1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Alenkasestr [34]
3 years ago
8

A fusible link should be how many wire sizes smaller than the actual circuit wire?

Engineering
1 answer:
V125BC [204]3 years ago
5 0

Answer:

.75mm

Explanation:

jak ammu chamba lng

You might be interested in
An FPC 4 m2 in area is tested during the night to measure the overall heat loss coefficient. Water at 60 C circulates through th
sp2606 [1]

Answer:

<em> - 14.943 W/m^2K  ( negative sign indicates cooling ) </em>

Explanation:

Given data:

Area of FPC = 4 m^2

temp of water = 60°C

flow rate = 0.06 l/s

ambient temperature = 8°C

exit temperature = 49°C

<u>Calculate the overall heat loss coefficient </u>

Note : heat lost by water = heat loss through convection

m*Cp*dT  = h*A * ( T - To )

∴ dT / T - To = h*A / m*Cp  ( integrate the relation )

In ( \frac{49-8}{60-8} ) =  h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )

In ( 41 / 52 ) = 0.0159*h

hence h = - 0.2376 / 0.0159

              = - 14.943  W/m^2K  ( heat loss coefficient )

7 0
3 years ago
In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and100.06 g after the soil had
kipiarov [429]

Answer:

  • Moisture/ water content w = 26%
  • Void ratio , e =  0.73

Explanation:

  • Initial mass of saturated soil w1 = mass of soil - weight of container

                                                 = 113.27 g - 49.31 g = 63.96 g

  • Final mass of soil after oven w2 = mass of soil - weight of container

                                                  = 100.06 g - 49.31 g = 50.75

Moisture /water content, w =   \frac{w1-w2}{w2} = \frac{63.96-50.75}{50.75} = 0.26 = 26%

Void ratio =  water content X specific gravity of solid

                  = 0.26 X 2.80 =0.728

5 0
3 years ago
A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find
tresset_1 [31]

The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

Torque is find out by

P = 2πNT/60

10000 = 2π x 2000 x T / 60

T =47.74 N.m

The gear ratio Ne / Ns =4/1

Ns =2000/4 = 500

Ts =Ps x 60/(2π x 500)

Ts =190.96 N.m

Maximum shear stress τ = 16/π x (T / (d₀⁴ - d₁⁴))

τ max =T/J x D/2
where d₁ = 30mm = 0.03 m

           d₀ = 30 +(2x 4) = 38mm =0.038 m

Substitute the values into the equation, we get

τ max = 16 x 190.96 x 0.038 /π x (0.038⁴ - 0.03⁴)

τ max = 28.98 MPa.

Thus, the maximum shear stress in the tube is 28.98 MPa.

Learn more about power.

brainly.com/question/13385520

#SPJ1

7 0
2 years ago
You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the
Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, \sigma /Strain, ∈

initial Strain, \epsilon_i = \frac{\sigma}{E}

\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}

\epsilon_i = 0.002

creep rate in the steady state

\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )

\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )

but Tinitial = 0

\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001

\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0
3 years ago
You guys are amazing :D
Sloan [31]
Ik i am thank you tho xoxo
3 0
3 years ago
Read 2 more answers
Other questions:
  • Annie has collected several items from around her house. She is using these objects to investigate which objects are attracted t
    12·1 answer
  • Types of technology include:
    8·1 answer
  • A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
    6·1 answer
  • What is engine knock? What cause the engine knock problem?
    5·1 answer
  • Which of the following correctly explains why it would be beneficial for an engineer to become a member of the Leadership in Ene
    15·2 answers
  • Air is compressed in a well insulated compressor from 95 kPa and 27 C to 600 kPa and 277 C. Use the air tables; assume negligibl
    11·1 answer
  • An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM sig
    5·1 answer
  • For each resource, list 3 examples of how it would be used to produce a hamburger. Think outside of the “hamburger” box!
    12·1 answer
  • In homes today, what is behind the reason for flashover fires occurring much more rapidly than in the past generations?
    8·1 answer
  • Which option identifies the requirement standard NOT met in the following scenario?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!