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alexira [117]
3 years ago
5

At what time of day would you be most likely to find that the air over water is significantly warmer than the air over land near

by?
Physics
2 answers:
MA_775_DIABLO [31]3 years ago
5 0
The answer is 5:00 AM
Brut [27]3 years ago
4 0

This would happen later at night or early in the morning.

The reason being land becomes warm and cold quicker than the water because of the heat capacity. So during the day water warms up because of sunlight but at night the land becomes a lot cooler as compared to the water which is still war. So the air over water is significantly warmer than the air over land.

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The orbits of ____ lie closer to the sun than does earth's orbit.
Eva8 [605]

Answer:

Mercury and Venus lie closer to Sun than position of Earth

Explanation:

As we know that all planets around the sun in the range of their distance can be arranged as following:

1). Mercury.

2). Venus.

3). Earth.

4). Mars.

5). Jupiter.

6). Saturn.

7). Uranus.

8). Neptune.

Since Earth lie at 3rd position from sun so two closer planets are Mercury and Venus

5 0
3 years ago
What did Bohr’s model of the atom include that Rutherford’s model did not have?
hammer [34]
Bohr explained that the electron revolve around a orbit of fixed energy level that Rutherford fails to explain !!!
8 0
3 years ago
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Explain why it is important to identify a reference point for any description of motion.
Leya [2.2K]

A reference is critical because it determines how much motion has occurred. You can't determine how much an objected has moved if you don't know where it came from.

3 0
2 years ago
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A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits
Katyanochek1 [597]

Answer:

Answer:

750 nm

Explanation:

= separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

= screen distance = 4.8 m

= position of first bright fringe =

= order = 1

Position of first bright fringe is given as

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

Explanation:

5 0
3 years ago
.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effec
stira [4]

Answer:

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

KEfin = 5.368 + 1.610 = 6.978 J

Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

5 0
3 years ago
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