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2 years ago

Comparing sound waves with different frequencies and wavelengths traveling through air,

Physics

1 answer:

aleksklad [387]2 years ago

4 0

Explanation:

The speed of sound wave only depends on the property of the medium like density and the bulk modulus of the medium particle. The speed of sound also depends on the temperature of the medium.

On comparing sound waves with different frequencies and wavelengths traveling through air, the speed of the wave doesn’t depend on the frequency or the wavelength. Hence, the correct option is (1).

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Look at the illustration , what concepts of heat are being shown in this pictures <br>

Phoenix [80]

<h2> (っ◔◡◔)っ ♥ Hey Honey Here<em>!</em> ♥</h2>

Answer:

♥ Convection.

♥ Conduction.

♥ Radiation.

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3 0

2 years ago

Point charge q1 of 30 nC is separated by 50 cm from point charge q2 of -45 nC. As shown in the diagram, point a is located 30 cm

Angelina_Jolie [31]

Answer:

E1 = 2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

=8.99 x 10^9 x 30 x 10^-9/(30x10^-2)^2

= 2996.667N/C

E2 = kQ2/r^2

= 8.99 x 10^9 x 50 x 10^-9/(20x10^-2)^2

= 11237.5N/C

The direction are towards the point a

6 0

2 years ago

A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition

Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

$\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}$

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

$\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\$

6 0

2 years ago

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

Learn More:

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0

3 years ago

Read 2 more answers

Can someone explain to me #4.

nalin [4]

A) We balance the masses: 4(1.00728) vs 4.0015 + 2(0.00055)4.02912 vs. 4.0026This shows a "reduced mass" of 4.02912 - 4.0026 = 0.02652 amu. This is also equivalent to 0.02652/6.02E23 = 4.41E-26 g = 4.41E-29 kg.
b) Using E = mc^2, where c is the speed of light, multiplying 4.41E-29 kg by (3E8 m/s)^2 gives 3.96E-12 J of energy.
c) Since in the original equation, there is only 1 helium atom, we multiply the energy result in b) by 9.21E19 to get 3.65E8 J of energy, or 365 MJ of energy.

4 0

2 years ago