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2 years ago

A train travels 650 meters in 25 seconds. What is the train's velocity?

Engineering

1 answer:

frosja888 [35]2 years ago

3 0

The train is traveling 26 meters A second .

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allsm [11]

Answer:

wow

Explanation:

that's so cute!!!!!!!!

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2 years ago

The level of water in a dam is 6 m. The rectangular gate ABC is pinned at point B so it can rotate freely about this point. When

olchik [2.2K]

Answer:

The reaction at support B

Rb= 235440N

The reaction at support C

RC= 29430N

Explanation : check attachment

6 0

3 years ago

Is Tesla French, American, German, or Russian?

ozzi

Answer:

I am gonna say it is AMERICAN

6 0

3 years ago

Read 2 more answers

A pumping test was made in pervious gravels and sands extending to a depth of 50 ft. ,where a bed of clay was encountered. The n

Vikki [24]

Answer:per minute from the pumping well, a steady state was attained in about 24 hr. The draw-down at a distance of 10 ft. was 5.5 ft. and at 25 ft. was 1.21 ft.

Explanation:

6 0

2 years ago

A lake has a carrying capacity of 10,000 fish. At the current level of fishing, 2,000 fish per year are taken with the catch uni

arlik [135]

Answer:

The population size would be $p' = 5000$

The yield would be $MaxYield = 2082 \ fishes \ per \ year$

Explanation:

So in this problem we are going to be examining the application of a population dynamics a fishing pond and stock fishing and objective would be to obtain the maximum sustainable yield and and the population of the fish at the obtained maximum sustainable yield, so basically we would be applying an engineering solution to fishing

So the current yield which is mathematically represented as

$\frac{dN}{dt} = \frac{2000}{1 \ year }$

Where dN is the change in the number of fish

and dt is the change in time

So in order to obtain the solution we need to obtain the rate of growth

For this we would be making use of the growth rate equation which is

$r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}$

Where N is the population of the fish which is given as 4,000 fishes

and K is the carrying capacity which is given as 10,000 fishes

r is the growth rate

Substituting these values into the equation

$r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]} =0.833$

The maximum sustainable yield would be dependent on the growth rate an the carrying capacity and this mathematically represented as

$Max Yield = \frac{rK}{4} = \frac{(10,000)(0.833)}{4} = 2082 \ fishes \ per \ year$

So since the maximum sustainable yield is 2082 then the the population need to be higher than 4,000 so in order to ensure a that this maximum yield the population size denoted by $p'$ would be

$p' = \frac{K}{2} = \frac{10,000}{2} = 5000\ fishes$

7 0

3 years ago

Read 2 more answers