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valina [46]
3 years ago
10

The ocean’s tides are not only affected by the shape of the coastlines, or slope of the ocean floor, but also by the gravitation

al tug of the Sun and the Moon. Write a 600 word report on the effect of the Moon, and Sun, on tidal reaction, and high tides and low tides. What orientation are the Sun and Moon in to create high tides? Low tides? There are two special types of tides. What are they called and what positions are the Sun and the Moon in during these times?
Physics
2 answers:
Genrish500 [490]3 years ago
5 0

Answer:a

Explanation:

dexar [7]3 years ago
4 0

Answer:

a) According to Newton's law of gravitation, as the distance between the Moon and the Earth decreases, the gravitational attraction increases and vice versa

The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon

The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it

Therefore, the region where the Moon is closest to the Earth we have a high tide as the water level rises and the region which is perpendicular to where the Moon is located has a  low tide

b) The two special types of tides are

1) The neap tide

2) The spring tide

Neap tide

Neap tide occurs when the Sun and Moon are 90° apart from each other when they are viewed by an observer from Earth

The gravitational pull of the Sun cancels (partially) the effect of the gravitational pull and tidal force of the Moon, resulting in minimum tidal range

Spring Tide

Spring tide occurs when the Earth, the Moon, and the Sun are simultaneously inline, such that the Sun reinforces the gravitational pull and tidal force of the Moon, resulting in a maximum tidal range

Explanation:

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dimulka [17.4K]

Answer:

-1.7908787542\times 10^{-9}\ C

3.4260289211\times 10^{-9}\ C

1.7908787542\times 10^{-9}\ C

1.6351501669\times 10^{-9}\ C

Explanation:

r = Radius

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

Electric field is given by

E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C

The charge is -1.7908787542\times 10^{-9}\ C

Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C

The charge is 3.4260289211\times 10^{-9}\ C

The charge inside will have the polarity changed

q=+1.7908787542\times 10^{-9}\ C

Outside the charge will be

3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C

3 0
3 years ago
How are the electric field lines around a positive charge affected when a second positive charge is near it?
Lesechka [4]

Answer: The field lines bend away from the second positive charge

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