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Len [333]
3 years ago
15

During which phase of the moon do neap tides occur?

Physics
2 answers:
Ostrovityanka [42]3 years ago
6 0

Answer:

The answer is during the moon's <u>Quarter Phases</u>                              

Explanation:

During the moon's quarter phases the sun and moon work at right angles, making the <em>bulges </em>to cancel each other. The result is a smaller variation between high and low tides. This is known as a <u>Neap tide</u>.

Neap tides are particularly weak tides. They occur when the gravitational forces of the Moon and the Sun are perpendicular to one another (with respect to the Earth). Neap tides occur during quarter moons.

What is a Bulge?

Earth's gravitational pull is so powerful that it creates a small bulge on the surface of the moon.

Cheers!

Fynjy0 [20]3 years ago
4 0

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

F = G\frac{m1\cdot m2}{r^{2}} (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

               

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

           

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

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Answer:

In general solids are easier to transport than liquids, but the above metal example is a valid one and the only other one that comes to mind is that of concrete. It is mixed as a liquid and transported as such, but then sprayed or laid down to dry and form a solid surface or filler. 

Explanation:

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3 years ago
Why is fusion important?
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Abundant energy: Fusing atoms together in a controlled way releases nearly four million times more energy than a chemical reaction such as the burning of coal, oil or gas and four times as much as nuclear fission reactions (at equal mass)

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2 years ago
an object with a mass of 5kg is moving with an initial velocity of 20m/s.the object accelerates at a rate of 15m/s/s for 8s.what
Alinara [238K]

It doesn't matter what the object's initial velocity is, or how long
the acceleration lasts.  All that matters is the object's mass and
acceleration.

Force = (mass) x (acceleration) =

                (5kg) x (15 m/s²) =

                         75 kg-m/s² = <em>75 newtons .</em>


5 0
3 years ago
Calculate delta g for the reaction if the partial pressures of the initial mixture are pcl5 = .0029 atm, pcl3 = .27 atm, and pcl
AnnyKZ [126]

Answer: equation for the reaction is given below

PCL2+PCL3=PCL5

Where pcl2=0.40atm,pcl3=0.27atm

Pcl5=0.0029atm

Using ∆G=-RTin(PCL5/PCl2*PCL3)

Where R=8.314J/K/mol and T=298K

∆G=-8.314*298in(0.0029/0.40*.27)

∆G=8962.6J/mol

Explanation:

7 0
3 years ago
The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

PE_1 = 9.05*10^{15}J

PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

8 0
3 years ago
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