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Dahasolnce [82]
3 years ago
6

This graph is accurate because A) mechanical waves cannot travel through a medium. B) electromagnetic waves cannot travel throug

h a vacuum. C) electromagnetic waves move faster than mechanical waves. D) mechanical waves move faster than electromagnetic waves.

Physics
2 answers:
aleksklad [387]3 years ago
6 0

Answer:

Option c

Explanation:

Electromagnetic waves are the waves which can travel in vacuum unlike mechanical waves which require a medium to travel. Electromagnetic waves have highest speed in vacuum. In fact, no thing can have speed greater than the speed of EM waves in vacuum. EM waves, thus move faster than the mechanical waves. EM waves carry energy by oscillations of magnetic and electric fields perpendicular to the direction of motion.

il63 [147K]3 years ago
4 0

Answer:

IT would be C.electromagnetic waves move faster than mechanical waves.

Explanation:

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Katarina [22]

Answer:

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Explanation:

5 0
2 years ago
Waves diffract the most when their wavelength is
djverab [1.8K]
<h2>Answer: about the same size of the gap  or slit</h2>

Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.

Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.

In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.

Therefore:

<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap </u></h2>

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5 0
3 years ago
A bimetallic strip (brass/steel), which is straight at room temperature, will be immersed in boiling water and allowed to equili
lakkis [162]

The thermal expansion of the materials allows to find the deflection of the bimetallist strip is Δy = 3.48 cm

given paramers

    * Bimetallic brass / steel tape

    * Initial temperature, room temperature T = 20ºC

    * Final temperature, boiling water  = 100ºC

    * initial length L₀ = 222mm (1cm / 10mm) = 22.2cm

    * thickness of bimetallic tape e = 0.036 inch (2.54 cm/1 inch) = 0.0914 cm

to find

    * perpendicular deviation or deflection (Δy)

Thermal expansion is the phenomenon of change in the length of a body due to the change in temperature, due to the increase in the length of the atomic and molecular bonds, macroscopically it is described by

        ΔL = α L₀ ΔT

ΔL and ΔT are the variation of the length and temperature respectively, L₀ is the initial length and α the coefficient of expansion ends.

In this case we have a strip formed by two materials with different coefficient of thermal expansion,

Brass       α_{brass}   = 19 10⁻⁶ ºC⁻¹

Steel       α_{steel}    = 11 10⁻⁶ ºC⁻¹

In the attached we can see a diagram of the process, as the temperature increases, the material with greater thermal expansion lengthens more, so the system must curve towards the center of the material with less

thermal expansion. Let's find the length of the strip for each material

brass          L_{f brass} - L₀ = α_{brass} L₀ ΔT

Steel           L_{f steel} - L₀ = \alpha_{steel} L₀ ΔT

Note that the initial length is the same for the two materials and that the strip is in thermal equilibrium at room temperature.

If we assume that we have an arc of circumference, we can write the length of the arc

        θ = L / r

where θ is the angle in radines, L the length of the arc and r the radius of curvature, let's write this equation for each material

brass     L_{f \ brass} =θ r₁

steel      L_{f \ steel} = θ r₂

we substitute in our equations

           θ r₁ - L₀ = α_{brass} L₀ ΔT

           θ r₂ - L₀ = α_{steel} L₀ ΔT

let's subtract the two equations

           θ (r₁- r₂) = L₀ ΔT (α_{brass} - α_{steel})

the thickness of the strip is

           e = r₁ -r₂

           θ = Lo \ \Delta T \ \frac{\alpha_{brass} - \alpha_{steel}}{e}

we calculate

           θ = 22.2 \ (100-20) \ \frac{(19-11) \ 10^{-6}}{0.0914}

           θ = 0.155 rad

Let's use trigonometry to find the perpendicular deflection

          tan θ = Δy / L₀

          Δy = L₀ tan θ

          Δy = 22.2 tan 0.155

          Δy = 3.48 cm

Using the thematic expansion of the two materials we find the deflection of the bimetallist strip is 3.38 cm

Learn more about thermal expansion here: brainly.com/question/18717902

7 0
3 years ago
I need a story about kepler's laws for a child, something a child would understand. And i'd like it to be okay for someone if I
andreyandreev [35.5K]

Answer:

(1) The orbits are ellipses, with focal points ƒ1 and ƒ2 for the first planet and ƒ1 and ƒ3 for the second planet. The Sun is placed in focal point ƒ1.

(2) The two shaded sectors A1 and A2 have the same surface area and the time for planet 1 to cover segment A1 is equal to the time to cover segment A2.

(3) The total orbit times for planet 1 and planet 2 have a ratio a13/2 : a23/2

8 0
3 years ago
Read 2 more answers
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Sergeeva-Olga [200]

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

P = \ro g h = 1000 * 10 * 20 = 200000 Pa

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

P_1V_1 = P_2V_2

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

V_1 = Ah

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

P_1Ah_1 = P_2Ah_2

h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m

b) If the temperatures changes, we can still reuse the ideal gas equation above:

\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}

h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m

3 0
3 years ago
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