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Juli2301 [7.4K]
2 years ago
9

For a small company it's usually best to keep the corporate and brand image as___ as possible​

Engineering
1 answer:
Elanso [62]2 years ago
8 0

Answer:

soon

Explanation:

the answer is soon can i have brainliest please

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Information such as tolerances and scale can be found in the _______________ of an engineering drawing.
nasty-shy [4]

Answer:

Information such as tolerance and scale can be found in the <u>title block</u> of an engineering drawing

Explanation:

The title block of an engineering drawing can normally be found on the lower right and corner of an engineering drawing and it carries the information that are used to specify details that are specific the drawing including, the name of the project, the name of the designer, the name of the client, the sheet number, the drawing tolerance, the scale, the issue date, and other relevant information, required to link the drawing with the actual structure or item

8 0
2 years ago
What does the simplify command do
uranmaximum [27]

Answer:The simplify command is used to apply simplification rules to an expression. The simplify routine searches the expression for function calls, square roots, radicals, and powers and invokes the appropriate simplification procedures. For detailed information on the simplify command, see simplify/details.

Explanation:

4 0
2 years ago
The function below takes a single argument: data_list, a list containing a mix of strings and numbers. The function tries to use
pychu [463]

Answer:

def filter_only_certain_strings(data_list):

   new_list = []

   for data in data_list:

       # fix here. change >= to >  

       # because we want to return strings longer than 5 characters in length.

       if type(data) == str and len(data) > 5:

           new_list.append(data)

   return new_list

Explanation:

def filter_only_certain_strings(data_list):

   new_list = []

   for data in data_list:

       # fix here. change >= to >  

       # because we want to return strings longer than 5 characters in length.

       if type(data) == str and len(data) > 5:

           new_list.append(data)

   return new_list

3 0
3 years ago
A 2-m3 rigid tank initially contains air at 100 kPa and 22°C. The tank is connected to a supply line through a valve. Air is flo
Finger [1]

Answer:

9.58 Kg of air has entered the tank.

heat entered=3483.76 Kilo.Joule

Explanation:

(A) R=287 Kilo.J/Kg.K

as per initial conditions P=100 Kilo.Pa ,V=2 cubic meter, T=22 C=295.15 K,

using the relation P*V=m*R*T

m=(100*1000*2)/(287*295.15)=2.36 Kg this is the mass that is already present in tank.

after filling tank at 600 Kilo.Pa.

P=600 Kilo Pa T=77 C=350.15 K

P*V=m*R*T

m=(600*1000*2)/(287*350.15)=11.94 Kg

mass that has entered=11.94-2.36=9.58 Kg

(b) using air psychometric  property table

specific heat content initial  100 KILO Pa and 22 C=295.576 Kilo.Joule/Kg

specific heat content final  600 Kilo Pa and 77 C=350.194 Kilo.Joule/Kg

heat at initial stage=295.576*2.36=697.56 Kilo.Joule

heat at final stage=350.194*11.94=4181.32 Kilo.Joule

heat entered=4181.32-697.56=3483.76 Kilo.Joule

3 0
3 years ago
The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals n
tekilochka [14]

Answer:

the required expression is E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - e^(-0.05t) ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w'  ------ let this be equation 1

w' is the net power on the system

w' = w_{elect - w_{shaft

w_{shaft = T × ω

we substitute

w_{shaft= 18 × 100

w_{shaft = 1800 W

w_{shaft = 1.8 kW  

so

w' = w_{elect - w_{shaft

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - e^{(-0.05t) ] + 0.2

dE/dt = -0.2 + 0.2e^{(-0.05t) ] + 0.2

dE/dt = 0.2e^{(-0.05t)

Now, the change in total energy, increment E, as a function of time;

ΔE = \int\limits^t_0}\frac{dE}{dt}  . dt

ΔE = \int\limits^t_0} 0.2e^{(-0.05t)} dt

ΔE = \int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0

E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

Therefore, the required expression is E_2 - E_1 = 4[ 1 - e^{(-0.05t)} ]

5 0
2 years ago
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