Answer:
symbolic machine code.
Explanation:
The instructions in the language are closely linked to the machine's architecture.
Answer:
The distance measure from the wall = 36ft
Explanation:
Given Data:
w = 10
g =32.2ft/s²
x = 2
Using the principle of work and energy,
T₁ +∑U₁-₂ = T₂
0 + 1/2kx² -wh = 1/2 w/g V²
Substituting, we have
0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²
170 = 0.15528V²
V² = 170/0.15528
V² = 1094.796
V = √1094.796
V = 33.09 ft/s
But tan ∅ = 3/4
∅ = tan⁻¹3/4
= 36.87°
From uniform acceleration,
S = S₀ + ut + 1/2gt²
It can be written as
S = S₀ + Vsin∅*t + 1/2gt²
Substituting, we have
0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)
19.85t - 16.1t² + 3 = 0
16.1t² - 19.85t - 3 = 0
Solving it quadratically, we obtain t = 1.36s
The distance measure from the wall is given by the formula
d = VCos∅*t
Substituting, we have
d = 33.09 * cos 36. 87 * 1.36
d = 36ft
Answer:
The rate of energy absorbed per unit time is 3500W.
Explanation:
From the question, we were given the following parameters;
Plane, opaque, gray, diffuse surface
â = 0.7
Surface area, A = 0.5m²
Incoming radiant energy, G = 10000w/m²
T = 500°C
Rate of energy absorbed is âAG;
âAG = 0.7 × 0.5 × 10000
âAG = 3500W.
The energy absorbed is measured in watts and denoted by the symbol W.
Answer:
n = 2r³/Rd²
Explanation:
See the attached file for the derivation.
