Answer:
magnesium (Mg), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table, and the lightest structural metal. Its compounds are widely used in construction and medicine, and magnesium is one of the elements essential to all cellular life.
Explanation:
hope it helps
From conservation of momentum, the ram force can be calculated similarly to rocket thrust:
F = d(mv)/dt = vdm/dt.
<span>In other words, the force needed to decelerate the wind equals the force that would be needed to produce it.
</span><span> v = 120/3.6 = 33.33 m/s
</span><span> dm/dt = v*area*density
</span> dm/dt = (33.33)*((45)*(75))*(1.3)
dm/dt = <span>
146235.375 </span><span>kg/s
</span><span> F = v^2*area*density
</span> F = (33.33)^2*((45)*(75))*(1.3) = <span>
<span>4874025 </span></span><span>N
</span> This differs by a factor of 2 from Bernoulli's equation, which relates velocity and pressure difference in reference not to a head-on collision of the fluid with a surface but to a fluid moving tangentially to the surface. Also, a typical mass-based drag equation, like Bernoulli's equation, has a coefficient of 1/2; however, it refers to a body moving through a fluid, where the fluid encountered by the body is not stopped relative to the body (i.e., brought up to its speed) like is the case in this problem.
Needs mass to be reasonably solvable, sorry
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