This question involves the concepts of the law of conservation of momentum and velocity.
The velocity of the eight ball is "5.7 m/s".
According to the law of conservation of momentum:
where,
m₁ = mass of number three ball = 5 g
m₂ = mass of the eight ball = 6 g
u₁ = velocity of the number three ball = 3 m/s
u₂ = velocity of the eight ball = - 1 m/s (negative sign due to opposite direction)
v₁ = final velocity of the three number ball = - 5 m/s
v₂ = final velocity of the eight ball = ?
Therefore,
(5 g)(3 m/s) + (6 g)(- 1 m/s) = (5 g)(- 5 m/s) + (6 g)(v₂)
<u>v₂ = 5.7 m/s</u>
<u></u>
Learn more about the law of conservation of momentum here:
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
Let d is the distance moved in 2.25 s. Using second equation of motion,
So, it will move 6.32 m from rest in 2.25 seconds.
Answer:
a = - 1.47 [m/s²], descending or going down
Explanation:
To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
∑F = m*a
where:
∑F = Forces applied [N]
m = mass = 10 [kg]
a = acceleration [m/s²]
Let's assume the direction of the upward forces as positive, just as if the movement of the box is upward the acceleration will be positive.
By performing a summation of forces on the vertical axis we obtain all the required forces and other magnitudes to be determined.
where:
g = gravity acceleration = 9.81 [m/s²]
N = normal force measured by the scale = 83.4 [N]
Now replacing:
The acceleration has a negative sign, this means that the elevator is descending at that very moment.