A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
<span>buena suerte mi amigo</span>
Explanation:
Show that the motion of a mass attached to the end of a spring is SHM
Consider a mass "m" attached to the end of an elastic spring. The other end of the spring is fixed
at the a firm support as shown in figure "a". The whole system is placed on a smooth horizontal surface.
If we displace the mass 'm' from its mean position 'O' to point "a" by applying an external force, it is displaced by '+x' to its right, there will be elastic restring force on the mass equal to F in the left side which is applied by the spring.
According to "Hook's Law
F = - Kx ---- (1)
Negative sign indicates that the elastic restoring force is opposite to the displacement.
Where K= Spring Constant
If we release mass 'm' at point 'a', it moves forward to ' O'. At point ' O' it will not stop but moves forward towards point "b" due to inertia and covers the same displacement -x. At point 'b' once again elastic restoring force 'F' acts upon it but now in the right side. In this way it continues its motion
from a to b and then b to a.
According to Newton's 2nd law of motion, force 'F' produces acceleration 'a' in the body which is given by
F = ma ---- (2)
Comparing equation (1) & (2)
ma = -kx
Here k/m is constant term, therefore ,
a = - (Constant)x
or
a a -x
This relation indicates that the acceleration of body attached to the end elastic spring is directly proportional to its displacement. Therefore its motion is Simple Harmonic Motion.
Answer:
48.4 km, 34.3° north of east
Explanation:
Let's say east is the +x direction and north is the +y direction.
Adding up the x components of the vectors:
x = 20 cos 60 + 30 + 0
x = 40 km
Adding up the y components of the vectors:
y = 20 sin 60 + 0 + 10
y = 27.3 km
The magnitude of the displacement is:
d = √(x² + y²)
d = 48.4 km
The direction is:
θ = atan(y/x)
θ = 34.3° north of east
Answer:
The velocity after 2 seconds can be found through:
V = u +a*t
Where V is final velocity, u is initial velocity, a is acceleration and t is time.
V = 0 + 2* 2= 4 meters/second
The distance (s) can be found through:
V^2= u^2 +2*a* s
Where V is final velocity, u is initial velocity, a is acceleration.
4^2= 0^2 + 2 *2*s
16= 0 + 4s
s= 4 meters
Distance (s) can also be found through:
s= ut + 1/2 at^2
s= 0+ 1/2 *2*2^2= 1 *2*2
s= 4 meters
Explanation:
