Answer:
That scenario can be explained by the idea of the contribution of dark matter on that point.
Explanation:
It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.
(1)
Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.
Notice, that If the distance increases the orbital speed decreases (inversely proportional).
<span>Mass of the ball is m = 0.10kg
Initial speed of the Ball v = 15m/s
a. When the ball is at maximum height the velocity is 0
Momentum of ball = mass x velocity
Momentum = 0.10kg x 0 = 0
b. Getting the maximum height,
Using the conservation of energy equation KEinitial = mgh
1/2mVin^2 = mgh => h = v^2/2g
h = 15^2/2x9.8 = 11.48m => Half Height h = 5.96m
Applying the conservation of energy equation at halfway V^2 = 2gh
V = square root of (2x9.8x5.96) => V = square root of (116.816)
So the velocity at the half way V = 10.81 m/s
Momentum M = m x V => M = 0.10 x 10.81 => M = 1.081kg-m/s</span>
Answer:
the buoyant force on the chamber is F = 7000460 N
Explanation:
the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.
Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume
mass of water displaced = density of seawater * volume displaced
m= d * V , V = 4/3π* Rext³
the buoyant force is the weight of this volume of seawater
F = m * g = d * 4/3π* Rext³ * g
replacing values
F = 1025 kg/m³ * 4/3π * (5.5m)³ * 9.8m/s² = 7000460 N
Note:
when occupied the tension force on the cable is
T = F buoyant - F weight of chamber = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N
Answer:
Nuclease is the answer I know
I hope this is the answer
You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.
