The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
Data Analysis and Conclusion
<span>Exothermic reaction is a chemical reaction that releases energy. This reaction releases heat energy or light .
In an endotermic reaction energy is used.
Enthaply is the heat energy change , delta H.
If the sum of the enthalpies of the reactans is greater than the products the reaction is exothermic. If the products side has a larger enthaply than the process is endothermic. So, if delta H is negative then the process is exothermic. If delta H is positive, than the process is endothermic.
Exothermic are: A+BC -> AB+C
A2+B2 -> 2AB
Endothermic are:AB+C -> AC+B
A2 + C2 -> 2AC
B2+C2 -> 2BC</span>
The answer to this question is The first option, Or what I should say "A.Thermal"
Your welcome!
Answer:
t = 1.62 h
Explanation:
A flat mirror fulfills the law of reflection where the incident angle is equal to the reflected angle.
θ_i = θ_r
If we use trigonometry to find the angles, the mirror is at a height of L = 1.87 m, and the reflected rays reach a distance x1 = 3.56 m
tan θ₁ = x₁ / L
tan θ₁ = 
θ₁ = tan⁻¹ 1.90
θ₁ = 62.29º
for the second case x₂ = 1.46 m
tan θ₂ = x₂ / L
θ₂ = tan⁻¹ 
θ₂ = 37.98º
the difference in degree traveled is
Δθ = θ₁- θ₂
Δθ = 62.29 - 37.98
Δθ = 24.31º
as in the exercise they indicate that every 15º there is an hour
t = 24.31º (1h / 15º)
t = 1.62 h