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
Answer:
See below
Explanation:
<u>I will use 3 x 10^8 m/s for speed or wave</u>
speed = wavelength * frequency
3 x 10^8 = w * 7.34 x 10^2 <====== are you sure this isn't KILO Hz ?
w = <u>408719. 3 meters </u>
Carbon atoms can form straight, and branched chains, and rings
usually gravity is what causes us to make us weigh more or less depending on which planet we are on
The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied, in the same time. In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage. Velocity ratio is sometimes called distance ratio.