1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
xxTIMURxx [149]
3 years ago
10

A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

frequency of the tuning fork by pulling out the "tuning joint" to lengthen her flute slightly.
Physics
2 answers:
laiz [17]3 years ago
7 0

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, \Delta f = 4\ beats/s

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

\Delta f = f - f_{i}

f_{i} = f \pm \Delta f

when

f_{i} = f + \Delta f = 523 + 4 = 527 Hz

when

f_{i} = f - \Delta f = 523 - 4 = 519 Hz

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

STatiana [176]3 years ago
4 0

Answer:

527 Hz

Explanation:

given,

flute beat = 4 beats/s

frequency of the tuning fork = 523 Hz

to find her initial frequency.

we know

f_{beat}=|f_1-f_2|

4=|f_1-f_2|

  f₁ - f₂ = ±4

  f₁ = 523 ± 4 Hz

now,

f₁ = 527 Hz    (or)  519 Hz

when flute length increase the frequency decreases.

hence, initial frequency is equal to 527 Hz  

You might be interested in
I need to write an acrostic poem on Globalisation and i need help. Does anyone have any ideas?
Aleonysh [2.5K]

Answer:

you could talk about the people and say something like change

Explanation:

5 0
3 years ago
A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 N. What must be done to f
amid [387]

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude <em>W</em> = <em>m g</em>

• the normal force, pointing perpendicular to the hill and away from the ground with mag. <em>N</em>

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector <em>F</em> pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. <em>F</em>.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = <em>W</em> (//) = <em>m</em> <em>a</em> (//)

∑ (⟂) = <em>W</em> (⟂) + <em>N</em> = <em>m </em><em>a</em> (⟂)

where, for instance, <em>W</em> (//) denotes the component of the sled's weight in the direction parallel to the hill, while <em>a</em> (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -<em>F</em> to the first equation.

If the hill makes an angle of <em>θ</em> with flat ground, then <em>W</em> makes the same angle with the hill so that

<em>W</em> (//) = -<em>m g </em>sin(<em>θ</em>)

<em>W</em> (⟂) = -<em>m g</em> cos(<em>θ</em>)

So we have

<em>-m g </em>sin(<em>θ</em>) = <em>m</em> <em>a</em> (//)   →   <em>a</em> (//) = -<em>g </em>sin(<em>θ</em>)

<em>-m g</em> cos(<em>θ</em>) + <em>N</em> = <em>m </em><em>a</em> (⟂)   →   <em>a</em> (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

<em>a</em> = <em>a</em> (//)

with magnitude

||<em>a</em>|| = <em>a</em> = <em>g </em>sin(<em>θ</em>).

6 0
3 years ago
A cruise ship is moving at constant speed through the water. The vacationers on the ship are eager to arrive at their next desti
MArishka [77]

Answer: 1. higher than it was before they started running

Explanation: As the vacationers run towards the back(stern) of the ship the exerting more pressure against the pressure exerted by the wave supporting the moving ship,the pressure exerted on the moving ship will be increased, leading to a slight increase in the speed of the ship compared to the speed before they started running towards the back(stern) of the ship.

6 0
3 years ago
Examine the nuclear reaction: Superscript 1 subscript 1 H plus superscript 1 subscript 0 n right arrow superscript 2 subscript 1
AysviL [449]

Answer:

C. A change has occurred in the nucleus.

Explanation:

6 0
3 years ago
Read 2 more answers
if the forces are moving in the same direction, ____ the forces. Please help i’m actually so confused!
jasenka [17]

Answer:

Explanation:

the directions may change

Or they will repel and become opposite sides

6 0
2 years ago
Other questions:
  • An element in group IIA would form a __________ ion while an element in group VIIA would form a(n) __________ ion.
    5·1 answer
  • Explain how changing protons, neutrons, and electrons affects atoms
    14·2 answers
  • A. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
    10·1 answer
  • Explain why the term “nuke it” is incorrect when referring to cooking food
    9·1 answer
  • the lever in the figure below accomplishes 5.0 newtons-meters of work raising a 25 newton weight. How far,in meters is the weigh
    7·1 answer
  • Why Ice Is Slippery?
    13·2 answers
  • What is the smallest thing in universe ?​
    8·1 answer
  • Need help now due in 2 minutes
    10·1 answer
  • What are 3 things that continental drifts have that plate tectonics dont
    9·1 answer
  • A passenger on a jet airplane claims to be able to walk at a speed in excess of 500 mph. Can this be true?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!