A mass is oscillating up and down on a spring. In the above graph of
1 answer:
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Answer:
The velocity of skaters after collision is 0.55 m/s and they are moving to the left
Explanation:
Consider m₁ and m₂ are the masses of two skaters and their initial velocity be v₁ and v₂ respectively.
Consider the velocity along right direction as positive while negative for left direction.
According to the problem,
Mass of first skater, m₁ = 66.0 kg
Mass of second skater, m₂ = 75.0 kg
Velocity of first skater, v₁ = + 2.00 m/s
Velocity of second skater, v₂ = -2.80 m/s
Since, the two skaters grab each other after collision. Hence, they are moving with same final velocity v.
Applying conservation of linear momentum,
Momentum before collision = Momentum after collision
m₁v₁ + m₂v₂ = (m₁+m₂)v
Substitute the suitable values in the above equation.
66 x 2 - 75 x 2.8 = (66 + 75 )v
v = -0.55 m/s
The negative sign denotes that both the skaters are moving in the left direction.
Answer:
Explanation:
The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.
where:
w = weight [N]
m = mass = 0.06 [kg]
g = gravity acceleration = 10 [N/kg]
Therefore:
By Hooke's law we know that the force in a spring can be calculated by means of the following expression.
where:
k = spring constant [N/m]
x = deformed distance = 6 [cm] = 0.06 [m]
We can find the spring constant.
Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.
Since this force is equal to the weight, we can now determine the gravitational acceleration.