When the effects of the moon and sun work together, this is what kind of tide? Explain

1 answer:

7 0

Around each new moon and full moon, the sun, Earth, and moon arrange themselves more or less along a line in space. Then the pull on the tides increases, because the gravity of the sun reinforces the moon’s gravity. In fact, the height of the average solar tide is about 50% the average lunar tide. Thus, at new moon or full moon, the tide’s range is at its maximum. This is the spring tide: the highest (and lowest) tide. Spring tides are not named for the season. This is spring in the sense of jump, burst forth, rise. So spring tides bring the most extreme high and low tides every month, and they happen around full and new moon. So spring tide occurs. Hope it cleared your doubt.

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An astronaut orbits planet Y in her spaceship. To remain in orbit at 410km above the planet's center, she maintains a speed of 6

choli [55]

Formula for orbital speed, v = √(GM/R)

Where G is the universal gravitational constant, M = Central Mass,

R = Distance between centers of Mass.

Given. v = 68 m/s, M = ? , R = 410 km = 410000 m., G = 6.674 * 10⁻¹¹ Nm²/kg²

68 = √(GM/R)

68 = √(6.674 * 10⁻¹¹ * M/410000)

68² = (6.674 * 10⁻¹¹ * M)/410000

(68² * 410000) / 6.674 * 10⁻¹¹ = M

2.84 × 10¹⁹ = M

Mass of Planet Y = 2.84 × 10¹⁹ kg

3 0

3 years ago

Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly

loris [4]

Answer: $36.86\°$

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: $d=2.4 m$

Ferdinand's initial velocity: $V_{o}=5 m/s$

Time it takes a jump: $t=0.6 s$

We need to find the angle $\theta$ at which Ferdinand jumps.

In order to do this, we first have to find the horizontal component (or x-component) of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

$V_{x}=\frac{d}{t}$ (1)

$V_{x}=\frac{2.4 m}{0.6 s}$ (2)

$V_{x}=4 m/s$ (3)

On the other hand, the x-component of the velocity is expressed as:

$V_{x}=V_{o}cos\theta$ (4)

Substituting (3) in (4):

$4 m/s=5 m/s cos\theta$ (5)

Clearing $\theta$ :

$\theta=cos^{-1} (\frac{4 m/s}{5 m/s})$

$\theta=36.86\°$ This is the angle at which Ferdinand the frog jumps between lily pads

4 0

3 years ago

Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l

goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

$\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}$