Answer:
Explanation:
El impulso aplicado a la pelota produce una variación en su momento lineal.
J = m (V -Vo)
Conviene elegir positivo el sentido de la velocidad final.
J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s
Saludos Herminio
Answer:
The work done is 205 kJ.
Explanation:
Hi there!
Work can be calculated using the following equation:
W = F · Δx
Where:
W = work
F = applied force
Δx = displacement
In this case, the force varies with the position, so we can divide the traveled distance in very small parts and calculate the work done over each part of the trajectory. Then, we have to sum all the works and we will obtain the work done from the initial position (xi) to the final position (xf). This is the same as saying:
W = ∫ F · dx
F = 3.6 N/m³ · x³ - 76 N
W = ∫ (3.6 x³ - 76)dx
W = 0.9 x⁴ - 76x
Evaluating from xi to xf:
W = 0.9 N/m³ (21.9 m)⁴ - 76 N · 21.9 m - 0.9 N/m³(5.41 m)⁴ + 76 N · 5.41 m
W = 205 kJ
Answer:
The velocity of the cart at the bottom of the ramp is 1.81m/s, and the acceleration would be 3.30m/s^2.
Explanation:
Assuming the initial velocity to be zero, we can obtain the velocity at the bottom of the ramp using the kinematics equations:
Dividing the second equation by the first one, we obtain:
And, since 
, then:
It means that the velocity at the bottom of the ramp is 1.81m/s.
We could use this data, plus any of the two initial equations, to determine the acceleration:
So the acceleration is 3.30m/s^2.
Answer:
Velocity, quantity that designates how fast and in what direction a point is moving.
Explanation:
