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3 years ago

Which type of wave would actually slow down when moving from the air into the ocean?

Physics

2 answers:

stiv31 [10]3 years ago

8 0

The correct answer to the question is : Light or any electromagnetic wave.

EXPLANATION :

Before going to answer this question, first we have to understand refraction.

Refraction is the optical phenomenon in which light ray will bend at the interface of the two media having different optical densities.

When light ray comes from optical rarer medium to optical denser medium, the ray will bend towards the normal. It is so because the velocity of light in rarer medium is more than velocity of light in denser medium.

When the light comes from optical denser medium to optical rarer medium, the light ray bends away from the normal. It is so because the velocity of light increases in the rarer medium.

As per the question, wave is moving from air to ocean, and its velocity slows down.

Hence, there is refraction. So, the wave must be light or any electromagnetic wave.

kvasek [131]3 years ago

5 0

Any electromagnetic wave, like light or heat.

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An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

3 years ago

A certain parallel-plate capacitor is filled with a dielectric for which Κ = 5.5 .The area of each plate is 0.034 m2 , and the p

Nesterboy [21]

Answer:

The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

Explanation:

Given that,

dielectric constant k = 5.5

the area of each plate, A = 0.034 m²

separating distance, d = 2.0 mm = 2 x 10⁻³ m

magnitude of the electric field = 200 kN/C

Capacitance of the capacitor is calculated as follows;

$C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F$

Maximum potential difference:

V = E x d

V = 200000 x 2 x 10⁻³ = 400 V

Maximum energy that can be stored in the capacitor:

E = ¹/₂CV²

E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²

E = 6.62 x 10⁻⁵ J

Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J

4 0

3 years ago

A harmonic wave on a string with a mass per unit length of 0.050 kg/m and a tension of 60 N has an amplitude of 5.0 cm. Each sec

Dennis_Churaev [7]

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec$

Velocity of the string wave is equal to $v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec$

Power of wave propagation is equal to $P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt$

So power of the wave will be equal to 5.464 watt

6 0

3 years ago

Zigmanuir [339]

In energy point of view, the larger stone had more potential energy before dropping. impacting the water, the larger one, having more kinetic energy which changed from potential energy, tranfered energy to the water and formed wave. the amplitude of the wave indicate the energy of the wave. more energy more amplitude.

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3 years ago

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

poizon [28]

Answer:

$4.048\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

E = Energy = $4.91\times 10^{-19}\ J$

Wavelength ejected is given by

$\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m$

The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is $4.048\times 10^{-7}\ m$

4 0

3 years ago