Answer:
check attachments for answers
Explanation:
Answer:
Weather Conditions
Explanation:
You cant see as well when driving through the fog
Answer:
(a) shear plane angle (∅) = 33.53°
(b) shear strain (y) = 1.989
(c) material removal rate (MRR) = 1404mm³/s
Explanation:
Cutting speed = 1.8m/s
Feed = 0.3mm
Depth = 2.6mm
Angle = 8°
Thickness = 0.49
(a) calculating the chip thickness ratio using the formula;
r = t₀/tc
= 0.3/0.49
= 0.6122
Calculating the shear angle using the formula;
∅ = tan⁻¹[rcosα/1-rsinα]
= tan⁻¹[(0.6122*cos8)/(1-0.6122sin8)]
= 33.53°
(b) Calculating the shear strain using the formula;
y = cot∅ + tan(∅-α)
= cot 33.53 + tan(33.53-8)
= 1.509 + 0.477
= 1.989
(c) Calculating the material removal rate using the formula;
MRR =f*d*V
= 0.3 * 2.6 *1.8 *1000mm
= 1404mm³/s
Answer:
v₀ = 2,562 m / s = 9.2 km/h
Explanation:
To solve this problem let's use Newton's second law
F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v
F dx = m v dv
We replace and integrate
-β ∫ x³ dx = m ∫ v dv
β x⁴/ 4 = m v² / 2
We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max
-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)
x_max⁴ = 2 m /β v₀²
Let's look for the speed that the train can have for maximum compression
x_max = 20 cm = 0.20 m
v₀ =√(β/2m) x_max²
Let's calculate
v₀ = √(640 106/2 7.8 104) 0.20²
v₀ = 64.05 0.04
v₀ = 2,562 m / s
v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)
v₀ = 9.2 km / h