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Rainbow [258]
3 years ago
11

Algorithm for sum 2+4+6+….+n

Engineering
1 answer:
gavmur [86]3 years ago
3 0
20 this is the sum hope this help
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A _____ satellite system employs many satellites that are spaced so that, from any point on the Earth at any time, at least one
Wittaler [7]

Answer:

d. low earth orbit (LEO)

Explanation:

This type of satellites form a constellation deployed as a series of “necklaces” in such a way that at any time, at least one satellite is visible by a receiver antenna, compensating the movement due to the earth rotation.

Opposite to that, a geostationary satellite is at an altitude that makes it  like a fixed point over the Earth´s equator, rotating synchronously with the Earth, so it is always visible in a given area.

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The moons light-colored highlights are
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The light colored material in areas of the mood is the earliest crust on the moon.
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3 years ago
Consider a refrigerator that consumes 320 W of electric power when it is running. If the refrigerator runs only one quarter of t
Semenov [28]

Answer:

B. $5.18

Explanation:

Cost of electricity per kWh = $0.09

Power consumption of refrigerator = 320W = 320/1000 = 0.32kW

In a month (30 days) the refrigerator works 1/4 × 30 days = 7.5 days = 7.5 × 24 hours = 180 hours

Energy consumed in 180 hours = 0.32kW × 180h = 57.6kWh

Cost of electricity of 57.6kWh energy consumed by the refrigerator = 57.6 × $0.09 = $5.18

3 0
4 years ago
Read 2 more answers
If increased productivity in the workplace is an example of the macroeconomic impact of the green economy and increased income f
Ivan

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7 0
3 years ago
A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and th
Dennis_Churaev [7]

Answer:

A)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate  - Q ( KJ )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

7 0
3 years ago
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