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3 years ago

What happens when a population exceeds its carrying capacity?

Physics

2 answers:

pashok25 [27]3 years ago

8 0

Answer:

If a population exceeds carrying capacity, the ecosystem may become unsuitable for the species to survive. If the population exceeds the carrying capacity for a long period of time, resources may be completely depleted.

Mademuasel [1]3 years ago

4 0

They either die from not enough food, they move to another place, or sometimes they adapt to the amount of organisms there. That is all I know

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A 2400W electric toaster is connected to a standard 120 V wall outlet. A family

jolli1 [7]

Answer:

1) energy = 15.6 kWh, 2) total_cost = $ 2.03

Explanation:

1) The energy dissipated is the product of the power and the time of use In a month it was used t = 6.5 h and the power of the toaster is

P = 2400 W = 2,400 kW

energy = P t

energy = 2,400 6.5

energy = 15.6 kWh

using rounding to a decimal

energy = 15.6 kWh

2) The cost of energy is unit_cost = $ 0.13 / kWh

so the total cost

total_cost = energy unit_cost

total_cost = 15.6 0.13

total_cost = $ 2.028

rounding to two decimal places

total_cost = $ 2.03

5 0

2 years ago

Explain what is the capacity of a battery in your own words

Vinil7 [7]

Battery capacity (AH) is defined as a product of the current that is drawn from the battery while the battery is able to supply the load until its voltage is dropped to lower than a certain value for each cell.

4 0

2 years ago

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the ap

Verizon [17]

Answer:

The answer is " $1.144 \times 10^{-4} \ m$ ".

Explanation:

$w=\frac{2.44 \lambda L}{D}\\\\D=\frac{2.44 \lambda L}{w}\\\\$

$=\frac{2.44 \times 633 \times 10^{-9}\times 4 }{0.054}\\\\=\frac{6178.08\times 10^{-9}}{0.054}\\\\=1.144 \times 10^{-4} \ m$

8 0

3 years ago

A spring is stretched 2.0 cm. If the same spring is stretched 8.0 cm the ratio of the second and first potential energies of the

jok3333 [9.3K]

Potential energy of spring equals K times X squared divided by 2 where X is displacement

4 times squared equals 16

choose 1st answer

6 0

2 years ago

At the intersection of Texas Avenue and University Drive,

Zielflug [23.3K]

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector $\hat{i}$ pointing north and $\hat{j}$ pointing east, the final velocity will be

$\vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )$

$\vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

The final linear momentum will be:

$\vec{P}_f = (m_{car}+ m_{truck}) * V_f$

$\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )$

$\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

As there are not external forces, the total linear momentum must be constant.

So:

$\vec{P}_0= \vec{P}_f$

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

$\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )$

so:

$\vec{P}_0= \vec{P}_f$

$( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )$

$\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.$

So, for the truck

$m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s}$

$v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg}$

$v_{truck} = 21.93 \frac{m}{s}$

And, for the car

$950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}$

$v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}$

$v_{car}=19.524 \frac{m}{s}$

5 0

2 years ago