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Oksi-84 [34.3K]
3 years ago
15

A compound machine contains three simple machines with IMAs of 2, 4 and 5, respectively. What is the overall ideal mechanical ad

vantage of the machine?
Engineering
1 answer:
anygoal [31]3 years ago
3 0

Answer:

Overall ideal mechanical advantage of the machine = 40

Explanation:

Given:

Ideal mechanical advantage of three machine = 2, 4, 5

Find:

Overall ideal mechanical advantage of the machine

Computation:

Overall ideal mechanical advantage of the machine = 2 × 4× 5

Overall ideal mechanical advantage of the machine = 40

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umka21 [38]

Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.

The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.

Learn more about the torque at brainly.com/question/28220969

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3 0
1 year ago
Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more ele
shutvik [7]

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

8 0
3 years ago
Water vapor at 100 psi, 500 F and a velocity of 100 ft./sec enters a nozzle operating at steady sate and expands adiabatically t
almond37 [142]

Answer:

a)exit velocity of the steam, V2 = 2016.8 ft/s

b) the amount of entropy produced is 0.006 Btu/Ibm.R

Explanation:

Given:

P1 = 100 psi

V1 = 100 ft./sec

T1 = 500f

P2 = 40 psi

n = 95% = 0.95

a) for nozzle:

Let's apply steady gas equation.

h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}

h1 and h2 = inlet and exit enthalpy respectively.

At T1 = 500f and P1 = 100 psi,

h1 = 1278.8 Btu/Ibm

s1 = 1.708 Btu/Ibm.R

At P2 = 40psi and s1 = 1.708 Btu/Ibm.R

1193.5 Btu/Ibm

Let's find the actual h2 using the formula :

n = \frac{h_1 - h_2*}{h_1 - h_2}

n = \frac{1278.8 - h_2*}{1278.8 - 1193.5}

solving for h2, we have

h_2 = 1197.77 Btu/Ibm

Take Btu/Ibm = 25037 ft²/s²

Using the first equation, exit velocity of the steam =

(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}

Solving for V2, we have

V2 = 2016.8 ft/s

b) The amount of entropy produced in BTU/ lbm R will be calculated using :

Δs = s2 - s1

Where s1 = 1.708 Btu/Ibm.R

At h2 = 1197.77 Btu/Ibm and P2 =40 psi,

S2 = 1.714 Btu/Ibm.R

Therefore, amount of entropy produced will be:

Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R

= 0.006 Btu/Ibm.R

3 0
3 years ago
Given the strings s1 and s2, not necessarily of the same length, create a new string consisting of alternating characters of s1
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Answer:

THE ANSA IZ A

Explanation:

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Bryan a project manager and his team have been assigned a new project. The team members have already started working on their as
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Answer:

there teams management is good

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