Why do we experience a high tide twice a day?

1 answer:

7 0

Coastal areas experience two low tides and two high tides every lunar day, or 24 hours and 50 minutes. The two tidal bulges caused by inertia and gravity will rotate around the Earth as the moons position changes. These bulges represent high tides while the flat sides indicate low tides.

Over the course of 1 day, the position of the moon does not change very much compared to the rotation of the earth. As the earth rotates below the moon, one point on the earth will go through all levels of tide as the day passes by. Strongly attracted, middling, weakly attracted, and then middleagain. From our perspective it looks like "high, low, high, low." Or equivalently you can think about how the points below the moon and opposite the moon will be high tide, and as the earth rotates, those areas will change.

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A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres

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Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

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An old millstone, used for grinding grain in a gristmill, is a solid cylindrical wheel that can rotate about its central axle wi

MAXImum [283]

Answer:

The answer is I=70,513kgm^2

Explanation:

Here we will use the rotational mechanics equation T=Ia, where T is the Torque, I is the Moment of Inertia and a is the angular acceleration.

When we speak about Torque it´s basically a Tangencial Force applied over a cylindrical or circular edge. It causes a rotation. In this case, we will have that T=Ft*r, where Ft is the Tangencial Forge and r is the radius

Now we will find the Moment of Inertia this way:

$Ft*r=I*a$ -> $(Ft*r)/(a) = I$