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Andrews [41]
3 years ago
7

9. An elevator on a construction site is being operated at rated capacity of 6 tons, and is supported by two standard steel cabl

es, each having a cross-section area of 1.25-in.2 and an effective modulus of elasticity of 12 * 106 psi. As the elevator is descending at a constant 500 ft/min, an accident causes the top of the cable, 200 ft above the elevator, to stop suddenly. (a) Calculate the Impact Factor. (b) Estimate the maximum elongation and maximum tensile stress developed in cable. (c) Estimate the maximum elongation and maximum tensile stress developed in cable if the cable length was 100 ft. (d) Comment on the effect of changing cable length on maximum tensile stress developed in the cable.
Engineering
1 answer:
aleksklad [387]3 years ago
8 0
Woah that is a longgg question
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You are given a noninverting 741 op-amp with a dc-gain of 23.6 dB. The input signal to this amplifier is;Vin(t) = (0.18)∙cos(2π(
Vsevolod [243]

Answer:

Output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Explanation:

Given:

dc gain A = 23.6 dB

Input signal V_{in} (t) = 0.18 \cos (2\pi (57000)t +18.3)

Now convert gain,

A = 10^{\frac{23.6}{20} } = 15.13

DC gain at frequency f = 0 is given by,

  A = \frac{V_{out} }{V_{in} }

V_{out} =AV_{in}

V_{out} = 15.13 \times   0.18 \cos (2\pi (57000)t +18.3)

At zero frequency above equation is written as,

V_{out} = 2.72 \times \cos 18.3

V_{out} = 2.72

Now we write output voltage as input voltage,

V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

Therefore, output voltage equation is V_{out} (t) = 2.72 \cos (2\pi (57000)t +18.3)

7 0
3 years ago
A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu
NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet P_1 = 96 kPa

the pressure at the exit P_2 = ???

the pressure \rho = 1000 kg/m³

∴

0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}

0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}

\dfrac{800000}{2} = P_2-96000

400000 = P₂ - 96000

400000 +  96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0
2 years ago
Impact strips may be designed into a bumper cover.<br> True<br> False
d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

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3 years ago
Find the value of L
KonstantinChe [14]

Answer:

the value is 356732 Volt

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Find the altitude of the right cylinder of maximum convex surface that can be inscribed in a given sphere.
strojnjashka [21]

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The radius 4 is maximum in convex surface

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