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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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Suppose that a constant force is applied to an object. Newton's Second Law of Motion states that the acceleration of the object
omeli [17]
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4 0
3 years ago
A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you
soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

5 0
3 years ago
Consult Interactive Solution 10.37 to explore a model for solving this problem. A spring is compressed by 0.0647 m and is used t
padilas [110]

Answer:

\omega=32.14\ rad/s

Explanation:

Given that,

The compression in the spring, x = 0.0647 m

Speed of the object, v = 2.08 m/s

To find,

Angular frequency of the object.

Solution,

We know that the elation between the amplitude and the angular frequency in SHM is given by :

v=\omega\times A

A is the amplitude

In case of spring the compression in the spring is equal to its amplitude

\omega=\dfrac{v}{A}

\omega=\dfrac{2.08\ m/s}{0.0647\ m}

\omega=32.14\ rad/s

So, the angular frequency of the spring is 32.14 rad/s.

4 0
3 years ago
A 10000 N net force is accelerating a car at a rate if 5.5 m/s^2. What is the cars’s mass?
lianna [129]

Answer:

1800 kg

Explanation:

Newton's second law:

F = ma

10000 N = m (5.5 m/s²)

m ≈ 1800 kg

8 0
3 years ago
Read 2 more answers
A lead bar 12 cm long, 2 cm wide, and 2 cm talk has a density of 11.3g/cm3. If you cut the bar in half, what is the density of o
VMariaS [17]

Answer:

The density of one halves  11.3 g / cm cube

Explanation:

Density remain same because cutting the bar in half , mass and volume will decrease to half so density will not change .

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5 0
3 years ago
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