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3 years ago

What does snow fall from?

Engineering

1 answer:

Klio2033 [76]3 years ago

4 0

Answer:

Clouds

Explanation:

It is created by trapped dust and water.

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A single phase molor is located

belka [17]

Answer:

hyguhttbjjgccd h jb

Explanation:

hcufugibibuguguv

5 0

3 years ago

Two sites are being considered for wind power generation. On the first site, the wind blows steadily at 7 m/s for 3000 hours per

kirill [66]

Solution :

Given :

$V_1 = 7 \ m/s$

Operation time, $T_1$ = 3000 hours per year

$V_2 = 10 \ m/s$

Operation time, $T_2$ = 2000 hours per year

The density, ρ = $1.25 \ kg/m^3$

The wind blows steadily. So, the K.E. = $(0.5 \dot{m} V^2)$

$= \dot{m} \times 0.5 V^2$

The power generation is the time rate of the kinetic energy which can be calculated as follows:

Power = $\Delta \ \dot{K.E.} = \dot{m} \frac{V^2}{2}$

Regarding that $\dot m \propto V$ . Then,

Power $\propto V^3$ → Power = constant x $V^3$

Since, $\rho_a$ is constant for both the sites and the area is the same as same winf turbine is used.

For the first site,

Power, $P_1= \text{const.} \times V_1^3$

$P_1 = \text{const.} \times 343 \ W$

For the second site,

Power, $P_2 = \text{const.} \times V_2^3 \ W$

$P_2 = \text{const.} \times 1000 \ W$

5 0

2 years ago

Physical properties of minerals

lord [1]

Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

- BRAINLIEST answerer ❤️✌

6 0

2 years ago

Read 2 more answers

A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat

Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

$Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ$