Responder:
Velocidad = 41.5m / s
Espacio recorrida = 352.5 metros
Explicación:
Dado lo siguiente:
Velocidad inicial (u) = 19.8 km / h
Aceleración (a) = 2.4m / s ^ 2
Tiempo de viaje (t) = 15 s
A.) velocidad después de 15 s
Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s
Usando la ecuación: v = u + at, donde v es la velocidad
v = 5.5 + 2.4 (15)
v = 5.5 + 36
v = 41.5m / s
Espacio recorrida:
v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida
41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S
1722.25 = 30.25 + 4.8S
1722.25 - 30.25 = 4.8S
1692 = 4.8S S = 1692 / 4.8 S = 352.5m
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One has a large enough sample set of data on which tobase inductions
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Answer:
T₂ = 305.17 K
Explanation:
Given that,
Heat, Q = 6000 J
Mass, m = 200 gram
Initial temperature, T₁ = 25° C
We need to find its final temperature. Let it is T₂.
We know that,
Where
c is the specific heat of water, c = 4.18 J/g°C
So,
So, the final temperature is equal to 305.17 K.
Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:
<u>v₀ = 16.55 m/s</u>
