D) How fast you are moving
Apply conservation of angular momentum:
L = Iw = const.
L = angular momentum, I = moment of inertia, w = angular velocity, L must stay constant.
L must stay the same before and after the professor brings the dumbbells closer to himself.
His initial angular velocity is 2π radians divided by 2.0 seconds, or π rad/s. His initial moment of inertia is 3.0kg•m^2
His final moment of inertia is 2.2kg•m^2.
Calculate the initial angular velocity:
L = 3.0π
Final angular velocity:
L = 2.2w
Set the initial and final angular momentum equal to each other and solve for the final angular velocity w:
3.0π = 2.2w
w = 1.4π rad/s
The rotational energy is given by:
KE = 0.5Iw^2
Initial rotational energy:
KE = 0.5(3.0)(π)^2 = 14.8J
Final rotational energy:
KE = 0.5(2.2)(1.4)^2 = 21.3J
There is an increase in rotational energy. Where did this energy come from? It came from changing the moment of inertia. The professor had to exert a radially inward force to pull in the dumbbells, doing work that increases his rotational energy.
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½
)/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Answer:
The impulse of the net force on the ball during its collision with the wall is 25N
Explanation:
Step one :
Given data
Mass =5kg
Velocity v1=10m/s
Velocity v2=5m/s
The equation for impulse is given as
P=m(v1-v2)
Where P =impulse
Substituting our values into the equation we have
P=5*(10-5)
P=5*5
P=25N
The impulse 25N
What is impulse?
Impulse is the change of momentum of an object when the object is acted upon by a force for an interval of time.
Answer:
6
Explanation:
everything after the decimal is counted
