Average speed = (total distance covered) / (total time to cover the distance)
= (2,742 km) / (4.33 hours)
= (2,742 / 4.33) km/hr
= 633 km/hr (rounded)
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:
A book on its side exerts a greater force.
Explanation:
Pressure = Force / Area
Assuming that 1kg = 10N
2kg = 20N
Area of book lying flat = 0.3m × 0.2m
= 0.6m²
Pressure of book lying flat = 20N / 0.6m²
= 30Pa (1 s.f.)
Area of book on its side = 0.2m × 0.05m
= 0.01m²
Pressure of book on its side = 20N / 0.01m²
= 2000Pa (1 s.f.)
Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.
Answer:
I think he would be dead poggers
Explanation:
Warm water<span> has more energy </span><span>than cold water</span>
