You drop a ball from a height of 1.7 m, and it bounces back to a height of 1.2 m.
1 answer:
Answer:
A) ΔEm = 0.29, B) v₁ = 5.8 m/s, c) v₂= 4.9 m / s D) the correct answer from 4
Explanation:
For this exercise we will use conservation of energy, taking care of how to choose our system
A) For this case we take two instants
starting point. When the ball goes out
Em₀ = U = m g y₁
Final point. When the ball reaches its maximum height
= U = m g y₂
In this case we see that there is a loss of mechanical energy at the moment of rebound, therefore the fraction of energy lost is
ΔEm = Em_{f} / Em₀
ΔEm = mg y₂ / mg y₁
ΔEm = y₂ / y₁
ΔEm = 1.2 / 1.7
the lost part of energy
ΔEm = 1 -0.706
ΔEm = 0.29
B) the velocity just before the bounce
starting point. When the ball is released
Em₀ = U = m g y₁
final punot. Just wide of the bounce
Em_{f} = K = ½ m v₁²
As it has not yet rebounded, it has no energy loss, therefore the mechanical energy is conserved
Em₀ = Em_{f}
m g y₁ = ½ m v₁²
v₁ = √ 2 g y₁
let's calculate
v₁ =√ (2 9.8 1.7)
v₁ = 5.77 m / s
v₁ = 5.8 m/s
C) the velocity just after the bounce
starting point, after bounce
Em₀ = K = ½ m v₂²
final point. Maximum height after bounce
Em_{f} = U = m g y₂
as it already bounced, the energy is conserved in this interval
Em₀ = Em_{f}
½ m v₂² = m g y₂
v₂ = √ (2 g y₂)
v₂ = √ (2 9.8 1.2)
v₂ = 4.85 m / s
v₂= 4.9 m / s
D) during the time that the bounce lasts, there is a strong change in energy, part of it is transformed into thermal energy, due to several processes: friction, change in the potential energy of the molecules of the ball, change in the internal energy of the balls. molecules.
Therefore we cannot specify a single process, consequently the correct answer from 4
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Answer:
t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s
Explanation:
Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso
y = y₀ + t – ½ g t²
en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)
0 = y₀ – ½ g t²
t= √ (2 y₀/g)
calculemos
t= √ ( 2 12 / 9,8)
t= 1,56 s
El alcance del proyectil es la distancia horizontal recorrida
x = v₀ₓ t
x = 80 1,56
x= 124,8 m
La velocidad de impacto cuando toca el suelo
vx = v₀ₓ = 80 ms
= v_{oy} – gt
v_{y} = - 9,8 1,56
v_{y} = - 15,288 m/s
la velocidad es
v = (80 i^ - 15,288 j ) m/s
Traducttion
This is a projectile launching exercise, let's start by finding the time it takes to reach the ground
y = y₀ + v_{oy} t - ½ g t²
in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( = 0)
0 =y₀I - ½ g t²
t = √ (2y₀ / g)
let's calculate
t = √ (2 12 / 9.8)
t = 1.56 s
Projectile range is the horizontal distance traveled
x = v₀ₓ t
x = 80 1.56
x = 124.8 m
Impact speed when it hits the ground
vₓ = v₀ₓ = 80 ms
v_{y} = v_{oy} - gt
v_{y} = - 9.8 1.56
v_{y} = - 15,288 m / s
the speed is
v = (80 i ^ - 15,288 j) m / s
Answer:
Force, |F| = 2100 N
Explanation:
It is given that,
Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s,
Initial speed, v = 42 m/s
The momentum is reduced to zero, final speed, v = 0
The relation between the force and the momentum is given by :
|F| = 2100 N
So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.