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4 years ago

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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In the atmosphere water vapor condenses to form clouds.

laiz [17]

True is the correct answer

4 0

3 years ago

SOH-CAH-TOA is used to solve for the ________

Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0

3 years ago

A vehicle travels from a 30m marker to a 100m marker. What is the change in distance

alexgriva [62]

70 meters is your answer

Note that the beginning is the 30 meter mark, and the ending is the 100 meter mark. Subtract the beginning amount from the end:

100 - 30 = 70

The vehicle traveled 70 meters.

8 0

3 years ago

Which list places different units of matter in the correct sequence from largest to smallest

weqwewe [10]

Answer:

What do you mean?

6 0

3 years ago

Read 2 more answers

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

shutvik [7]

Answer:

4.0 m/s

Explanation:

The motion of the diver is the motion of a projectile: so we need to find the horizontal and the vertical component of the initial velocity.

Let's consider the horizontal motion first. This motion occurs with constant speed, so the distance covered in a time t is

$d=v_x t$

where here we have

d = 3.0 m is the horizontal distance covered

vx is the horizontal velocity

t = 1.3 s is the duration of the fall

Solving for vx,

$v_x = \frac{d}{t}=\frac{3.0 m}{1.3 s}=2.3 m/s$

Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

$y(t) = h + v_y t - \frac{1}{2}gt^2$

where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

So now we can find the magnitude of the initial velocity:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(2.3 m/s)^2+(3.3 m/s)^2}=4.0 m/s$

4 0

4 years ago