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Mariana [72]
3 years ago
13

What are the most used electronic tools for electronic works?​

Engineering
1 answer:
Zanzabum3 years ago
6 0

Answer:

  • Needle-nose pliers.
  • Hemostats.
  • A spool of 30-gauge wire.
  • A spool of solder wick.
  • A spool of solder.
  • A loupe.
  • Three sets of tweezer

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DaniilM [7]

Answer:

DESCULPA MAS EU NÃO ENTENDI

8 0
3 years ago
What is the composition, in atom percent, of an alloy that consists of 4.5 wt% Pb and 95.5 wt% Sn?
jeka57 [31]

Answer: Option A is correct -- 2.6 at% Pb and 97.4 at% Sn.

Explanation:

Option A is the only correct option -- 2.6 at% Pb and 97.4 at% Sn. While option B, which is 7.6 at% Pb and 92.4 at% Sn. and option C, which is 97.4 at% Pb and 2.6 at% Sn. and option D, which is 92.4 at% Pb and 7.6 at% Sn. are wrong.

6 0
4 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

7 0
4 years ago
Which physical quantity measures fraction of the input energy a machine actually concerts into output energy ?
Dvinal [7]
The answer is Kinetic Power.
4 0
3 years ago
Steam enters an adiabatic turbine at 10 MPa and 500°C and leaves at 10 kPa with a quality of 90 percent. Neglecting the changes
Anna35 [415]

Answer:

The mass flow rate of steam m=5.4 Kg/s

Explanation:

Given:

  At the inlet of turbine P=10 MPa  ,T=500 C

 AT the exit of turbine  P=10 KPa   ,x=0.9

 Required power=5 MW

From steam table

<u> At 10 MPa and 500 C:</u>

  h=3374 KJ/Kg  ,s=6.59 KJ/Kg-K  (Super heated steam table)

<u>At 10 KPa:</u>

h_g=2675.1 KJ/Kg, h_f=417.51  KJ/Kg

s_g= 7.3  KJ/Kg-K                ,s_f=1.3   KJ/Kg-K

So enthalpy of steam at the exit of turbine

h= h_f+x(h_g- h_f)

Now by putting the values

h= 417.51+0.9(2675.1- 417.51) KJ/Kg

h=2449.34  KJ/Kg

Lets take m is the mass flow rate of steam

So 5\times 10^3=m\times (3374-2449.34)

m=5.4 Kg/s

So the mass flow rate of steam m=5.4 Kg/s

8 0
4 years ago
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