which it was dropped;  [tex]h = 2.5m[/tex]Acceleration due to gravity; [tex]g = 9.8 m/s^2[/tex]Coefficient of restitution a frozen puck; [tex]0.35[/tex]

First we will find the velocity of the Puck when it hits the ground

From the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

We substitute in our value and find "v"

[tex]v^2 = 0 + (2 \ *\ 9.8m/s^2\ *\ 2.5m )\\\\v^2 = 47.04m^2/s^2\\\\v= \sqrt{47.04m^2/s^2}\\\\v = 6.85857m/s[/tex]

Now, Velocity of the hock puck after it hits the ground and bounce back;

We know that; Coefficient of restitution [tex]= \frac{Relative\ velocity\ after\ collision}{Relative\ velocity\ before\ collision}[/tex]

Hence, Relative Velocity after collision = Coefficient of restitution × Relative Velocity before collision

we substitute in our values;

Relative Velocity after collision [tex]= 0.35 \ *\ 6.85857m/s[/tex]

Relative Velocity after collision [tex]= 2.4 m/s[/tex]

Now, to determine how high should the puck bounced back

We use the Third Equation of Motion:

[tex]v^2 = u^2 + 2as[/tex]

Where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity and s is the distance.

Since the pluck is under gravity, we will have:

[tex]v^2 = u^2 + 2gh[/tex]

Now, since the hockey puck bounces back, it is experiencing a negative acceleration

Hence, the equation becomes

[tex]v^2 = u^2 - 2gh[/tex]

We substitute our values into the equation and find "h"

[tex](0m/s)^2 = (2.4m/s)^2 - ( 2*9.8m/s^2*h)\\\\0 = 5.76m^2/s^2 - (19.6m/s^2*h)\\\\(19.6m/s*h) = 5.76m^2/s^2 \\\\h= \frac{ 5.76m^2/s^2 }{19.6m/s^2}\\\\h = 0.3m[/tex]

Therefore, If its is a frozen hockey puck, it bounce off the ground after collision to a height of 0.3m.

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find charge and charge density on the surface of a conducting sphere of radius 15.2cm where potential at 215 v

Answers

this is the correct answer

Which statement describes friction?

Answers

Answer:

include the statements pls so i can choose wich one it is and tell you

Explanation:

The image shows mountains in Alaska.



Which describes the main feature of the circled area of these mountains?

A syncline is visible.
An anticline is visible.
These mountains show no evidence of folding.
These mountains likely formed from normal faults

Answers

Answer:

A syncline is visible.

Explanation:

A syncline is visible.

Answer:

These mountains show no evidence of folding.

() The two forces F_{1} and F_{2} shown in Fig. 4- 39a and b (looking down) act on a 27.0-kg object on a frictionless tabletop . If F_{1} = 10.2N and F_{2} = 16.0N find the net force on the object and its acceleration for each situation , (a) and (b) .

Answers

It looks like 4.78 inches + 56x

A 500 kg car is at rest at the top of a 72 m high hill. The car rolls to the bottom of the hill. At the bottom of the hill, the car has a speed of 25.6 m/s. Calculate the mechanical energy of the car at the top and bottom of the hill. (Assume the bottom of the hill has a height of 0 m, g=9.80 ms2/).

Answers

Explanation: Solution

1.

Gravitational potential energy

U=mgh=500*9.8*50

U=245000 J

2.

Kinetic energy is present at bottom of the hill

K=(1/2)mV2=(1/2)*500*27.82

K=193210 J

3.

Work done by friction

W=193210-245000=-51790 J

The mechanical energy at the top and bottom of the hill is equal to 352800 J and 163840 J respectively.

What is the kinetic energy and potential energy?

Kinetic energy (KE) can be described as the energy possessed by a moving object due to its motion. Work by a body will be done to change the kinetic energy. The kinetic energy is represented as K.E = ½mv².

Potential energy (P.E) can be described as the energy that is stored by an object due to its position and is represented in the equation as P.E = mgh, where ‘m’ is the mass, ‘g’ is the acceleration due to gravity and ‘h’ is the height.

The mechanical energy = Kinetic energy + potential energy

Given, the mass of the car, m = 500 Kg

The height of the hill, h = 72 m

The velocity of the car, v = 25.6 m/s

At the top of the hill, the mechanical energy = potential energy

The potential energy at the top of the hill, = mgh

P. E. = 500 × 9.8 ×72

P.E. = 352800 J

At the bottom of the hill, the mechanical energy = kinetic energy

The kinetic energy of the car at the bottom of the hill,

K.E. =  ½ × 500 (25.6)²

K.E. = 163840 J

Learn more about kinetic energy and potential energy, here:

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#SPJ2

A dry cell gives static electricity true or false?

Answers

Answer:

False

Explanation:

What is the kinetic energy of a 36N toy car which is moving at 5 m/s?

Answers

Correct option is
C
y
max

=11 m
Work done =mgh

2
1

×base×height=mgh
(area under the curve is work done)

2
1

×11×100=5×10×h

or h=11m

A bowling ball, basketball, and tennis ball are all raised to the same height above the ground. Give the order of objects from the least potential energy to the most

Answers

Answer:

bowling ball, basket ball, tennis ball

Explanation:

How does the mass of the bob affect the number of swings of a pendulum?

Answers

Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period

Which of the following is the current best hypothesis for the formation of the solar system?
A. Formed by an exploding super nova star which then collapsed and coalesced into a spinning
disk forming Sun and planets
B. Our solar system has always been here and has never changed
C. Formed from the Sun’s explosion releasing particles into space forming planets and other
objects
D. Our solar system was formed by a great collision of other stars with one another

Answers

Answer:

A

Explanation:

all galaxies exploded in order to create the sun/stars

Most gasoline engines in today's automobiles are belt driven. This means that the crankshaft, a rod which rotates and drives the
pistons, is timed to the camshaft, the mechanism which actuates the valves, by means of a belt. Starting from rest, assume it
takes t = 0.0320 s for a crankshaft with a radius of r = 3.75 cm to reach 1250 rpm. If the belt does not stretch or slip, calculate
the angular acceleration ay of the larger camshaft, which has a radius of r2 = 7.50 cm, during this time period.

Answers

The angular acceleration of the larger camshaft is 995.72 rad/s².

The given parameters;

initial angular velocity, [tex]\omega _i[/tex] = 0time of motion, t = 0.032 sradius of the crankshaft, r = 3.75 cm final angular speed, [tex]\omega _f[/tex] = 1250 rpm

The angular acceleration of the 3.75 cm camshaft is calculated as follows;

[tex]\omega _f = \omega _i + \alpha t\\\\\omega _f =0 + \alpha t\\\\\omega _f = \alpha t\\\\(1250 \ \frac{rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1\min}{60 \ s} ) = 0.032 \alpha \\\\130.92 = 0.032\alpha \\\\\alpha = \frac{130.92}{0.032} = 4091.25 \ rad/s^2[/tex]

The angular momentum of the camshaft is calculated as follows;

[tex]I_1 \alpha _1 = I_2 \alpha_2 \\\\\frac{1}{2} mr_1^2 \alpha _1 = \frac{1}{2}m R^2 \alpha_2\\\\r_1^2 \alpha _1 = R^2 \alpha_2\\\\\alpha_2 = \frac{r_1^2 \alpha _1 }{R^2} \\\\\alpha_2 =\frac{(0.037)^2 \times (4091.25)}{(0.075)^2} \\\\\alpha _2 = 995.72 \ rad/s^2[/tex]

Thus, the angular acceleration of the larger camshaft is 995.72 rad/s².

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Show your workikkkkkkkk

Answers

Answer:

Explanation:

F = ma

Assuming the 20° is angle θ measured to the horizontal

mgsinθ - μmgcosθ = ma

g(sinθ - μcosθ) = a

at constant velocity, a = 0

g(sinθ - μcosθ) = 0

   sinθ - μcosθ = 0

                 sinθ = μcosθ

μ = sinθ/cosθ

μ = tanθ

μ = tan20

μ = 0.3639702342...

μ = 0.36

What happened to an enzyme’s structure as it exceeds the typical human body temperature

Answers

Enzymes. ... This is because heat energy causes more collisions, with more energy, between the enzyme molecules and other molecules. However, if the temperature gets too high, the enzyme is denatured and stops working. A common error in exams is to write that enzymes are killed at high temperatures.

How can wind, coiled wire, and magnets be used together to generate electricity?

Answers

Answer:

b

Explanation:

hope helpful

have a great day

Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther from the center? Explain your answer.​

Answers

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

giúp em toàn bộ nhé ............

Answers

Explanation:

nnnnkgcbchg of bbjbjk oh jjzjjzkedkkdkfjjsjdsjfkjfdkkdowkronqfojofj see j FC

Find the angle between two equal forces (p) when their resultant is equal to (p)

Answers

Answer:

60°

Explanation:

It will be equilateral triangle, you can prove with the cosine theorem.

1)P^2=P^2+P^2 - 2×P×P× cos A

2) cos A= P^2 / 2× P^2 = 1/2

cos A = 1/2

A = 60°

If you wrote this question right,it will be solved like this :)

Toe toe toe toe toe toe
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