Ok, I’ll try to help, but I’d need to see the picture of the positions of the sun, earth, and moon to help you fully.
So, the first thing to note is that gravity is an attractive force, meaning that; something that has mass, call the m1 will “pull toward” another mass, call it m2. The two objects pull on each other, mutually.
If an object has more mass it pull more, and if an object has less mass, it pulls less.
Another thing to note is that distances matter. The closer the objects are to each other, the more pull they’ll “feel”.
So, the ocean tides are the effect of ocean water responding to a gravitational gradient, the moon plays a larger role in creating tides than the sun does. But the sun's gravitational gradient across the earth is significant and it does contribute to tides as well.
So, when the bulge of the ocean caused by the sun’s gravity, partially cancels out the bulge of the ocean caused by the moon gravity. This produces moderate tides known as the neap tides, meaning that high tides are a little lower and low tides are a little higher than average.
I hope that helps.
Answer:
t = 0.2845Nm (rounded to 4 decimal places)
Explanation:
The disk rotates at a distance of an arc length of 28cm
Arc length = radius × central angle × π/180
28cm = 10cm × central angle × π/180
Central angle = 
× 180/π ≈ 160.4°
Torque (t) = rFsin(central angle) , where F is the applied force
Radius in meters = 10/100 = 0.1m
t = 0.1m × 16N × sin160.4°
t = 0.2845Nm (rounded to 4 decimal places)
The displacement of the train after 2.23 seconds is 25.4 m.
<h3>Resultant velocity of the train</h3>
The resultant velocity of the train is calculated as follows;
R² = vi² + vf² - 2vivf cos(θ)
where;
- θ is the angle between the velocity = (90 - 51) + 37 = 76⁰
R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)
R² = 129.75
R = √129.75
R = 11.39 m/s
<h3>Displacement of the train</h3>
The displacement of the train is the change in position of the train after a given period of time.
The displacement is calculated as follows;
Δx = vt
Δx = 11.39 m/s x 2.23 s
Δx = 25.4 m
Thus, the displacement of the train after 2.23 seconds is 25.4 m.
Learn more about displacement here: brainly.com/question/2109763
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