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Yuki888 [10]
2 years ago
8

Twice each each lunar month, all year long, these tides occur. Whenever the Moon, Earth and Sun are aligned, the gravitational p

ull of the sun _____ to that of the moon causing _____ tides. A) adds; neap B) adds; maximum C) interferes; neap D) interferes; minimum

Physics
2 answers:
LenKa [72]2 years ago
5 0
Hi there!

Twice each lunar month, all year long, these tides occur. Whenever the Moon, Earth, and Sun are aligned, the gravitational pull of the sun A. Adds to that of the moon causing A. Neap tides.

As a result of the extra gravity lent by the sun to the moon, it's pull is strengthened causing what we know as Neap tides.

Hope this helps!
Zarrin [17]2 years ago
4 0

That was a lucky pick.

Twice each each lunar month, all year long, whenever the Moon,
Earth and Sun are aligned, the gravitational pull of the sun adds
to that of the moon causing maximum tides.

This is the setup at both New Moon and Full Moon.  It doesn't matter
whether the Sun and Moon are both on the same side of the Earth,
or one on each side.  As long as all three bodies are lined up, we
get the biggest tides.

These are called "spring tides", when there is the greatest difference
between high and low tide.

At First Quarter and Third Quarter, when the sun, Earth, and Moon form a
right angle, there is the least difference between high and low tide.  Then
they're called "neap tides".
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How large a force is necessary to stretch a 4.0-mm-diameter steel wire from its original length by 1.0%?
jekas [21]

The force needed to stretch the steel wire by 1% is 25,140 N.

The given parameters include;

  • diameter of the steel, d = 4 mm
  • the radius of the wire, r = 2mm = 0.002 m
  • original length of the wire, L₁
  • final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
  • extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
  • the Youngs modulus of steel, E = 200 Gpa

The area of the steel wire is calculated as follows;

A = \pi r^2\\\\ A= 3.142 \times (0.002)^2\\\\ A= 1.257 \times 10^{-5} \ m^2

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;

E = \frac{stress}{strain} = \frac{F/A}{e/L} = \frac{FL}{Ae} \\\\F = \frac{EAe}{L}

F = \frac{200 \times 10^9\  \times\  1.257\times 10^{-5}\  \times \ 0.01l_1}{l_1} \\\\F = 25,140\ N

Thus, the force needed to stretch the steel wire by 1% is 25,140 N.

Learn more here: brainly.com/question/21413915

4 0
2 years ago
You are directed to set up an experiment in which you drop, from shoulder height, objects with similar surface areas but differe
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Explanation:

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So if you drop two objects with the same volume but different mass, because the acceleration is the same for both of them, they will hit the ground at the same time, this means that the density of the object has no impact in how much time the object needs to reach the floor.

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