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elena-s [515]
3 years ago
12

A string with tension Ft= 10 N, with one end pinned and the other end free, is used to produce a wave supplied by a driving freq

uency. (a) If the frequency with which you produce waves is f= 10KHz and the wavelength of the wave is a = 0.01 m, what is the velocity with which the wave? (5 points) (b) What is the period of the wave? (5 points) (c) What is the linear mass density of the string? (5 points)
Physics
1 answer:
Hitman42 [59]3 years ago
5 0

To solve the problem it is necessary to apply the theory of sine waves. The wavelength λ of a sinusoidal waveform traveling at constant speed v is given by

\lambda = \frac{v}{f}

Where,

v= velocity

f = frequency

Re-arrange to find v, we have

v = f\lambda

PART A ) Replacing with our values we have,

v = (10*10^3)(0.01)

v = 100m/s

PART B) In the case of the period we know that it is defined as a function of frequency as,

T= \frac{1}{f}

That is to say that using the previously given values we have that the period in seconds is,

T=1*10^{-4}s

PART C) Finally the transverse wave velocity is given by,

v=\sqrt{\frac{T}{\mu}}

Where,

T= Period

\mu = Linear mass density

Re-arrange to find \mu,

\mu = \frac{T}{v^2}

\mu = \frac{1*10^{-4}}{100^2}

\mu = 1*10^{-8}kg/m

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Give two ways of reversing the direction of the forces on the coil in the electric motor?​
nikklg [1K]

Answer:

Interchanging the poles of the magnet

Reversing the direction of the applied current

Explanation:

  1. The working of the electric motor is associated with Fleming's left-hand rule.
  2. It states that if a current-carrying conductor is placed inside a magnetic field, it experiences a force in the direction perpendicular to the direction of the electric current and magnetic field.
  3. These three physical quantities are placed in a mutually perpendicular direction.
  4. So, in order to reverse the direction of force, you have to either change the direction of the current or magnetic field.
4 0
3 years ago
A 22.0 kg child slides down a slide that makes a 37.0° angle with the horizontal. (a) What is the magnitude of the normal force
patriot [66]

Answer:

(a) 172.185 N

(b) 53^{\circ}

Solution:

As per the question:

Mass of the child, m = 22.0 kg

Angle, \theta = 37.0^{\circ}

Now,

(a) The magnitude of the normal force exerted by the slide on the child:

F_{N} = mgcos\theta

F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N

Now,

(b) The angle from the horizontal at which the force is directed is:

90^{\circ} - 37^{\circ} = 53^{\circ}

6 0
4 years ago
what were two weakness of america government under the articals of confederation did the philadelphia mutiny highlight
marusya05 [52]

The weaknesses of the America government under the articals of confederation in which the Philadelphia mutiny highlighted include:

  • Inability to regulate commerce.
  • Inability to form a military.

<h3>What is a Mutiny?</h3>

This is a rebellious act among group of people in order to oppose a certain authority or government.

The government couldn't give soldiers a good working condition due to lack of funds which led to mutiny.

Read more about Mutiny here brainly.com/question/1157738

#SPJ1

4 0
2 years ago
During the compression stroke of an internal combustion engine, _____
Leona [35]

easy, The fuel is ignited

6 0
3 years ago
A confined aquifer with a porosity of 0.15 is 30 m thick. The potentiometric surface elevation at two observation wells 1000 m a
AlekseyPX

Answer:

Part (a) The flow rate per unit width of the aquifer is 1.0875 m³/day

Part (b) The specific discharge of the flow is 0.0363 m/day

Part (c) The average linear velocity of the flow is 0.242 m/day

Part (d) The time taken for a tracer to travel the distance between the observation wells is 4132.23 days = 99173.52 hours

Explanation:

Part (a) the flow rate per unit width of the aquifer

From Darcy's law;

q = -Kb\frac{dh}{dl}

where;

q is the flow rate

K is the permeability or conductivity of the aquifer = 25  m/day

b is the aquifer thickness

dh is the change in th vertical hight = 50.9m - 52.35m = -1.45 m

dl is the change in the horizontal hight = 1000 m

q = -(25*30)*(-1.45/1000)

q = 1.0875 m³/day

Part (b) the specific discharge of the flow

V = \frac{Q}{A} = \frac{q}{b} = -K\frac{dh}{dl}\\\\V = -(25 m/d).(\frac{-1.45 m}{1000 m}) = 0.0363 m/day

V = 0.0363 m/day

Part (c) the average linear velocity of the flow assuming steady unidirectional flow

Va = V/Φ

Φ is the porosity = 0.15

Va = 0.0363 / 0.15

Va = 0.242 m/day

Part (d) the time taken for a tracer to travel the distance between the observation wells

The distance between the two wells = 1000 m

average linear velocity = 0.242 m/day

Time = distance / speed

Time = (1000 m) / (0.242 m/day)

Time = 4132.23 days

        = 4132.23 days *\frac{24 .hrs}{1.day} = 99173.52, hours

4 0
3 years ago
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