The correct answer is <span>The car has both potential and kinetic energy, and it is moving at 24.6 m/s.</span>
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
Answer:
The right solution is:
(a) 2.87 eV
(b) 1.4375 eV
Explanation:
Given:
Wavelength,
= 433 nm
Potential difference,
= 1.43 V
Now,
(a)
The energy of photon will be:
E = ![\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}](https://tex.z-dn.net/?f=%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B433%5Ctimes%2010%5E%7B-9%7D%7D)
= ![4.59\times 10^{-19} \ J](https://tex.z-dn.net/?f=4.59%5Ctimes%2010%5E%7B-19%7D%20%5C%20J)
or,
= ![\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=%5Cfrac%7B4.59%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
= ![2.87 \ eV](https://tex.z-dn.net/?f=2.87%20%5C%20eV)
(b)
As we know,
⇒ ![Vq=\frac{hc}{\lambda}-\Phi_0](https://tex.z-dn.net/?f=Vq%3D%5Cfrac%7Bhc%7D%7B%5Clambda%7D-%5CPhi_0)
By substituting the values, we get
⇒ ![1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0](https://tex.z-dn.net/?f=1.43%5Ctimes%201.6%5Ctimes%2010%5E%7B19%7D%3D%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-34%7D%5Ctimes%203%5Ctimes%2010%5E8%7D%7B433%5Ctimes%2010%5E%7B-9%7D%7D-%5CPhi_0)
⇒ ![\Phi_0=2.3\times 10^{-19} \ J](https://tex.z-dn.net/?f=%5CPhi_0%3D2.3%5Ctimes%2010%5E%7B-19%7D%20%5C%20J)
or,
⇒ ![=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2.3%5Ctimes%2010%5E%7B-19%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
⇒ ![=1.4375 \ eV](https://tex.z-dn.net/?f=%3D1.4375%20%5C%20eV)
The answer would be D, electromagnetic waves