Miguel Hidalgo was a Priest so I think it would be 1.
Hope I Helped ( :
Answer:
Speed of light
Explanation:
The famous Einstein's equation is:

where
E is the energy
m is the mass
is the speed of light
In this equation, Einstein summarized the following fact: mass can be converted into energy, and the amount of energy released in such a process is given by the equation.
An example of application of this equation is the nuclear fusion process. In a nuclear fusion, two lighter nuclei combine into a heavier nucleus. However, the mass of the heavier nucleus is slightly less than the sum of the masses of the two original nuclei: some of the mass of the original nuclei has been converted into energy, accorging to the previous equation.
Answer:
b.
Explanation:
It's b I sis thebksqjwnsx0qkqnsnd991isnd
The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.
Em = K + U
Let's write the energy in two points.
Starting point. Highest part of the oscillation
Em₀ = U = m g h
Final point. Lower part of the movement
= K = ½ m v²
Energy is conserved.
Emo =
m g h = ½ m v²
v² = 2 gh
Let's use trigonometry to find the height, see attached.
h = L - L cos θ
h = L (1- cos θ)
They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.
h = 3.7 (1- cos 48)
h = 1.22 m
this is the maximum height of the movement.
Let's calculate the velocity.
v = 4.89 m / s
In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;
a) The maximum speed is v = 4.89 m / s
b) The maximum height is h = 1.22 m
Learn more here: brainly.com/question/13010190
Answer:
about 602 milliseconds
Explanation:
The motion can be approximated by the equation ...
y = -4.9t^2 -22.8t +15.5
where t is the time since the arrow was released, and y is the distance above the ground.
When y=0, the arrow has hit the ground.
Using the quadratic formula, we find ...
t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))
= (22.8 ± √823.64)/(-9.8)
The positive solution is ...
t ≈ 0.60195193
It takes about 602 milliseconds for the arrow to reach the ground.