You are looking straight down on a magnetic compass that is lying flat on a table. A wire is stretched horizontally under the table, parallel to and a short distance below the compass needle. The wire is then connected to a battery so that a current I flows through the wire. This current causes the north pole of the compass needle to deflect to the left. The questions that follow ask you to compare the effects of different actions on this initial deflection.
If the wire is lowered farther from the compass, how does the new angle of deflection of the north pole of the compass needle compare to its initial deflection?

Solution :

Given :

Length of the propeller rods, L =0.30 m

Mass of each, M = 0.040 kg

Moment of inertia of one propeller rod is given by  

[tex]$I=\frac{1}{3}\times M \times L^2$[/tex]

Therefore, total moment of inertia is

[tex]$I=3 \times \frac{1}{3}\times M \times L^2$[/tex]

[tex]$I=M\times L^2$[/tex]

[tex]$I=0.04\times (0.3)^2$[/tex]

  [tex]$0.0036 \ kg \ m^2$[/tex]

Now energy required is given by

[tex]$E=\frac{1}{2}\times I \times \omega^2 $[/tex]

where, angular speed, ω = 5800 rpm

[tex]$\omega = 5800 \times \frac{2 \pi}{60} $[/tex]

 = 607.4 rad/s

Therefore energy,

[tex]$E=\frac{1}{2}\times 0.0036 \times (607.4)^2 $[/tex]

   = 664.1 J

The moment of inertia of the propeller is 0.0036 kgm² and the energy required is 663.21 J

Given that the mass of the propellers is m = 0.040kg,

and their length is L = 0.30m

The moment of inertia of a rod with the rotation axis at one end is given by :

[tex]I = \frac{1}{3}m L^2[/tex]

so for 3 propellers:

[tex]I=3\times\frac{1}{3}\times(0.04)\times(0.3)^2[/tex]

I = 0.04 × 0.09

I = 0.0036 kgm²

Now, the frequency is given f = 5800 rpm

so anguar speed, ω = 5800×(2π/60)

ω = 607 rad/s

Energy required:

E = ¹/₂Iω²

E = 0.5 × 0.0036 × (607)² J

E = 663.21 J

Learn more about moment of inertia:

https://brainly.com/question/15248039?referrer=searchResults

Answer:

f = 29605.2 [N]

Explanation:

To solve this problem we must use the following equation of kinematics, then use Newton's second law.

[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]

where:

Vf = final velocity = 300 [m/s]

Vo = initial velocity = 450 [m/s]

a = acceleration [m/s²]

x = distance = 7.6 [cm] = 0.076 [m]

Now replacing and clearing a

2*a*0.076 = (450² - 300²)

a = 740131.57 [m/s²]

Now using Newton's second law which tells us that the force on a body is equal to the product of mass by acceleration.

f = m*a

where:

f = friction force [N]

m = mass = 40 [g] = 0.04 [kg]

f = 0.04*740131.57

f = 29605.2 [N]