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MaRussiya [10]
4 years ago
11

When the availability of resoures in an area is limited organisms may leave a population to move to an area with more resoures .

What is a movement called
Physics
1 answer:
disa [49]4 years ago
6 0

Answer:

Migration

Explanation:

Migration is the movement of organisms from one region or place of habitat to another region. Organisms migrate from one region to another to increase their rates of survival. Migration of organisms from one habitat to another may be due to unavailability of insufficiency of resources that are required for their survival.

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A transverse wave has a frequency of 200 Hz with a wavelength of 1.0 m. Determine the speed
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Answer:

200 m\ s Ans .....

Explanation:

Data:

f = 200 Hz

w = 1.0 m

v = ?

Formula:

v = f w

Solution:

v = ( 200)(1.0)

v = 200 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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When red light and green light shine on the same place on a piece of white paper, the spot appears t obe?
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4 years ago
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What is the relationship between the force of gravity between two objects in regard to their mass and the distance between them?
Mashcka [7]

Answer:

hi

Explanation:

5 0
3 years ago
In which situation are unbalanced forces acting on an object? (1 point)
Natalka [10]

Answer:

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Explanation:

3 0
3 years ago
A tank contains 150 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
nikitadnepr [17]

Answer:

The number A(t) of grams of salt in the tank at time t is A(t) = 150 - 110 e^{-\frac{t}{50} }

Explanation:

Knowing

\frac{dA}{dt} = Rin - Rout

First we have to find the Rin and Rout

Rin = (concentration of the salt inflow) * (input rate of brine)

Rin = 1 g/L * 3 L/min = 3 g/L

Rout = (concentration of the salt outflow) * (output rate of brine)

Rout = (\frac{A(t)}{150} g/L) * (3 L/min) = \frac{A(t)}{50} g/min

Substituting this results

\frac{dA}{dt} = 3 - \frac{A(t)}{50} --> \frac{dA}{dt} + \frac{1}{50} A(t)  = 4

Thus, integration factors is

e^{ \int\limits^._. {\frac{1}{50} } \, dt } = e^{\frac{t}{50} }

e^{\frac{t}{50} } \frac{dA}{dt} + \frac{1}{50} e^{\frac{t}{50} A(t) = 4 e^{\frac{t}{50} }

e^{\frac{t}{50} } A(t) = \int\limits^._. {3 e^{\frac{t}{50} } } \, dt\\ \\A(t) = 150 + c e^{-\frac{t}{50} }

Applying the initial conditions

A(0) = 40

c = 150 - 40 = 110

Now, substitute this result in the solution to get

A(t) = 150 - 110 e^{-\frac{t}{50} }

8 0
4 years ago
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