A water balloon slingshot launches its projectiles essentially from ground level at a speed of 27.5 m/s . (You can ignore air re
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La longitud <em>final</em> del puente de acero es 100.018 metros.
Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (), en metros:
(1)
Donde:
- Longitud inicial del puente, en metros.
- Coeficiente de dilatación, sin unidad.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si tenemos que ,
,
y
, entonces la longitud final del puente de acero es:
La longitud <em>final</em> del puente de acero es 100.018 metros.
Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416
See the graph in attachment
Explanation:
In this problem we have to draw a velocity-time graph for an object travelling initially at -3 m/s, then slowing down and turning around.
In the graph, we see that the initial velocity at time t = 0 is
and it is negative, so below the x-axis.
Later, the object slows down: this means that the magnitude of its velocity increases, therefore (since the velocity is negative) the curve must go upward, approaching and reaching the x-axis (which corresponds to zero velocity).
After that, the object's velocity keep increasing, but now it is positive: this means that the object is travelling in a direction opposite to the initial direction, so it has turned around.
Learn more about velocity:
#LearnwithBrainly
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)