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jonny [76]
3 years ago
8

HELP ASAP PLEASE - Which best describes how the arrangement of the sun, moon, and the earth affect the range of the tides during

a spring tide?
A. Because the earth, moon, and sun are all out of alignment, the pull of the moon and sun act in the opposite direction causing the highest high tides and the lowest low tides.


B. Because the earth, moon, and sun are in alignment, the pull of the moon and sun act in the same direction, causing the highest high tides and the lowest low tides.


C. Because the earth, moon, and sun are all out of alignment, the pull of the moon and sun act in the opposite direction, causing the lowest high tides and the highest low tides.


D. Because the earth, moon, and sun are all in alignment, the pull of the moon and sun act in the same direction causing the lowest high tides and the highest low tides.
Physics
2 answers:
koban [17]3 years ago
3 0
B. When the sun and the moon are aligned, their combined gravitational force attracts the water, causing super high tides on one side of the planet, and super low tides on the opposite side.
Simora [160]3 years ago
3 0

Answer: B. Because the earth, moon, and sun are in alignment, the pull of the moon and sun act in the same direction, causing the highest high tides and the lowest low tides.

Explanation:

The rise in level of waves due to gravitational pull of the sun and the moon are tides. Spring tides occur during new moon or full moon when the Sun, the Moon and The Earth all align in one straight line. The strong gravitational pull causes the highest high tides and the lowest low tides. Thus, the correct option is B.

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Answer:

very small solid particles called interstellar dust.

Explanation:

In the space between the stars there is gas and dust, which represent at least 20% of the mass of our galaxy. In the Milky Way it is considered that there is a gas density of approximately 0.2 to 0.5 atoms / cm3 in the surroundings of the Sun; with respect to the dust an average of 1 g / cm3 is estimated.

Gas is about atoms and molecules, mainly hydrogen; In order of abundance, helium, carbon, oxygen, nitrogen and iron follow. On the other hand, the dust is tiny particles, generally smaller than 10 microns; the dust does not shine and therefore it is only distinguished when it is projected on bright regions (nebulae or clusters).

Interstellar matter is mainly concentrated towards the plane of the galaxy, in the strip corresponding to the Milky Way; there you can see bright nebulas of diffuse character called nebulas. These nebulae are classified according to three types: (a) bright or emission nebulae, (b) reflection nebulae and (c) planetary nebulae.

Hydrogen appears both ionized and neutral; The bright nebulae are composed of ionized hydrogen and other ionized elements. Non-ionized (neutral) hydrogen is found in the spiral arms of the Milky Way and can be detected through radio waves.

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3 years ago
Sohan said to Geeta I have done my work change the following sentence into indirect​
12345 [234]

Answer:

Sohan told Geeta that i had done my work

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A liquid has a volume of 100 cm and a mass of 85g.
djyliett [7]

Answer:

B. Its density is lower than that of water

Explanation:

density = mass / volume

density of the liquid = 85 / 100 = 0.85 g/cm^3

now,

density of water is 1 g/cm^3 which is greater than the density of the given liquid ( 0.85 g/cm^3 )

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Lourdes mixes several ingredients in a bowl, creating a cake batter. She holds the bowl up and turns it upside down, causing the
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A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w
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Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = ?

v_{1i} = initial velocity of the first body before collision = v

v_{2i} = initial velocity of the second body before collision = 0 m/s

v_{1f} = final velocity of the first body after collision =

using conservation of momentum equation

m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}

Using conservation of kinetic energy

m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35

b)

m_{1} = mass of first body = 2.7 kg

m_{2} = mass of second body = 1.35 kg

v_{1i} = initial velocity of the first body before collision = 4 ms⁻¹

v_{2i} = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67 ms⁻¹

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