Answer:
a) The velocity of rock at 1 second, v = 9.8 m/s
b) The velocity of rock at 3 second, v = 29.4 m/s
c) The velocity of rock at 5.5 second, v = 53.9 m/s
Explanation:
Given data,
The rock is dropped from a bridge.
The initial velocity of the rock, u = 0
a) The velocity of rock at 1 second,
Using the first equation of motion
v = u + gt
v = 0 + 9.8 x 1
v = 9.8 m/s
b) The velocity of rock at 3 second,
v = u + gt
v = 0 + 9.8 x 3
v = 29.4 m/s
c) The velocity of rock at 5.5 second,
v = u + gt
v = 0 + 9.8 x 5.5
v = 53.9 m/s
placing a magnetically hard material in a strong magnetic field
AnswerNo
However, the acceleration is the opposite of the movement:
Explanation:
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
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
