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3 years ago

Give some example acostic using "ELEMENTS"

Physics

1 answer:

Jet001 [13]3 years ago

5 0

E=ERBIUM
L=LITHIUM
E=EINSTEINIUM
M=MAGNESIUM
E=EUROPIUM
N=NITROGEN
T=TITANIUM
S=SULPHURDIOXIDE

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What effects do coil of wire on a compass after it has been connected to a battery

Yanka [14]

Explanation:

The compass needle moved when the wire was connected to the battery. The important point here is that the needle is affected by the wire only when both ends of the wire are connected to the battery because only at this time is current flowing through the circuit.

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3 years ago

Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pres

irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

Learn more about pressure here: brainly.com/question/28012687

#SPJ1

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1 year ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

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

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is: