Se deja caer una piedra A en reposo desde un acantilado muy alto. Cuando ha caído 5 m, se deja caer una piedra B. A. Explicar ¿c
ómo cambia la distancia entre ambas piedras a medida que caen? B. Determina la velocidad de la piedra B cuando ha recorrido 5 m. C. En una sola gráfica velocidad vs tiempo, bosque las gráficas de las dos piedras cuando la piedra B ha recorrido 10 m. NO necesita colocar valores en las gráficas.
1 answer:
Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.
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Answer:
51.2 mi/h
Explanation:
Total distance, d = 100 miles
First 60 miles with speed 55 mi/h
Next 40 miles with speed 75 mi/h
Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h
Time taken for 40 miles, t2 = 40 / 75 = 0.533 h
Time spent to get stuck, t3 = 20 min = 0.33 h
Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h
The average speed is defined as the ratio of total distance traveled to the total time taken.
Average speed =
Thus, the average speed of the journey is 51.2 mi/h.
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here is the velocity after the collision.
Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by
where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched
Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components