Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
Energy is released in the reaction
Explanation:
In a given where the energy of the products is greater than that of the reactants, we can infer that energy is released in the reaction.
This indicates that the reaction is an exothermic or exergonic reaction.
These reaction types are accompanied by release of energy.
- In an exothermic change energy is released to the surroundings.
- The surrounding becomes hotter at the end of the change.
- This applies in exergonic reaction which leaves a reaction having more energy than it originally started with.
Learn more:
Exothermic process brainly.com/question/10567109
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Answer:
1047 miles
Explanation:
The radius of the Earth is
(miles)
So its circumference, which is the total length of the equator, is given by

Now we know that the Earth rotates once every 24 hours. So the distance through which the equator moves in one hour is equal to its total length divided by the number of hours, 24:

Answer:
boron
aluminum
gallium
indium
thallium
Explanation:
Any of these could work. Nitrogen has 5 valence electrons so you just needed to pick an element that has 3 valence electrons that nitrogen could borrow. This periodic table shows valence electron counts:
Answer:
k = 11,564 N / m, w = 6.06 rad / s
Explanation:
In this exercise we have a horizontal bar and a vertical spring not stretched, the bar is released, which due to the force of gravity begins to descend, in the position of Tea = 46º it is in equilibrium;
let's apply the equilibrium condition at this point
Axis y
W_{y} - Fr = 0
Fr = k y
let's use trigonometry for the weight, we assume that the angle is measured with respect to the horizontal
sin 46 =
/ W
W_{y} = W sin 46
we substitute
mg sin 46 = k y
k = mg / y sin 46
If the length of the bar is L
sin 46 = y / L
y = L sin46
we substitute
k = mg / L sin 46 sin 46
k = mg / L
for an explicit calculation the length of the bar must be known, for example L = 1 m
k = 1.18 9.8 / 1
k = 11,564 N / m
With this value we look for the angular velocity for the point tea = 30º
let's use the conservation of mechanical energy
starting point, higher
Em₀ = U = mgy
end point. Point at 30º
= K -Ke = ½ I w² - ½ k y²
em₀ = Em_{f}
mgy = ½ I w² - ½ k y²
w = √ (mgy + ½ ky²) 2 / I
the height by 30º
sin 30 = y / L
y = L sin 30
y = 0.5 m
the moment of inertia of a bar that rotates at one end is
I = ⅓ mL 2
I = ½ 1.18 12
I = 0.3933 kg m²
let's calculate
w = Ra (1.18 9.8 0.5 + ½ 11,564 0.5 2) 2 / 0.3933)
w = 6.06 rad / s