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Gnom [1K]
4 years ago
11

The sound waves used for sonar are _____.

Physics
2 answers:
lakkis [162]4 years ago
6 0

<u>Ultrasound </u>''Although humans cannot hear ultrasound, scientists have developed technology to make use of these high-frequency sounds. For example, sonar technology uses ultrasound.''

Stella [2.4K]4 years ago
4 0

Acoustic Energy


Sonar uses the echo principle by sending out sound waves underwater or through the human body to locate objects

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Specific Heat and Thermal energy question. 30 points, will give brainliest. Please answer. Pic attached
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Answer:

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Explanation:

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3 years ago
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What kind of waves are present during an earthquake? ...
Brrunno [24]

Answer:

1. The two main types of waves are body waves and surface waves. Body waves can travel through the earth's inner layers, but surface waves can only move along the surface of the planet like ripples on water. Earthquakes radiate seismic energy as both body and surface waves.

2. potential energy

3. Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

4. Refraction is the bending of light

5. Density uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). Density is defined as mass per unit volume.

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3 years ago
A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra
son4ous [18]

Answer:

The answer is v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}.

Explanation:

From the law of gravity,

F = G \frac{Mm}{r^2}

considering F as a conservative force, F = - \nabla U,

the general expression for gravitational potential energy is

U = -G \frac{Mm}{r},

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

g = -G \frac{M}{r^2},

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

U = g M_e R,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

U_p = g M_e(R- r_s),

then, the kinetik energy when the rock hits the surface is

U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s),

so

v = \sqrt{2g(R-r_s)}

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g = G \frac{M_s}{R^2}.

8 0
3 years ago
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

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3 years ago
It is important that your muscles are very warm when doing this type of stretching.
MaRussiya [10]

Answer:

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