<span>Since there is no friction, conservation of energy gives change in energy is zero
Change in energy = 0
Change in KE + Change in PE = 0
1/2 x m x (vf^2 - vi^2) + m x g x (hf-hi) = 0
1/2 x (vf^2 - vi^2) + g x (hf-hi) = 0
(vf^2 - vi^2) = 2 x g x (hi - hf)
Since it starts from rest vi = 0
Vf = squareroot of (2 x g x (hi - hf))
For h1, no hf
Vf = squareroot of (2 x g x (hi - hf))
Vf = squareroot of (2 x 9.81 x 30)
Vf = squareroot of 588.6
Vf = 24.26
For h2
Vf = squareroot of (2 x 9.81 x (30 – 12))
Vf = squareroot of (9.81 x 36)
Vf = squareroot of 353.16
Vf = 18.79
For h3
Vf = squareroot of (2 x 9.81 x (30 – 20))
Vf = squareroot of (20 x 9.81)
Vf = 18.79</span>
Probably false. but correct me if i’m wrong
Answer:
18%
Explanation:
There are two equal and opposite forces on a floating object: weight and buoyancy.
W = B
The weight of an object is its mass times gravity: W = mg
Buoyancy is the weight of the displaced fluid: W = mf g
Plugging in:
mg = mf g
m = mf
Mass is density times volume:
ρV = ρf Vf
Solving for the ratio of Vf / V:
Vf / V = ρ / ρf
Given that ρ = 0.82 g/mL and ρf = 1.00 g/mL:
Vf / V = 0.82
That means 82% of the object's volume (and therefore, 82% of its mass, assuming uniform density) is submerged. Which means that 18% is above the water line.
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Answer:
We know that the torque can be calculated as follows:
T = rpsinα
With r being the distance of the body from the center of the circumference he has as trajectory, p being the momentum of the body and sinα being the sine of the angle between the 2 vectors: r and p.
It's pretty obvious that T is directly proportional to the momentum, that can be written as p = m·v, with m being the mass of the object and v the velocity of the object.
