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
wariber [46]
3 years ago
6

A horizontal wire is stretched with a tension of 98.4 N, and the speed of transverse waves for the wire is 406 m/s. What must th

e amplitude of a traveling wave of frequency 60.5 Hz be for the average power carried by the wave to be 0.391 W? Give an answer in mm. Pay attention to the number of significant figures.
Physics
1 answer:
Anika [276]3 years ago
4 0

Answer:

Amplitude = 4.725 mm

Explanation:

We have got the following values :

T = 98.4 N

T is the wire tension

v = 406 m/s

v is the transverse waves speed

f = 60.5 Hz

f is the frequency

ΔU = 0.391 W

ΔU is the average power carried by the wave

If μ is  the mass per unit length of the wire ⇒ μ = \frac{T}{v^{2}}

If ω is the angular frequency of the wire ⇒ ω = 2π.f

So ω = 2π(60.5 Hz) and  ω units are s^{-1}

The equation that relates all is :

ΔU = (1/2).(T/v^2).(ω^2).(A^2).v

Where A is the wave amplitude

0.391 W=\frac{1}{2}.\frac{98.4 N}{(406\frac{m}{s})^{2}}.(2.pi.(60.5Hz))^{2}.A^{2}.406\frac{m}{s}

This will give us A in meters

A=4.7253(10^{-3})m\\A=4.725 mm

You might be interested in
A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,
Natasha_Volkova [10]

Answer:

the average force 11226 N  

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest  (Vo=0)

             

           Vf² = Vo² - 2gY

            Vf² = 0 - 2 9.8 7.69 = 150.7

            Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam.  They also give us the distance it travels before stopping, let's calculate the time

         

            Vf = Vo - g t

             0 = Vo - g t

             t = Vo / g

             t = 12.3 / 9.8

             t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

                 I = Δp

                F t = pf-po

The amount of final movement is zero because the system stops

                F = - po / t

                F = - mv / t

                F = - 1150 12.3 / 1.26

                F = -11226 N

This is the average force exerted by the stack on the vean

7 0
3 years ago
Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of
GalinKa [24]

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) =  Charge inside/ ∈₀

4 0
3 years ago
A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =
frutty [35]

Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

equation is given as

                   x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

                   x = A sin θ

           θ is the phase angle

      θ = π t + π/3

at   t = 2 s

      \theta =\pi \times 2 +\dfrac{\pi}{3}

      \theta =\dfrac{7\pi}{3}\ rad

Phase of the motion at t =2 s is θ = (7π / 3) rad

7 0
3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
Which best describes the way a sound wave is sent trough the radio?
horsena [70]
Sound wave > electric signal > radio wave > sound wave
4 0
3 years ago
Read 2 more answers
Other questions:
  • Under what conditions will the projectile have the greatest velocity when it hits the ground?
    9·1 answer
  • Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.
    11·1 answer
  • A gas is compressed from 600cm3 to 200cm3 at a constant pressure of 450kPa . At the same time, 100J of heat energy is transferre
    14·1 answer
  • 10-cm-long wire is pulled along a U-shaped conducting rail in a perpendicularmagnetic field. The total resistance of the wire an
    6·1 answer
  • The primary coil of a transformer has N1 = 275 turns, and its secondary coil has N2 = 2,200 turns. If the input voltage across t
    8·1 answer
  • What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface
    12·1 answer
  • How do you make a sound wave’s pitch lower?
    13·1 answer
  • Please help.................​
    9·2 answers
  • How do organisms use communication to survive?
    10·1 answer
  • PLEASE HELP a current of 0.5 A flows with a resistance of 50 ohms what is the e.m.f applied NEED HELP ASAP AND BRAINLIEST​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!