A vertical tube one meter long is open at the top. It is filled with 50 cm of water. If the velocity of sound is 344 m/s, what will the fundamental resonant frequency be (in Hz)?

Answers

Answer 1

Answer:

The fundamental resonance frequency is 172 Hz.

Explanation:

Given;

velocity of sound, v = 344 m/s

total length of tube, Lt = 1 m = 100 cm

height of water, hw = 50 cm

length of air column, L = Lt - hw = 100 cm - 50 cm = 50 cm

For a tube open at the top (closed pipe), the fundamental wavelength is given as;

Node to anti-node (N ---- A) : L = λ / 4

λ = 4L

λ = 4 (50 cm)

λ = 200 cm = 2 m

The fundamental resonance frequency is given by;

[tex]f_0 = \frac{v}{\lambda}\\\\f_0 = \frac{344}{2}\\\\f_0 = 172 \ Hz\\\\[/tex]

Therefore, the fundamental resonance frequency is 172 Hz.


Related Questions

A 100 kg. football player and a 20 kg. child sit on rolling carts and push off
from each other. Which person applies more force?*20 kg. child applies more force
100 kg. football player applies more force
Same force is applied

Answers

Answer:

the forces are the same

Explanation:

A. A piece of paper near a magnet
B. An aluminum nail near a magnet
C. An iron nail, not near a magnet
D. An iron nail near a magnet

Answers

Answer:

it’s c not d

Explanation:

took the test

Answer: D!!!

Explanation: jus got it wrong from the other answer.

The Slingshot is a ride for two people. It consists of a single passenger cage, two towers, and two elastic bands. Potential energy is stored in the elastic bands and the passenger cage is released. On the way up, this potential energy in the elastic bands is converted into the kinetic energy of the cage. At the maximum height of the ride, the energy has been converted into gravitational potential energy of the cage. The slingshot has two towers of height h = 76 m. The towers are a distance d = 31 m apart. Each elastic band has an unstretched length of L0 = 41 m and a spring constant of k = 310 N/m. The total mass of the passengers and cage is m = 410 kg. The car is pulled down to the ground in the middle of the two towers.

Calculate the maximum height, in meters, of the ride.

Answers

my bad I was in a herie last time can you please answer my question , I am going to give you the 5 points back for this question and extra 30 points ,

Explanation:

first its going to say 10 points for my question but after that I well make answer a small question and give you 30points. like whats your favorite color . stay tuned .

Answer:

The maximum height reached by the ride after it was pulled to the ground is 51.6 m.

The given parameters;

Distance between the two towers, d = 31 mHeight of the tower, h = 76 mUnstretched length of the band, L₀ = 41 mElastic constant of the band, k = 310 N/m

The distance half-way between the bands;

[tex]\frac{d}{2} = \frac{31}{2} = 15.5 \ m[/tex]

The maximum length of the band when stretched is calculated as;

[tex]c^2 = 15.5^2 + 76^2\\\\c^2 = 6016.25\\\\c = \sqrt{6016.25} \\\\c = 77.57 \ m[/tex]

The extension of the elastic band;

x = 77.57 m - 41 m

x = 36.37 m

The elastic potential energy stored in the band;

[tex]E = \frac{1}{2} kx^2\\\\E = \frac{1}{2} \times 310 \times (36.57)^2\\\\E = 207,291.56 \ J[/tex]

The elastic potential energy of the elastic band will be converted into kinetic energy of the ride and the speed of the ride is calculated as;

[tex]E = \frac{1}{2} mv^2\\\\207,291.56 = \frac{1}{2} \times 410 \times v^2\\\\v^2 = \frac{207,291.56}{(0.5\times 410)} \\\\v^2 = 1011.178\\\\v = \sqrt{1011.178} \\\\v = 31.8 \ m/s[/tex]

The maximum height reached by the ride is calculated as;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{(31.8)^2}{(2\times 9.8)} \\\\h = 51.6 \ m[/tex]

Thus, the maximum height reached by the ride is 51.6 m.

Learn more here: https://brainly.com/question/156316

3) A 10kg object rests on a frictionless surface when it is struck by a 300N force. At what rate will it accelerate?

3m/s/s
30m/s/s
0.3m/s/s
300m/s/s

Answers

Answer: 0.3m/s/s

(i'm really sorry if i'm wrong)

:(

A sled is pulled with a force of 540 N at an angle of 40° with the horizontal. What are the horizontal and vertical components of this force?

Answers

Answer:

Fx = 467.65N

Fy = 270N

Explanation:

Given

Force = 540N

angle of inclination = 40 degree

Horizontal component Fx = Fcos 30

Fx = 540cos30

Fx = 540(0.8660)

Fx = 467.65N

Hence the horizontal component is 467.65N

Vertical component Fy = Fsin 30

Fy = 540sin30

Fy = 540(0.5)

Fy = 270N

Hence the vertical component is 270N

in the case shown below, the 1 kg rock rides on a horizontal disk that rotates at constant speed 5m/s about its vertical axis. the radius of the disk is 1 meter. What is the magnitude of the friction?

Answers

Answer:

25

Explanation:

8x = -6. What does x equal?

Answers

Answer:

x=-3/4

Explanation:

3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterwards, the 3.00 kg block moves 1.71 m/s left. find the velocity of the 2.22 kg block afterwards

Answers

Momentum is conserved, so the total momentum before collision is equal to the total momentum after collision. Take the right direction to be positive. Then

(3.00 kg) (2.09 m/s) + (2.22 kg) (-3.92 m/s) = (3.00 kg) (-1.71 m/s) + (2.22 kg) v

where v is the velocity of the 2.22 kg block after collision. Solve for v :

6.27 kg•m/s - 8.70 kg•m/s = -5.13 kg•m/s + (2.22 kg) v

(2.22 kg) v = 2.70 kg•m/s

v = (2.70 kg•m/s) / (2.22 kg)

v ≈ 1.22 m/s

i.e. a velocity of about 1.22 m/s to the right.

Hey guys this is Ap physics please help I need this to pass i will mark brainliest for a good attempt

Answers

Split up the forces into components acting parallel to and perpendicular to the slope. See the attached picture for the reference axes.

The box stays on the surface of the plane, so that the net force acting perpendicular to it is 0, and the only acceleration is applied in the parallel direction.

Let m be the mass of the box, θ the angle the plane makes with the ground, and a the acceleration of the box. By Newton's second law, we have

• net parallel force

∑ Force (//) = W (//) - F = m a

(that is, the net force in the parallel direction is the sum of the parallel component of the weight W and the friction F which acts in the negative direction)

• net perpendicular force

∑ Force (⟂) = W (⟂) + N = 0

Notice that

W (//) = W sin(θ) … … … which is positive since it points down the plane

W (⟂) = -W cos(θ) … … … which is negative since it points opposite the normal force N

So the equations become

W sin(θ) - F = m a

-W cos(θ) + N = 0

Solving for a gives

a = (W sin(θ) - F ) / m

which is good enough if you know the magnitude of the friction force.

If you don't, you can write F in terms of the coefficient of kinetic friction between the box and plane, µ, as

F = µ N

so that

a = (W sin(θ) - µ N ) / m

and the normal force itself has a magnitude of

N = W cos(θ)

so that

a = (W sin(θ) - µ W cos(θ) ) / m

The weight W has magnitude m g, where g is the magnitude of the acceleration due to gravity, so

a = (m g sin(θ) - µ m g cos(θ) ) / m

a = g (sin(θ) - µ cos(θ))

What is the wavelength of a wave whose velocity is 12m/s and has a frequency of .75 Hz?

Answers

Answer:

16 meters

Explanation:

Use the formula that relates frequency velocity and wavelength:

velocity = wave-length x frequency

in our case:

12 m/s = wave-length * 0.75 Hz

wave-length= 12/0.75  m = 16 meters

A 100kg couch is being pushed with 196N of force. As it slides along the ground it experiences a coefficient of friction of 0.1. What is the net force in this situation?

A 300N
B 202N
C 398N
D 98N

Answers

Answer:98

Explanation:hope this helps!

what is the initial position of the object?
a. 2m b. 4m c. 6m d. 8m e. 10 m
what is the velocity of the object?
a. -10 m/s b. -5 m/s c. 0 m/s d. 5 m/s e. 10 m/s
which of the following is true?
a. the object increase its velocity.
b. the object decrease its velocity.
c. the objects velocity stays unchanged.
d. the object stays at rest.
e. more information is required.​

Answers

Answer:

e:10 m   b: -5 m/s b.The object decreases its velocity.

Explanation:

Ask For
6.
A horizontal 100 N force is applied to a 50 kg classmate resting on a level tile
floor. The coefficient of kinetic friction is 0.15.
a. Draw a force diagram to represent this situation.
b. What is the acceleration of the classmate?
c. Suppose the classmate was resting on a carpet where the coefficient of static
friction is 0.25. Is the horizontal 100 N force sufficient to cause the classmate to
accelerate? Draw a force diagram, and then explain why or why not.

Answers

I think c I’m not sure tho

There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N. The bell has ____________
energy.

Answers

Answer:

The bell has a potential energy of 8550 [J]

Explanation:

Since the belt is 45 [m] above ground level, only potential energy is available. And this energy can be calculated by means of the following equation.

[tex]E_{p}= W*h\\E_{p} = 190*45\\E_{p}=8550[J][/tex]

How high does a rocket have to go above the earth's surface to be subject to a gravitational field from the earth that is 50.0 percent of its value at the earth's surface?

A) 2.650 km
B) 3,190 km
C) 9.020 km
D) 12.700 km​

Answers

Answer:

A) 2.650 km

Explanation:

The relationship between acceleration of gravity and gravitational constant is:

[tex]g = \frac{Gm}{R^2}[/tex] ---- (1)

Where

[tex]R = 6,400 km[/tex] -- Radius of the earth.

From the question, we understand that the gravitational field of the rocket is 50% of its original value.

This means that:

[tex]g_{rocket} = 50\% * g[/tex]

[tex]g_{rocket} = 0.50 * g[/tex]

[tex]g_{rocket} = 0.5g[/tex]

For the rocket, we have:

[tex]g_{rocket} = \frac{Gm}{r^2}[/tex]

Where r represent the distance between the rocket and the center of the earth.

Substitute 0.5g for g rocket

[tex]0.5g = \frac{Gm}{r^2}[/tex] --- (2)

Divide (1) by (2)

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}/\frac{Gm}{r^2}[/tex]

[tex]\frac{g}{0.5g} = \frac{Gm}{R^2}*\frac{r^2}{Gm}[/tex]

[tex]\frac{1}{0.5} = \frac{1}{R^2}*\frac{r^2}{1}[/tex]

[tex]2 = \frac{r^2}{R^2}[/tex]

Take square root of both sides

[tex]\sqrt 2 = \frac{r}{R}[/tex]

Make r the subject

[tex]r = R * \sqrt 2[/tex]

Substitute [tex]R = 6,400 km[/tex]

[tex]r = 6400km * \sqrt 2[/tex]

[tex]r = 6400km * 1.414[/tex]

[tex]r = 9 049.6\ km[/tex]

The distance (d) from the earth surface is calculated as thus;

[tex]d = r - R[/tex]

[tex]d = 9049.6\ km - 6400\ km[/tex]

[tex]d = 2649.6\ km[/tex]

[tex]d = 2650\ km[/tex] --- approximated

A skier is moving 8.33 m/s when
he starts to slide UP a 8.44°
frictionless slope. How much time
does it take him to come to a stop?
(Unit = s)

Answers

Answer:

5.79

Explanation:

Vf=Vi+at

0=8.33+(-9.8)sin8.44t

t=8.33/(9.8sin8.44)=0.83/sin 8.44

=5.79

Carter's favorite ride at playland amusement park is the rollercoaster. The roller coaster
car and passengers have a combined mass of 1620kg and they descend the first hill at
an angle of 45.0 degrees to the horizontal. With what force is the rollercoaster pulled
down the hill?

Answers

Answer:

F = 11226.02 N

Explanation:

The roller coaster  car and passengers have a combined mass of 1620kg.

They descend the first hill at  an angle of 45.0 degrees to the horizontal.

We need to find force is the rollercoaster pulled  down the hill.

We firstly find the rectangular component of the downward. The force acting in the downward direction is mgsinθ such that,

F = mgsinθ

= 1620 × 9.8 × sin(45)

= 11226.02 N

So, the roller coaster is pulled down the hill with a force of 11226.02 N.

A hare can run at a rate of 15 m/s, while a turbocharged tortoise can now crawl at a rate of 3 m/s, how much of a head-start (time-wise) does the tortoise need in order to tie the hare in a 250 meter race?

A.
16.7 seconds

B.
66.7 seconds

C.
83.3 seconds

D.
100 seconds

Answers

Answer:

t = 66.7 s

Explanation:

Given that,

Speed of a hare, v = 15 m/s

Speed of a turbocharged tortoise, v' = 3 m/s

The hare in a 250 meter race

Let the Hare takes time t. It can be calculated as follows :

[tex]t=\dfrac{250}{15}=16.67\ s[/tex]

Let a turbocharged tortoise takes t'. It can be calulated as follows :

[tex]t'=\dfrac{250}{3}= 83.33\ s[/tex]

To tie the race, required time is given by :

[tex]\Delta t = t'-t\\\\=83.33-16.67\\\\=66.66\ s\\\\\approx 66.7\ s[/tex]

Hence, the correct option is (b) i.e. 66.7 seconds.

Calculate the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.

Answers

Answer:

RMS velocity, [tex]v_{rms}=5748.75\ m/s[/tex]

Explanation:

We need to find the RMS speed of helium atoms near the surface of the Sun at a temperature of about 5300 K.

The formula for RMS speed of a gas is given by :

[tex]v_{rms}=\sqrt{\dfrac{3RT}{m}}[/tex]

Where

R is ideal gas constant, R = 8.314 J /mol K

T = 5300 K

m is molar mass of Helium, [tex]m = 4\times 10^{-3}\ Kg/mol[/tex]

Substituting all the values in above formula :

[tex]v_{rms}=\sqrt{\dfrac{3\times 8.314\times 5300}{4\times 10^{-3}}}\\\\=5748.75\ m/s[/tex]

So, the RMS speed Helium atoms 5748.75 m/s.

If two exactly the same cars are driving down a road, which one would have the most kinetic energy. The one that is moving faster, the one that is moving downhill, the one that is moving uphill, or the one that is moving slower.

Answers

Answer: the car that is moving downhill

Explanation:

HELPP physics final will give brainliest

Answers

0.5 m/s^2

Please give me brainliest! You don’t have to though :3

Which scenario is an example of the transfer of thermal energy by radiation?

A. Water boils in a pan.

B. Hot air circulates in an oven.

C. An ice cube melts in a person's hand.

D. A frozen lake melts under the Sun.

Correct answer is D

Answers

Answer:

its D: A frozen lake melts under the sun.

Explanation:

Radiation is the transfer of heat energy through space by electromagnetic radiation. Most of the electromagnetic radiation that comes to the earth from the sun is invisible. Only a small portion comes as visible light. Light is made of waves of different frequencies.

d i hope this helps the person above me told you the right answer!!

A rocket will move upward as long as which condition applies?

Answers

The force of thrust is greater than the force if gravity !
Answer found on quizlet !

what is the force of an egg that is thrown at a brick wall if the egg has a mass of 0.3 kg and an acceleration of 50 m/s/s

Answers

Answer:

15N

Explanation:

F=ma so F=.3*50 therefore F=15N

The force of an egg that is thrown at a brick wall is equal to 15 N.

What is force?

Force can be defined as the influence or effect that changes the state of the body of from motion to rest or vice versa. The S.I. unit of force is Newton (N) as well as force is a vector quantity. Force can change the direction or the speed of the moving object.

The force acting on an object can be calculated from the multiplication of the mass(m) and acceleration(a). The mathematical form of the second law of motion for force can be written as follows:

F = ma

Given, the mass of the egg, m = 0.3 Kg

The acceleration of the egg with which it is thrown on the wall, a = 50 m/s²

The force of an egg that is thrown at a brick wall can be calculated as:

F = ma = 50 ×0.3 = 15 N

Learn more about force, here:

brainly.com/question/13191643

#SPJ2

Find the state of Georgia in the southeastern U.S. What can you say about the weather in this state?

Answers

The weather in Georgia isn’t like Florida, it follows along with spring, summer, autumn, winter.
I went there in March, expecting it to be warm, but it was actually fairly cold.
So it has a temperate climate.

(1-dimension) A fish has a mass of 6 kg and is moving at a speed of 4m/s to the right. What is its momentum?

Answers

Answer:

24 kg m/s

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 6 × 4

We have the final answer as

24 kg m/s

Hope this helps you

Allure of the seas is one of the most expensive cruise ships around the world with a length of 362 meters(1,187 ft) and a height of 72 meters(236 ft) above water line. On her first day of operation she moves with a uniform acceleration of 83.5 km/hr2 from rest has gone 10 nautical miles. How many seconds she is in motion? Note: 1 nautical mile = 1.852 km (help 3 mins left​

Answers

Answer:

  about 2398 seconds

Explanation:

The relation between time, distance, and acceleration is ...

  d = (1/2)at²

  t = √(2d/a) = √(2·10·1.852 km/(83.5 km/h²)) ≈ √0.4436 h ≈ 0.6660 h

That is about ...

  (0.6660 h)(3600 s/h) ≈ 2397.7 s

The cruise ship takes about 2397.7 seconds to cruise 10 nautical miles, accelerating all the way.

d. If a dog has a mass of 12 kg, what is its weight on Neptune?
11.7N/kg

Answers

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

a mass of 2.00 kg rest on a rough horizontal table. The coefficient of static friction between the block and the table is 0.60. The block is attached to a hanging mass by a string that goes over a smooth pulley,as shown in the diagram. Determine the largest mass that can hang in this way without forcing the block to slide.

Answers

Answer:

1.2 kg

__________________________________________________________

We are given:

Mass of the block = 2 kg

Coefficient of Static Friction = 0.6

__________________________________________________________

Friction Force on the Block:

Finding the Normal Force:

We know that the normal force will be equal and opposite to the weight of the 2 kg block

So, Normal Force = mg

replacing the variables with the given values

Normal Force = (2)(9.8)                    [Taking g = 9.8]

Normal Force = 19.6 N

Friction force on the Block:

We know that:

Coefficient of Static Friction =  Static Friction Force/Normal Force

replacing the variables

0.6 = Static Friction force / 19.6

Static Friction force = 0.6*19.6 N                 [Multiplying both sides by 19.6]

Static Friction force = 11.76 N

__________________________________________________________

Largest Mass that can Hang:

We know that the Static Friction force is 11.76 N, this means that a force of 11.76 N will be applied to keep the object at rest

So, if the weight of the second block is less than the static friction force, it will hang

Weight of the second block ≤ 11.76

We know that weight = mg

mg ≤ 11.76

m(9.8) ≤ 11.76                                                   [since g = 9.8]

m ≤ 1.2 kg                                                        [dividing both sides by 9.8]

From this, we can say that the maximum mass of the second block is 1.2 Kg

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular acceleration of 0.311 rad/s2 for 4.13 s. What is the drill's angular displacement during that time interval?

Answers

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, [tex]\omega _i[/tex] = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

[tex]\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad[/tex]

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

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