Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
The force exerted by a magnetic field on a wire carrying current is:
where I is the current, L the length of the wire, B the magnetic field intensity, and
the angle between the wire and the direction of B.
In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is
, so we can re-arrange the formula and substitute the numbers to find B:
Answer:
Heat Input = Work Output (at 100% efficiency)
ΔQ = ΔW
(you cannot get something for nothing)
Answer: D <u>(chemical</u> -> <u>heat</u> -> <u>mechanical</u>)
In automobile engines the petrol/diesel fuel enter in to the engine cylinder, due to spark at the end of the compression, fuel burnt increase the temperature and pressure, develops heat <em>(chemical energy -> heat energy). </em><em>This heat energy acts on a piston develops the work on the crankshaft </em><em>( Heat energy -> Mechanical energy)</em><em>. </em>
If it is just a reseach investigatory project, you could monitor it by using different variables (controlled), acurate measuring devices and camera for documentations
