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baherus [9]
3 years ago
13

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500m3/sat a location 90 m above the lake sur

face. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.
Physics
1 answer:
olga_2 [115]3 years ago
3 0

Answer:

mechanical energy per unit mass is 887.4 J/kg

power generated is 443.7 MW

Explanation:

given data

average velocity = 3 m/s

rate = 500 m³/s

height h = 90 m

to find out

total mechanical energy and power generation potential

solution

we know that mechanical energy is sum of potential energy and kinetic energy

so

E = \frac{1}{2}×m×v² + m×g×h    .............1

and energy per mass unit is

E/m =  \frac{1}{2}×v² + g×h

put here value

E/m =  \frac{1}{2}×3² + 9.81×90

E/m = 887.4 J/kg

so mechanical energy per unit mass is 887.4 J/kg

and

power generated is express as

power generated = energy per unit mass ×rate×density

power generated = 887.4× 500× 1000

power generated = 443700000

so power generated is 443.7 MW

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A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300, what are its inventories?
Anna007 [38]

A firm current ratio is 1. 0 and its quick ratio is 1. 0. If current liabilities are 12300 then its inventories will be 12300

Inventory is the accounting of items, component parts and raw materials that a company either uses in production or sells

The quick and current ratios are liquidity ratios that help investors and analysts gauge a company's ability to meet its short-term obligations. The current ratio divides current assets by current liabilities. The quick ratio only considers highly-liquid assets or cash equivalents as part of current assets.

current ratio = current assets / current liabilities

current assets = current ratio  * current liabilities

                        = 1 * 12300 = 12300

since , inventory is a current asset for accounting purpose , hence inventories will be 12300

To learn more about current ratios

brainly.com/question/19579866?referrer=searchResults

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4 0
2 years ago
After you enlarge a map, which one of the following scale remains correct?
9966 [12]

Answer:

None

Explanation:

An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:

  • Verbal scale use of words to represent scale information on the map.  The distance or linear units are used for depicting this scale on the map.  For example: 1 inch = 1 Kilo meter.
  • Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio.  Example: 1: 10,000 or 1/10,000
  • In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given.  Example: 1:500
  • Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map.  Example: 1:25,000
  • A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of
Lana71 [14]

Electric field due to a point charge is given as

E = \frac{kq}{r^2}

here we know that

q = 1.6 \times 10^{-19} C

also the distance is given as

r = 5.29 \times 10^{-11} m

now we will have

E = \frac{(9\times 10^9)(1.6 \times 10^{-19})}{(5.29 \times 10^{-11})^2}

so we will have

E = 5.14 \times 10^{11} N/C

so above is the electric field due to proton

5 0
3 years ago
The escape speed from the moon is much smaller than from earth. True or False
Lisa [10]

Answer:

True

The escape speed from the Moon is much smaller than from Earth.

Explanation:

The escape speed is defined as:

v_{e} = \sqrt{\frac{2GM}{r}}  (1)

Where G is the gravitational constant, M is the mass and r is the radius.

The mass of the Earth is 5.972x10^{24}kg and its radius is 6371000m

Then, replacing those values in equation 1 it is gotten.

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(5.972x10^{24}kg)}{(6371000m)}}  

v_{e} = 11.18m/s

For the case of the Moon:

v_{e} = \sqrt{\frac{(2)(6.67x10^{-11}N.m^{2}/kg^{2})(7.347x10^{22}Kg)}{(1737000m)}}  

v_{e} = 2.38m/s

Hence, the escape speed from the Moon is much smaller than from Earth.

Since it has a smaller mass and smaller radius compared to that from the Earth.

4 0
3 years ago
Who was the first president of kenya​
Damm [24]

Answer: Jomo Kenyatta

Explanation: Jomo Kenyatta was an anti-colonial activist and politician and was the first Prime Minister of Kenya. He then served as president of the country from 1964 to his death in 1978

7 0
3 years ago
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