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Shkiper50 [21]
2 years ago
6

3. Assertive communication means:

Engineering
1 answer:
natta225 [31]2 years ago
7 0

Answer:

A. Confidently and directly communicating what you think and feel, without being aggressive or passive.

Explanation:

Assertive communication is the ability to express both positive and negative ideas in a way that it is a honest and direct message. It doesn't mean to say things to the other person with the intent of causing harm or making them feel better, it's just saying things as they are.

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What is the best way to collaborate with your team when publishing Instagram Stories from Hootsuite?
inn [45]
Implement a robust approval process with multiple touch points so every team member has input
Save the Story as a draft, then share it with your team via Hootsuite Inbox to gather feedback before publishing
Schedule Stories in the Planner so they can be reviewed and edited by the approver directly
Add specific publishing notes in the composer to help guide the teammate ultimately publishing the Stories
6 0
3 years ago
9. A piece of Cherry wood is 5/4 x 4" X 4'<br> What is the length in inches?
Firlakuza [10]

Answer:

48 inches

Explanation:

the 4' means 4 feet. the " means inches but the ' means feet. the longest measurment means its the length, so 4 feet converted to inches are 48 inches, giving you your answer. i hope this helps :D

8 0
3 years ago
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of
True [87]

Answer:

None of these

Explanation:

There are different types of amplifiers, and each has different characteristics.

  • Voltage amplifier needs high input and low output  resistance.
  • Current amplifier needs Low Input and High Output  resistance.
  • Trans-conductance amplifier Low Input and High Output resistance.
  • Trans-Resistance amplifier requires High Input and Low output  resistance.

Therefore, the correct answer is "None of these "

3 0
3 years ago
A long aluminum wire of diameter 3 mm is extruded at a temperature of 280°C. The wire is subjected to cross air flow at 20°C at
Musya8 [376]

Answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}C

Temperature of air, T_{\infinity}=20^{o}C

Velocity of air flow V=5.5m/s

The film temperature is determined as:

T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}C

Thermal conductivity, K = 0.03443 W/m^oC; kinematic viscosity v=2.860 \times 10^{-5} m^2/s; Prandtl number Pr=0.70275

The reynolds number for the flow is determined as:

Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92

sice the obtained reynolds number is less than 2\times10^5, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}

Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}

Rewrite and solve for h

h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/m

6 0
3 years ago
What is the tolerance of number 4?
Kamila [148]

Answer:

Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)

Explanation:

says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance

(based on image sent in other post)

5 0
3 years ago
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