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2 years ago

What is the necessary conditions for the production of sound?

Physics

1 answer:

TEA [102]2 years ago

8 0

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

Answer: Something that's vibrating, and you also need medium for those vibrations to start in.

I hope this helped!

<!> Brainliest is appreciated! <!>

- Zack Slocum

*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆*――*☆**☆*――*☆*――*☆*――*☆

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3.5-m-diameter merry-go-round is rotating freely with an angular velocity of 0.78 rad/s . Its total moment of inertia is 1710 kg

Monica [59]

I dont get the question your asking

4 0

3 years ago

A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught

meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

$y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}$ (Eq. 1)

Where:

$y_{o}$ - Initial height of the softball, measured in meters.

$y$ - Final height of the softball, measured in meters.

$v_{o}$ - Initial velocity of the softball, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

If we know that $y = y_{o}$ , $t = 3.56\,s$ and $g = -9.807\,\frac{m}{s^{2}}$ , the initial velocity of the softball is:

$v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0$

$3\cdot v_{o} -44.132\,m= 0$

$v_{o} = 14.711\,\frac{m}{s}$

The initial velocity of the softball is 14.711 meters per second.

8 0

3 years ago

The capacitor can withstand a peak voltage of 590 volts. If the voltage source operates at the resonance frequency, what maximum

kirill115 [55]

Answer:

The maximum voltage is 41.92 V.

Explanation:

Given that,

Peak voltage = 590 volts

Suppose in an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20×10^{-2}\ \mu F$ .

We need to calculate the resonance frequency

Using formula of frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.88\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi\times f\times C\times V_{c}$

$I=2\pi\times2356.88\times1.20\times10^{-8}\times590$

$I=0.1048\ A$

Impedance of the circuit is

$z=\sqrt{R^2+(X_{L}^2-X_{C}^2)}$

At resonance frequency $X_{L}=X_{C}$

$Z=R$

We need to calculate the maximum voltage

Using ohm's law

$V=I\times R$

$V=0.1048\times400$

$V=41.92\ V$

Hence, The maximum voltage is 41.92 V.

4 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

$E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}$

Substitute x=a and R=a

Then, we get

$E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}$

$E=\frac{KQa}{2\sqrt 2a^3}$

$E=\frac{KQ}{2\sqrt 2a^2}$

Where K= $9\times 10^9 Nm^2/C^2$

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center= $\frac{KQ}{2\sqrt 2a^2}$

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3 years ago

A constant power is supplied to a rotating disc .the relationship of angular velocity of disc and number of rotations made by th

Bogdan [553]

A rotating disc supplied with constant power where the relationship of the angular velocity of the disc and the number of rotations made by the disc is governed by Newton's second law for rotation. This law is specially made for rotating bodies which is extracted from Newton's second law of motion.

6 0

3 years ago