Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
A convex mirror makes a reflected light rays spread out.
Energy E of EM radiation is given by the equation E=hf, where h is Planck's constant and f is frequency. It means energy E and frequency f are proportional so as we increase the frequency, energy also increases. Also, the relationship between the wavelength and frequency is c=λ*f where λ is the wavelength and f is frequency and c is the speed of light. This tells us the wavelength and frequency are inversely proportional. So as we increase the frequency the wavelength is getting smaller. So as we go from left to right the frequency increases, energy also increases and the wavelength is decreasing. Or, on the left side we should have low frequency, low radiant energy, and long wavelength. On the right side we should have high frequency, high radiant energy and low wavelength. That is the third graph.
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ = 
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt = ![\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%7D%20%5C%20%5B%28v_%7Bo%7D%20-%20v%29%20%20-%20%28%20-%20v_%7Bo%7D%20%20-%20v%29%20%5D)
Δt = 2 v₀ / g
