Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
Given data in the problem :-
- Acceleration (a) = 4.0 m/s^2
- Initial velocity (u) = 0 m/s
- Final velocity (v) = 34 m/s
- Distance travelled by aircraft (S) = ?
From Newton's Laws of Motion we know that ,
v = u + at [t = Time taken by aircraft to cover the distance]
⇒ 34 = 0 + 4t
⇒ t = 34/4 s
∴ t = 8.5 s
From Newton's Laws of Motion we also know that ,
S = u.t + 1/2a.t^2
⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m
∴ S = 144.50 m
Thus the distance travelled by the aircraft while accelerating is 144.50 meter .
Mass is indirectly proportional to acceleration, so, lighter the object greater would be it's acceleration...
A) 0.10 kg is lightest among them, so it's your answer
<em>H</em> is the location of the Japanese nuclear power plant at Fkushima and <em>I</em> is the location of the Ukraine accident in Chernobyl. At the time, Chernobyl was part of the USSR, but that country no longer exists. These meltdowns will leave the area radioactive for many more years.
v initial = 0m/s
v final = 40m/s
t = 5s
d = ?
The kinematic equation which relates these quantities:
d = 1/2(vinitial + vfinal)t
d = 0.5(0+40)(5)
d = 100m
