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zloy xaker [14]
4 years ago
10

How are the climates of coastal regions affected by the specific heat capacity of water?

Physics
1 answer:
Katen [24]4 years ago
5 0

Answer:

The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.

This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows  from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.

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What is a lunar month
Ira Lisetskai [31]

Answer:

It's the duration between successive new moons. Also called a lunation or synodic month, it has a mean period of 29.53059 days (29 days 12 hours and 44 minutes).

7 0
3 years ago
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Why does matter increase in volume when it heats up?
ArbitrLikvidat [17]

Answer:All three states of matter (solid, liquid and gas) expand when heated. ... Heat causes the molecules to move faster, (heat energy is converted to kinetic energy ) which means that the volume of a gas increases more than the volume of a solid or liquid.

Explanation:well I tried lol she just copied and pasted faster than I could

8 0
3 years ago
Displacement vectors of 4 km north, 2 km south, 5 km north, and 5 km south combine to a total displacement of
valentinak56 [21]

south = -(north)

Displacement = (4 km north) + (2 km south) + (5 km north) + (5 km south)

Displacement = (4 km north) - (2 km north) + (5 km north) - (5 km north)

Displacement = (4 - 2 + 5 - 5) km north

<u>Displacement = 2 km north </u>

6 0
3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a 6 iron. The acceleration due to gravity on t
kozerog [31]

Answer:

a) 6 times farther.  b) 6 times longer.

Explanation:

Once released, in the horizontal direction, no other forces act on the ball, so it continues moving at the same initial velocity, which is given by the projection of the velocity vector in the horizontal direction, as follows:

vₓ = v* cos (25º) = 23 m/s * 0.906 = 20.8 m/s

In the vertical direction, the initial velocity is the projection of the velocity vector along the vertical axis, as follows:

vy = v* sin (25º) = 23 m/s * 0.422 = 9.72 m/s

Assuming that the acceleration is constant, and equal to 1/6*g, we can calculate the total time of flight, with the following kinematic equation for the vertical displacement:

y = voy*t - (\frac{1}{2}*\frac{g}{6} * t^{2} )

If the total displacement in the vertical direction is 0 (which means  that the time if the total time of flight), we can solve for t, as follows:

t = \frac{voy*12}{g} = \frac{9.72 m/s*12}{9.8m/s2} = 11. 9 s

On earth, this time could be calculated in the same way:

t = \frac{voy*12}{g} = \frac{9.72 m/s*2}{9.8m/s2} = 1.98 s

As the time is defined by the vertical movement, we can find the horizontal distance travelled on the moon, as follows:

Δx = v₀ₓ * t = 20.8 m/s * 11. 9 s = 248.1 m

On earth, the distance travelled had been as follows:

Δx = v₀ₓ * t = 20.8 m/s * 1.98 s = 41.3 m

⇒ Δx(moon) / Δx(earth) = 248.1 / 41.3 = 6.00

b) As we have just found, the time of flight on the moon and on the earth are as follows:

tmoon = 11. 9 s

tearth = 1.98 s

⇒ t(moon) / t(earth) = 11.9 / 1.98 = 6.0

8 0
3 years ago
What are the output waveforms of the following waves, after passing through a transformer?
Ber [7]
The output waveforms after passing through the transformer actually depend on the type of transformer used. It could either be a step-up transformer (steps voltage up), or a step-down transformer (steps voltage down). Both transformers have an output voltage in a form of a sine wave.
8 0
3 years ago
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