You are correct it is translucent because opaque let’s no light through but transparent lets almost all light through
Answer:
Explanation:
Impulse of a force is measured by force x time or F X t
Impulse also equals change in momentum or
F x t = m v₂ - m v₁
The given case is as follows
in the first case
F x t = mv - o = mv
F = mv / t
in the second case
F₁ x 4 t = mv
F₁ = 1/4 x mv /t
F₁ = F / 4
option a) is correct .
iii )
In the last case
F₂ X t = m v/2 -0
F₂ = 1/2 x mv / t
= 1/2 x F
F₂ = F/2
Option e ) is correct.
<span>The primary reason a light bulb emits light is due to the heating of the resistance in the filament of the light bulb. In fact, the power dissipated in a resistor is given by
</span>
<span>where I is the current and R the resistance. The larger the resistance or the current in the resistor, the larger the power dissipated. Due to this dissipation of power, the temperature of the filament becomes very high, and the resistance becomes incandescent, emitting light.</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
The magnitude of charge on a proton and electron is the same, 1.602 x 10-19 C. Protons are +, and electrons -.
