I think D x=vxt because it's equation finding change of x (displacement) and using time
Answer:
the acceleration of harry is equal to 66.126 m/s²
Explanation:
given,
harry is 35 m behind Draco
speed of Draco = 40 m/s
original speed of harry = 50 m/s
acceleration = ?
time taken by the Draco
t =
t = 1.875 s
distance covered by Harry
d = 35 + 175 = 210 m
to calculate the acceleration of harry
a × 3.516 × 0.5 = 116.25
a = 66.126 m/s²
hence, the acceleration of harry is equal to 66.126 m/s²
The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
<h3>What is period of oscillation?</h3>
This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.
The period of oscillation is given as T
T = 2 * pi * sqrt ( m / k )
where
m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:
m1 = mass of the block
m2 = mass pf the spring
k = force constant of the spring
including the two masses to the period gives
T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
Read more on period of oscillation here: brainly.com/question/22499336
#SPJ4
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
