Answer:
5.76 round off to 6
Explanation:
wall 1 = 15 × 12 = 180
wall 2 = 9 × 12 = 108
now 1 roll covers 50 square feet
formula = wall 1 + wall 2 / 50
= 180 + 108 / 50
= 288÷ 50
= 5.76
Answer:
a. 
b. 
Explanation:
The inertia can be find using
a.
now to find the torsion constant can use knowing the period of the balance
b.
T=0.5 s
Solve to K'
Momentum question. This is an inelastic collision, so
m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s / 180kg
Vf=0.67m/s
0.67m/s
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
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.
