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3 years ago

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The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

is string has a frequency of 300 hz and a wavelength of 0.76 m. what is the length of the string?

1 answer:

bagirrra123 [75]3 years ago

4 0

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

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A compact disc (CD) stores music in a coded pattern of tiny pits 10⁻⁷m deep. The pits are arranged in a track that spirals outwa

Serggg [28]

a) Angular speed of the innermost part: 50 rad/s, outermost part: 21.6 rad/s

b) The length of the track would be 5,550 m

c) The average angular acceleration is $-6.4\cdot 10^{-3}rad/s$

Explanation:

a)

For an object in uniform circular motion, the relationship between angular speed and linear speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circle

Here the linear speed of the track is constant:

v = 1.25 m/s

For the innermost part of the disk,

r = 25.0 mm = 0.025 m

So the angular speed is

$\omega_i = \frac{v}{r_i}=\frac{1.25}{0.025}=50 rad/s$

For the outermost part,

r = 58.0 mm = 0.058 m

So the angular speed is

$\omega_o = \frac{v}{r_o}=\frac{1.25}{0.058}=21.6 rad/s$

b)

The maximum playing time of the disk is

$t=74.0 min \cdot (60 s/min)=4440 s$

If the track was stretched out in a straight line, the motion of the track would be a uniform motion (since it is moving at constant speed), so we can use the equation

$v=dt$

where

v is the linear speed

d is the length coverted in a straight line

t is the time

Substituting v = 1.25 m/s and solving for d,

$d=vt=(1.25)(4440)=5550 m$

c)

The angular acceleration is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular speed

$\omega_i$ is the initial angular speed

t is the time

A CD is read from the innermost part to the outermost part, so we have:

$\omega_i = 50 rad/s$ (at the innermost part)

$\omega_f = 21.6 rad/s$ (at the outermost part)

t = 74.0 min = 4440 s is the time

Substituting, the average angular acceleration is

$\alpha = \frac{21.6-50.0}{4440}=-6.4\cdot 10^{-3}rad/s$

where the negative sign means the CD is decelerating.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

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2 years ago

) A 73-mH solenoid inductor is wound on a form that is 0.80 m long and 0.10 m in diameter. A coil having a resistance of is tigh

Aleonysh [2.5K]

Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil

Answer:

6.169 μA

Explanation:

Formula for induced EMF is given by the equation;

EMF = M(di/dt). We are given;

di/dt = 2.5 A/s

M = 19μH = 19 × 10^(-6) H

Thus;

EMF = 19 × 10^(-6) × 2.5.

EMF = 47.5 × 10^(-6) V

Formula for current is;

i = EMF/R. R is resistance given as 7.7 ohms.

Thus; i = 47.5 × 10^(-6)/7.7

i = 6.169 μA

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2 years ago

In order to walk barefoot on hot coals without hurting your feet

siniylev [52]

Before a person walks through burning coal, the person will make sure their feet are very wet. When they start walking on the coal, this moisture will evaporate and form a protective gas layer underneath the person's feet. You can see examples of this if you happen to drip some water on a hot stove or any very hot surface. The water will very easily glide around on top of a newly formed layer of air underneath it -- like air hockey pucks on an air hockey table. Note that when someone walks through burning coal, typically this is also done very quickly to prevent a great deal of exposure to possible harm. By walking quickly, thinking positively, and letting the water cushion you from immediate danger over a short distance, such a task is possible. You may have also heard of physics teachers demonstrating how this principle works by sticking their hand first in a bucket of water and then quickly in a bucket of boiling molten lead. In the lead, their hand is protected briefly by a layer of gas from the evaporated water (the water vapor). I'm fairly sure that there is a name for this particular layer of gas, but I'm afraid the name is beyond me at the moment. In other words, water vapor has a low heat capacity and poor thermal conduction. Very often, the coals or wood embers that are used in fire walking also have a low heat capacity. Sweat produced on the bottom of people's feet also helps form a protective water vapor. All of this together makes it possible, if moving quickly enough, to walk across hot coals without getting burned. WARNING: Do not attempt to perform any of the actions described above. You can seriously injure yourself. Answered by: Ted Pavlic, Electrical Engineering Undergrad Student, Ohio St. (citing my source)

5 0

3 years ago

What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc

Murljashka [212]

Answer:

The Value is $V_{AB} = 2.825V$

Explanation:

The explanation is shown on the first uploaded image

5 0

3 years ago

An electromagnetic flowmeter is useful when it is desirable not to interrupt the system in which the fluid is flowing (e.g. for

AURORKA [14]

Answer:

<em>2 m/s</em>

<em></em>

Explanation:

The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as

$E = Blv$

where $E$ is the induced voltage = 2.88 mV = 2.88 x 10^-3 V

$l$ is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m

$v$ is the velocity of the fluid through the field = ?

$B$ is the magnetic field = 0.120 T

substituting, we have

2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x $v$

2.88 x 10^-3 = 1.44 x 10^-3 x $v$

$v$ = 2.88/1.44 = <em>2 m/s</em>

8 0

3 years ago