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
bearhunter [10]
4 years ago
11

A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can t

he blade of a screwdriver be placed against the string without disturbing the amplitude of the vibrationa. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelengthe. The length of the string is equal to four times the wavelength
Physics
1 answer:
mixas84 [53]4 years ago
5 0

Answer:

d. The length of the string is equal to one-half of a wavelength

Explanation:

A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration

a. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelength

e. The length of the string is equal to four times the wavelength

A stretched string of length L fixed at both ends is vibrating in its third harmonic H

How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration

d. The length of the string is equal to one-half of a wavelength

There are two points during vibration , the node and the antinode

the node is the point where the amplitude is zero.

from the third harmonics, there are two nodes. The first node is half of the wavelength which is the closest to the fixed point.

for third harmonics=3/2lamda

You might be interested in
A radio signal has a frequency of 1.023 x 108 HZ. If the speed of the signal in air is 2.997 x 108m/s, what is the wavelength of
Sladkaya [172]

Answer:

2.93 m  (which agrees with answer "C" on the list)

Explanation:

Recall that the speed of the wave equals the product of the wave's length times its frequency. Therefore, the wavelength is going to be the quotient of the speed of the signal divided its frequency:

Wavelength = 2.997  10^8 / 1.023  10^8 =  2.93 m

5 0
3 years ago
What is a simple pendulum?
soldier1979 [14.2K]

Answer:

A pendulum is weight suspended from a pivot so that it can swing freely

8 0
4 years ago
1. Use Newton's first law of motion to explain how wearing a seat belt in a moving car could help prevent injury
hjlf
The diver is plummiting to earth and if he does not put it soon he is going to die
4 0
3 years ago
What must be the distance (in meters) between a point charge q1 = 16 μC and a point charge q2 = 32 μC for the electrostatic forc
adoni [48]

Answer:

d = 0.71 meters

Explanation:

It is given that,

Charge 1, q_1=16\ \mu C=16\times 10^{-6}\ C

Charge 2, q_2=32\ \mu C=32\times 10^{-6}\ C

Electrostatic force between charges, F = 9 N

Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :

F=k\dfrac{q_1q_2}{d^2}

d=\sqrt{\dfrac{kq_1q_2}{F}}

d=\sqrt{\dfrac{9\times 10^9\times 16\times 10^{-6}\times 32\times 10^{-6}}{9}}

d = 0.71 meters

So, the distance between the charges is 0.71 meters. Hence, this is the required solution.

6 0
4 years ago
A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it
lora16 [44]

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

5 0
3 years ago
Other questions:
  • In a parallel circuit, if bulb #2 were to blow out, bulb #1 would stay lit or go out?
    5·1 answer
  • Compared to its mass on earth, the mass of a 10-kg object on the moon is
    10·1 answer
  • Many compounds of living things, such as carbon dioxide, are formed by which kind of bond?
    7·2 answers
  • The motion map shows an object’s position and velocity at given times.
    8·2 answers
  • A 2.51 kg ball is attached to a ceiling by a 1.19 m long string. The height of the room is 3.45 m. The acceleration of gravity i
    9·1 answer
  • A 100-coil spring has a spring constant of 540 N/m. It is cut into four shorter springs, each of which has 25 coils. One end of
    14·1 answer
  • While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
    5·2 answers
  • Add the following vectors using head-to-tail method and verify your results using the component method.
    7·1 answer
  • A baseball bat exerts a 1500 N force on a baseball for 50 ms. What is the magnitude of the impulse that the bat exerts on the ba
    6·1 answer
  • Isotopes _____ at very different rates.
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!