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Evgen [1.6K]
3 years ago
14

f the rope is drawn toward the motor M at a speed of vM= (5t3/2) m/s, where t is in seconds, determine the speed of the cylinder

when t = 1.5
Engineering
1 answer:
SCORPION-xisa [38]3 years ago
4 0

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3 0
3 years ago
B is the desired product, and X and Y are foul pollutants that are expensive to get rid of. The specific reaction rates are at 2
AURORKA [14]

Answer:

See attachment for detailed answer.

Explanation:

Download pdf
3 0
3 years ago
One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
Jlenok [28]
False i think it would be
3 0
3 years ago
The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
leva [86]

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

5 0
3 years ago
Read 2 more answers
One kilogram of "as received" yard trimmings is made up of approximately 620 g moisture, 330 g of decomposable organic matter (r
Evgesh-ka [11]

Answer:

Explanation:

(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄

Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14

1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54

=  153.12 + 21.28 + 148.16 + 7.56

= 330.12 g/mol

which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable

The moles of methane produced will be given as

m = (4a + b -2c - 3d)/8

= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8

= (51.04 + 21.28 - 18.52 - 1.62)/8

= 52.18/8

= 6.5225

(b) Volume of methane V is given as

V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)

= 0.1461 m³ CH₄/kg lawn trimmings

(c) Energy will be given as

CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol

= 5805.025

≈ 5805 kJ/kg

5 0
3 years ago
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