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Lunna [17]
2 years ago
14

What is the best way to collaborate with your team when publishing Instagram Stories from Hootsuite?

Engineering
1 answer:
inn [45]2 years ago
6 0
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
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Can some help me with this !!! Is 26 points!!
Aleonysh [2.5K]
Third one
15,000,000 ohms because M=10^6
8 0
3 years ago
Calculate the amount of current flowing through a 75-watt light bulb that is connected to a 120-volt circuit in your home.
vodomira [7]

Answer:

I = 0.625 A

Explanation:

Given that,

Power of the light bulb, P = 75 W

Voltage of the circuit, V = 120 V

We need to find the current flowing through it. We know that, Power is given by :

P=V\times I

I is the electric current

I=\dfrac{P}{V}\\\\I=\dfrac{75\ W}{120\ V}\\\\I=0.625\ A

So, the current is 0.625 A.

5 0
2 years ago
Can someone explain it to me!
Darina [25.2K]

Answer:

It will be B

Explanation:

Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.

7 0
2 years ago
Is an isothermal process necessarily internally reversible? Explain your answer with an example
torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0
2 years ago
The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su
VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK  ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

                      R_i2 = 1 / h_i + L_1 / k_1

                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0
3 years ago
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