Answer:
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Explanation:
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 original length of the specimen is found to be 76.093 mm.
Explanation:
From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:
Initial Volume = Final Volume
πd1²L1/4 = πd2²L2/4
d1²L1 = d2²L2
L1 = d2²L2/d1²
where,
d1 = initial diameter = 19.636 mm
d2 = final diameter = 19.661 mm
L1 = Initial Length = Original Length = ?
L2 = Final Length = 75.9 mm
Therefore, using values:
L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²
<u>L1 = 76.093 mm</u>
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium conversion is 1.562 where lt denotes liter
The final pressure is 3895.94 Kpa
The plot of the composition of the reactor volume as a function of time is on the second uploaded image
Explanation:
The explanation is shown on the third , fourth and fifth images