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3 years ago

Select the statement that correctly completes the description of phase difference.

Physics

1 answer:

KengaRu [80]3 years ago

3 0

Answer:

The difference in the phase angle between any two waves at any given position along the waves.

Explanation:

Phase difference describes the difference in the phase angle between any two waves at any given position along the waves. The equation of wave is given by :

Wave 1, $y=A\ sin(\omega t)$

Wave 2, $y=A\ sin(\omega t+\phi)$

Where

A is the amplitude of wave

$\omega$ is the angular frequency

$\phi$ is the phase difference between two waves

If $\Delta d$ is the path difference, it is given by :

$\Delta d=d\ sin\theta$

Hence, the correct option is (a) "the difference in the phase angle between any two waves at any given position along the waves".

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Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

Alina [70]

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
Uncertainty in momentum (∆P) = ?
Planck's constant (h) = 6.26 × 10⁻³⁴ Js

$\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi}$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192}$

$\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21}$

$\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21}$

$\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}$

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2 years ago

A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i

guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

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3 years ago

A 8 g object displaces 50ml of water, what is its specific gravity?

Igoryamba