Answer:
Explanation:
A plane flies due north (90° from east) with a velocity of 100 km/h for 2 hours.
With no wind, it will be 100*2 = 200 km north of its starting point.
But a steady wind blows southeast at 30 km/h at an angle of 315° from due east.
So the wind itself will blow the plane 30*2 = 60km at an angle of 315° from due east.
That is the same as 60*cos315° = 42.43km due east and 60*sin315° = -42.43km north.
Combining, the plane is at 42.43km due east and 200-42.43 = 157.57km due north from its starting point.
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:
80 m/s
Explanation:
Given:
a = -5 m/s²
v = 0 m/s
Δx = 640 m
Find: v₀
v² = v₀² + 2a(x − x₀)
(0 m/s)² = v₀² + 2(-5 m/s²) (640 m)
v₀ = 80 m/s
The difference between the two is, well for one
Spectrum: The entire range that the "<em>waves" </em>could be such, as visible light, x-ray's and so on.
Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.
<em>It may confuse you but it makes sense to me (Sorry)</em>
Build walls around the coast