Because the box keeps going straight at the same speed, while the seat under it speeds up, slows down, or changes direction.
So that there isn't too much force restricting the structure of the circus and cause disaster
Answer:
6.1 x 10^-8 newtons
Explanation:
F = 8.98 *109 *1*1/3845000002
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
Answer:
a) v = 4.4 m/s
b) F = 400 N
Explanation:
a) ½kx² = ½mv²
v = √(kx²/m)
F = kx
v = √(Fx/m)
v = √(800(0.012) / 0.5) = √19.2 = 4.3817...
b) Fd = ½mv²
F = mv²/2d
F = 0.5(19.2) / (2(0.012) = 400 N