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
zhenek [66]
3 years ago
11

True or False? Duties of the company officials should be discussed

Engineering
1 answer:
Svet_ta [14]3 years ago
7 0

Answer:

F

Explanation:

You might be interested in
A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla
creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

so here Ф is very less  N_{\gamma } should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0
3 years ago
How many hours should I charge a 4.8v 600mah battery (I need it by today please)
prohojiy [21]

The number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

<h3>What is a battery?</h3>

It should be noted that an electric battery simply means a source of electric power that consist of one or more electrochemical cells that are with external connections that are important for powering electrical devices.

It should be noted that when a battery is supplying power, then the positive terminal is the cathode while the negative terminal is the anode.

In conclusion, the number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

Learn more about battery on:

brainly.com/question/16896465

#SPJ1

5 0
2 years ago
A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
3 years ago
You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19
Pepsi [2]

Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

3 0
3 years ago
A large increase in elevation can cause a carbureted engine to run ________ if not properly adjusted for the altitude. a Rich b
mash [69]

Answer:

B - Poor

Explanation:

As you get higher up, There is less oxygen which causes the engine to create less power.

3 0
3 years ago
Other questions:
  • An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th
    11·1 answer
  • A ballistic pendulum consists of a 3.60 kg wooden block on the end of a long string. From the pivot point to the center‐of‐mass
    6·1 answer
  • I have a question for you guys
    13·2 answers
  • The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the
    5·1 answer
  • In order to lift a lighter object on the other side, a boy placed 155 N of
    6·1 answer
  • Which organisms are consumers in this food chain? List all that apply. *
    5·1 answer
  • What is the name of the type of rocker arm stud that does not require a valve adjustment?
    12·1 answer
  • 2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
    8·1 answer
  • At the coast on a summer day, the land is hotter than the ocean. Warm air over the land rises and is replaced by cooler air, cau
    14·2 answers
  • How to plot 0.45 gradation chart for sieve analysis ?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!