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3 years ago

5

How is amplitude changed in an instrument or tuning fork

1 answer:

azamat3 years ago

7 0

Amplitude is affected by the energy wave in the instrument. High energy wave means high amplitude and low energy wave means low amplitude.

<u>Explanation:</u>

The amplitude of a periodic variable is a measure of its change over a single period. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amount of energy carried by a wave is related to the amplitude of the wave. Amplitude of an instrument is directly affected by the wave of the energy in the instruments. High energy wave means high amplitude and low energy wave means low amplitude in the instrument.

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fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

$\alpha+\rho+\tau=1$

where:

$\alpha=$ absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

$\rho=$ reflectivity is defined as the ratio of reflected radiation to the total irradiation

$\tau=$ transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

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3 years ago

A golfer hits her tee-shot due north towards the fairway. Her shot has an initial velocity of 60 m/s. A 15 m/s wind is blowing i

vovangra [49]

Answer:

$c=71.4m/s$

$\theta=8.54\textdegree$

Explanation:

From the question we are told that

Initial velocity of 60 m/s

Wind speed $V_w= 15 m/s \angle 45 \textdegree$

Generally Resolving vector mathematically

$sin(45\textdegree)15=10.6\\cos(45\textdegree)15=10.6$

Generally the equation Pythagoras theorem is given mathematically by

$c^2=a^2+b^2$

$c^2=10.6^2 +(10.6+60)^2$

$c=\sqrt{10.6^2 +(10.6+60)^2}$

Therefore Resultant velocity (m/s)

$c=71.4m/s$

b)Resultant direction

Generally the equation for solving Resultant direction

$\theta=tan^-1(\frac{y}{x})$

Therefore

$\theta=tan^-1(\frac{10.6}{70.6})$

$\theta=8.54\textdegree$

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2 years ago

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IRINA_888 [86]

Answer:

A

Explanation:

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2 years ago

Motions need an unbalanced net force to maintain.<br> True or False?

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Answer:

The answer is False

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3 years ago

Read 2 more answers

The escape speed from an object is v2 = 2GM/R, where M is the mass of the object, R is the object's starting radius, and G is th

Rom4ik [11]

Answer:

Approximate escape speed = 45.3 km/s

Explanation:

Escape speed

$v=\sqrt{\frac{2GM}{R}}$

Here we have

Gravitational constant = G = 6.67 × 10⁻¹¹ m³ kg⁻¹ s⁻²

R = 1 AU = 1.496 × 10¹¹ m

M = 2.3 × 10³⁰ kg

Substituting

$v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.3\times 10^{30}}{1.496\times 10^{11}}}=4.53\times 10^4m/s=45.3km/s$

Approximate escape speed = 45.3 km/s

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3 years ago