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3 years ago

Based on the following passage on construction technology during the Middle Ages, why might a worker not be allowed

Engineering

2 answers:

Alex777 [14]3 years ago

6 0

Answer:

He was not born into a family of skilled laborers

Explanation:

KonstantinChe [14]3 years ago

4 0

Answer:

A guild are an association of skilled merchants, craftsmen and artisans that provide support and protection for their members and push their professional interest forward and further their craft

Because the guild guild members mainly gained their skill through knowledge passed to them by their family bloodlines, other unskilled workers were by not being schooled on the craft and by family tradition not allowed to join a guild based on the nature knowledge transfer the guild members have which is mainly several generations of family trade trainings

Explanation:

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A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to

Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 = (16/3) * 10^2 V

c) E = 64/15 J

d) dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the ﬁnal potential difference across each capacitor

(d) the decrease in energy when the capacitors are connected.

Solution:

- The initial charge in the circuit is the one carried by the first charged capacitor.

Q_initial = C_1*V

Q_initial = 20*10^-6 * 800

Q_initial = 16 * 10^-3 C

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

Q_2 = Q_1*C_2/C_1

Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

Q_initial = 1.5*Q_1

Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

V_2 = V_1 = Q_1 / C_1

= 10.67*10^-3 / 20*10^-6

= (16/3) * 10^2 V

- The energy in the system is:

E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

E = 64/15 J

- The decrease in energy of the capacitors is:

dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

dE = 0.5*10^-6*20*800^2 - (64/15)

dE = 32/5 - 64/15 = 32/15 J

5 0

3 years ago

Convert the unit Decimeter (dm) into Micrometer (um).

oksian1 [2.3K]

Answer:

86701 Micrometers.

Explanation:

Multiply 0.86701 dm by 100,000 to get 86701 um.

7 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 $mm^4$

So the shear stress at point A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0

2 years ago

Click this link to view O*NET’s Wages and Employment section for Film and Video Editors.

vova2212 [387]

Answer:

much faster than average

Explanation:

did it on edge (2022-2032)

6 0

2 years ago

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

lianna [129]

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

5 0

3 years ago