A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?

Answers

Answer 1
Answer:

Answer:

ratio of the piccolo's length to the flute's length is 0.4916

Explanation:

given data

frequency of piccolo = 522.5 Hz

frequency of flute = 256.9 Hz

to find out

ratio of the piccolo's length to the flute's length

solution

we get here length of tube that is express as

length of tube = velocity of sound ÷ fundamental frequency .......................1

so here ratio of Piccolo length to flute that is

(L\ picco)/(L\ flute)  = (f\ flute)/(f \ piccolo)

(l \ piccolo)/(L\ flute) = (256.9)/(522.5)  =  0.4916

so ratio of the piccolo's length to the flute's length is 0.4916


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A ride at an amusement park moves the riders in a circle at a rate of 6.0 m/s. If the radius of the ride is 9.0 meters, what is the acceleration of the ride?4.0 m/s2
0.67 m/s2
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54 m/s2

Answers

4.0 m/s2


it's 9 squared divided by 6
the answer is B 4.0 m/s2

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .

Answers

Complete question:

An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: F₁ = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force F₂.

Answer:

(a) The net force of the electron, ∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ = 4.8 x 10⁻¹⁷ N

Explanation:

Given;

initial velocity of the electron, v_0 = +6.18 x 10⁵ m/s

final velocity of the electron, v_f = 2.59 x 10⁶ m/s

the distance traveled by the electron, d = 0.0708 m

The first electric force, F_1 = 8.87*10^(-17) \ N

(a) The net force of the electron is given as;

∑F = F₁ - F₂ = ma

where;

a is the acceleration of the electron

a = (v_f^2 -v_0^2)/(2d) \n\na = ((2.59*10^6)^2 -(6.18*10^5)^2)/(2(0.0708))\n\na = 4.468*10^(13) \ m/s^2

∑F = ma = (9.11 x 10⁻³¹ kg)(4.468 x 10¹³)

∑F = 4.07 x 10⁻¹⁷ N

(b) the electric force, F₂ is given as;

∑F = F₁ - F₂

F₂ = F₁ - ∑F

F₂ = 8.87 x 10⁻¹⁷ - 4.07 x 10⁻¹⁷

F₂ = 4.8 x 10⁻¹⁷ N

Final answer:

The problem involves calculating the acceleration of an electron, then using Newton's second law to find the net force on the electron. This is used to find the magnitude of a second electric force acting on the electron.

Explanation:

First, we can calculate the acceleration of the electron using the formula a = Δv/Δt, where 'a' is acceleration, 'Δv' is the change in velocity, and 'Δt' is the change in time. In this case, Δv = vf - vi = 2.59 x 106 m/s - 6.18 x 105 m/s = 1.972 x 106 m/s. The time taken by the electron to travel 0.0708 m can be found using the equation d = vi t + 0.5 a t₂. We use these values to get Δt which we use to find 'a'.

Next, let's use Newton's second law F = ma to find the net force acting on the electron. The only forces acting on the electron are electric forces, and we know one them is 8.87 x 10-17 N. If we designate this known force as F₁ then the total force F total = F₁ + F₂ where F₂ is the unknown electric force.

Finally, we can find F₂ = F total - F₁. This gives the magnitude of the second electric force.

Learn more about Physics - Electric Forces here:

brainly.com/question/33422584

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Energy can be transferred from a closed system to the surroundings by: (A) Internal chemical reactions (B) Heat (C) Shaft work (D) Change in pressure without changing volume (E) Mass transfer

Answers

Answer:

option the correct is B

Explanation:

Let's analyze the different options, for a closed system

- an internal reaction changes the system, but does not affect the surrounding environment

- Heat, is a means of transfer that occurs when two bodies are in contact, one of the body can be a closed system since the only thing that happens is thermal transfer, without movement of the system itself. This is the correct result.

- Work implies a movement whereby the system must be mobile, it is not an option

- Pressure change. change in the system, but does not affect the environment

- Mass transfer is not possible in a closed system

After analyzing each option the correct one in B

A boy and a girl are pulling a heavy crate at the same time with 7 units of firce each. What is the net force acts on the ibject? Is the object balanced or unbalanced?

Answers

Answer:

Net force= 14 units

The object is unbalanced

Explanation:

The net force refers to the sum of all forces applied to an object. However, the direction of force applied determine the net force. In this question, a boy and girl is pulling a heavy crate at the same time.

This means that the force is in the same direction, hence, the net force will be:

F(N) = 7 + 7 = 14 unit

However, since the pull is occuring at the same direction. This means that the object has a net force, therefore, will move in a particular direction. This means that the OBJECT IS UNBALANCED

How long does it take a wheel that is rotating at 33.3 rpm to speed up to 78.0 rpm if it has an angular acceleration of 2.15 rad/ s 2?

Answers

initial angular speed is given by 33.3 rpm

w_0 = 2\pi (33.3)/(60)

w_0 = 3.49 rad/s

final angular speed is given by 78 rpm

w_f = 2\pi (78)/(60)

w_f = 8.17 rad/s

now by using kinematics we will have

w_f = w_0 + \alpha * t

8.17 = 3.49 + 2.15 * t

t = 2.17 seconds

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).