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2 years ago

The ultrasound is reflected from the seafloor back to the submarine.

Physics

1 answer:

lesantik [10]2 years ago

7 0

Answer: c

Explanation:

The total distance the ultrasound pulse travelled is, twice the distance of the sea floor

Because it went back and forth

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Which vector below goes from (0,0) to (1,-3)?

vitfil [10]

I think D. It starts at (0.0) and goes to the correct points so it makes sense

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3 years ago

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The temperature of a black body is 500 and its radiation is of wavelength 600 . If the number of oscillators with energy is 100

stiks02 [169]

Answer: An equation is missing in your question below is the missing equation

a) ≈ 8396

b) 150 nm/k

Explanation:

<u>A) Determine the number of Oscillators in the black body</u>

number of oscillators = 8395

attached below is the detailed solution

<u>b) determine the peak wavelength of the black body </u>

Black body temperature = 20,000 K

applying Wien's law / formula

λmax = b / T ------ ( 1 )

T = 20,000 K

b = 3 * 10^6 nm

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2 years ago

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S_A_V [24]

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3 years ago

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Why do people tend to cup their hands around listeners' ears when whispering to them? A) A cupped hand makes the sound echo so t

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C. Hope this helps you!

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3 years ago

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A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$

The final velocity is:

$v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$

b) The change in the kinetic energy of the 13.5 g coin is:

$\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]$

$\Delta K = 9.559\times 10^{-5}\,J$

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3 years ago

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