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NeX [460]
3 years ago
6

Please help ASAP!!!!!!!!

Physics
1 answer:
liq [111]3 years ago
7 0

Answer:

the answer is STELLAR

Explanation:

im learning this rn in schhol

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olga nikolaevna [1]

Answer: 7.89141414

Explanation:

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3 years ago
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A microphone is attaced to a spring that is suspended from
kirill115 [55]

To solve this problem we will apply the concept related to the amplitude and the Doppler effect. The difference between the maximum and minimum frequency detected by the microphone would be given by the mathematical function

f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})

Here,

f_s= Source Frequency

v_0 = Speed of microphone

v = Speed sound

f_{max} - f_{min} = 2 f_s (\frac{v_0}{v})

Maximum speed of the microphone is

v_0 = f_{max} - f_{min} * \frac{v}{2f_s}

v_0= \frac{2.1Hz*343m/s}{2*440Hz}

v_0= 0.8185 m/s

Now the amplitude is

A = \frac{v_0}{2\pi/T}

Here T means the Period, then

A= \frac{0.8185 * 2.0}{2\pi}

A= 0.2605m

Therefore the amplitude of the simple harmonic motion is 0.2605m

3 0
4 years ago
A sanding disk with rotational inertia 3.8 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit
Elis [28]

Answer:

(a) Angular momentum of disk is 1.343\, \frac{kg.m^{2}}{s}

(b) Angular velocity of the disk is 353\frac{rad}{s}

Explanation:

Given

Rotational inertia of the disk , I=3.8\times 10^{-3}kg.m^{2}

Torque delivered by the motor , \tau =17N.m

Torque is applied for duration , \Delta t=79ms=0.079s

(a)

Magnitude of angular momentum of the disk = Angular impulse produced by the torque

\therefore L=\tau \Delta t=17\times 0.079\frac{kg.m^{2}}{s}

=>L=1.343\, \frac{kg.m^{2}}{s}

Thus angular momentum of disk is 1.343\, \frac{kg.m^{2}}{s}

(b)

Since Angular momentum , L=I\omega

where \omega= Angular velocity of the disk

=>\omega =\frac{L}{I}=\frac{1.343}{3.8\times 10^{-3}}\frac{rad}{s}

\therefore \omega =353\frac{rad}{s}

Thus angular velocity of the disk is 353\frac{rad}{s}

5 0
3 years ago
Drag the positive or negative feedback loop on the left to each process on the right. terms may be used once, more than once, or
slamgirl [31]

The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

<h3>What is a feedback loop?</h3>

A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

Both negative and positive feedback loops are possible. Insofar as they stay within predetermined bounds, negative feedback loops are self-regulating and helpful for sustaining an ideal condition. One of the most well-known examples of a self-regulating negative feedback loop is an old-fashioned home thermostat that turns on or off a furnace using bang-bang control.

To learn more about feedback loop, visit:

brainly.com/question/11312580

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5 0
2 years ago
A flywheel is a solid disk that rotates about an axis that is perpendicular to the disk at its center. Rotating flywheels provid
Sergeeva-Olga [200]

Answer:

8.37*10^5 rpm

Explanation:

Given that rotational kinetic energy = 4.66*10^9J

Mass of the fly wheel (m) = 19.7 kg

Radius of the fly wheel (r) = 0.351 m

Moment of inertia (I) = \frac{1}{2} mr ^2

Rotational K.E is illustrated as (K.E)_{rt} = \frac{1}{2} I \omega^2

\omega = \sqrt{\frac{2(K.E)_{rt}}{I} }

\omega = \sqrt{\frac{2(KE)_{rt}}{1/2 mr^2} }

\omega = \sqrt{\frac{4(K.E)_{rt}}{mr^2} }

\omega = \sqrt{\frac{4*4.66*10^9J}{19.7kg*(0.351)^2} }

\omega = 87636.04

\omega = 8.76*10^4 rad/s

Since 1 rpm = \frac{2 \pi}{60}  rad/s

\omega = 8.76*10^4(\frac{60}{2 \pi})

\omega = 836518.38

\omega = 8.37 *10^5 rpm

3 0
3 years ago
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