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slava [35]
3 years ago
14

What is name for grandmother in German?

Engineering
1 answer:
Alona [7]3 years ago
7 0

Answer:

Grandmother in German is "Oma."

Explanation:

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Low.

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This is encountered in sound waves propagation.

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I need solution for this question please<br>Select the right answer ​
Ostrovityanka [42]

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the 7th one answer is beacause mercury is bad at sharing electrons

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Define the Artist class in Artist.py with a constructor to initialize an artist's information. The constructor should by default
mina [271]

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Explanation:

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6 0
3 years ago
The annual average of solar photovoltaic energy in Phoenix is 6,720
Dovator [93]

Answer:

The amount of money saved is $22,075.2

Explanation:

The solar voltaic energy in Phoenix, S_E = 6,720 Wh/m²/day

The area on the roof available for solar panels, A = 500 m²

The percentage of the energy received by the solar panels converted into electricity, η = 15% = 0.15

The cost of electricity, C = $0.12/kWh

The number of days in a year, t = 365 days

Therefore, the maximum area covered by the solar panel = A = 500 m²

The total solar photovoltaic energy received by the solar panels, 'E', in a year is given as follows;

E = S_E × A × t

By plugging in the values of the variables, we get;

E = 6,720 Wh/m²/day × 500 m² × 365 days = 1,226,400,000 Wh

The electrical energy generated, E_e = η × E

∴ E_e = 0.15 × 1,226,400,000 Wh = 183960000 Wh = 183,960 kWh

The amount of money saved = The cost of electricity generated = C × E_e

C × E_e = $0.12/kWh × 183,960 kWh = $22,075.2

∴ The amount of money saved = C × E_e = $22,075.2

3 0
2 years ago
A composite wall is composed of 20 cm of concrete block with k = 0.5 W/m-K and 5 cm of foam insulation with k = 0.03 W/m-K. The
wariber [46]

Answer:

4.8°C

Explanation:

The rate of heat transfer through the wall is given by:

q=\frac{Ak}{L}dT

\frac{q}{A}=\frac{k}{L}dT

Assumptions:

1) the system is at equilibrium

2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point

3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side

{k_{fi}= 0.03 W/m.K

{L_{fi}= 5 cm = 0.05 m

{T_{fi}= 25 \°C

{k_{cb} = 0.5 W/m.K

{L_{cb}= 20 cm = 0.20 m

{T_{cb}= 0 \°C

{T_{m}= ? \°C = temperature at the interface

Solving for {T_{m} will give the temperature at the interface:

\frac{q}{A}=\frac{k_{fi} }{L_{fi} }(T_{fi} -T_{m})=\frac{k_{cb} }{L_{cb} }(T_{m} -T_{cb})

\frac{0.03}{0.05 }(25 -T_{m})=\frac{0.5}{0.2}(T_{m} -0})

15 -0.6T_{m}=2.5T_{m}

3.1T_{m}=15

T_{m}=4.8

3 0
3 years ago
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