Answer:
Vf = 39.2 m/s
h = 78.4 m
Explanation:
Using first equation of motion for vertical motion:
Vf = Vi + gt
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
g = 9.8 m/s²
t = time taken = 4 s
Therefore,
Vf = 0 m/s + (9.8 m/s²)(4 s)
Vf = 39.2 m/s
Now, we use second equation of motion
h = Vi t + (1/2)gt²
where,
h = height covered = ?
Therefore,
h = (0 m/s)(4 s) + (1/2)(9.8 m/s²)(4 s)²
h = 78.4 m
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?
Answer:
It is smaller
Explanation:
The engine of a model airplane must both spin a propeller and push air backward to propel the airplane forward. Model the propeller as three 0.30-m-long thin rods of mass 0.040 kg each, with the rotation axis at one end.
What is the moment of inertia of the propeller?
How much energy is required to rotate the propeller at 5800 rpm? Ignore the energy required to push the air.
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
Energy required for propeller: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:
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